plus.maths.org
https://plus.maths.org/content
enNew ways of seeing with the INTEGRAL project
https://plus.maths.org/content/new-ways-seeing-integral-project
<div class="field field-name-field-remote-encl field-type-file field-label-hidden"><div class="field-items"><div class="field-item even"><audio controls="controls" controlsList=""><source src="https://plus.maths.org/content/sites/plus.maths.org/files/podcast/2022/Integral/pluspodcast_integral.mp3" type="audio/mpeg" /></audio></div></div></div><div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/icon_62.jpg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><div class="rightimage" style="max-width:300px"><img src="/content/sites/plus.maths.org/files/podcast/2022/Integral/integralpodcastweb.jpg" alt="some of the members of the INTEGRAL team" ><p>Some of the members of the INTEGRAL team who spoke to us over zoom. From top left: Carola-Bibiane Schönlieb, James Woodcock, Angelica Aviles-Rivero, Saurabh Pandey, Sanjay Bisht, Debmita Bandyopadhyay, Rihuan Ke, David Coomes.</p></div>
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It's amazing what you can see now thanks to remote imaging technology! Visiting <a href="https://earth.google.com/web/@-34.93325267,117.35326855,43.49890635a,30721.29457938d,35y,360h,0t,0r" target="blank">far away landscapes</a> via satellite images or watching <a href="https://www.abbeyroad.com/Crossing" target="blank">live feeds</a> from a famous street is fun, but remotely gathered images offer exciting opportunities to map and observe the world. The problem is that the vast amount of remotely gathered data now available is useless on it's own – we need to have the means to analyse and extract information from those images.
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This is exactly what the members of the <a href="https://sites.google.com/view/remote-sens-research-for-india/home?authuser=0" target="blank">INTEGRAL project</a>, researchers based at the University of Cambridge and researchers and industry partners in India, are working on. This is an innovative collaboration between people collecting remote sensing data – such as satellite images of forests and video from traffic cameras – and researchers developing the technology to analyse those remotely gathered images to answer meaningful questions.
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In this podcast we talk to some of the members of the INTEGRAL team about the innovative machine learning approaches they are developing to understand remotely gathered images, and the significant impact these technologies can have on the world. Our thanks to <a href="https://www.damtp.cam.ac.uk/user/cbs31/Home.html" target="blank">Carola-Bibiane Schönlieb</a>, <a href="https://www.mrc-epid.cam.ac.uk/people/james-woodcock/" target="blank">James Woodcock</a>, <a href="https://www.maths.cam.ac.uk/person/ai323" target="blank">Angelica Aviles Rivero</a>, <a href="http://www.damtp.cam.ac.uk/person/db887" target="blank">Debmita Bandyopadhyay</a>, <a href="http://www.damtp.cam.ac.uk/person/rk621" target="blank">Rihuan Ke</a> and <a href="https://www.plantsci.cam.ac.uk/directory/david-coomes" target="blank">David Coomes</a>, all from the University of Cambridge, and to Saurabh Pandey from <a href="https://kritikalsolutions.com/" target="blank">KritiKal Solutions</a> and Sanjay Bisht from <a href="https://ioraecological.com/" target="blank">IORA Ecological Solutions</a>, both based in India.
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You can read more about the INTEGRAL team's work in <a href="/content/seeing-traffic-through-new-eyes" target="blank"><em>Seeing traffic through new eyes</em></a> and about their new AI approaches in <a href="/content/maths-minute-semi-supervised-machine-learning" target="blank"><em>Maths in a minute: Semi-supervised machine learning</em></a>. And you can find much more information about <a href="/content/maths-minute-machine-learning-and-neural-networks" target="blank">machine learning</a> and <a href="/content/tags/image-analysis" target="blank">image analysis</a> on <em>Plus</em>.
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<p><em>You can listen to the podcast using the player above, and you can subscribe to our <a href="https://plus.maths.org/content/podcast-feed/rss.xml" target="blank">podcast feed</a> in your podcast aggregator of choice, or directly through <a href="https://podcasts.apple.com/us/podcast/plus-podcast-maths-on-the-move/id263456080" target="blank">Apple Podcasts</a> or <a href="https://open.spotify.com/show/5ZiMqmLdf1iZN3aoMabEgR" target="blank">Spotify</a>.</em></p>
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</div></div></div>Thu, 27 Jan 2022 10:24:55 +0000Rachel7577 at https://plus.maths.org/contenthttps://plus.maths.org/content/new-ways-seeing-integral-project#commentsProof by picture!
https://plus.maths.org/content/proof-picture
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/cake_icon_2.jpg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>Drawing pictures is incredibly useful when doing maths. They help build your intuition, allow you to have fun playing around, and sometimes pictures can even serve as proof. Below is a selection of articles exploring proofs in which pictures play an important role. So get out your coloured pencils and enjoy!</p>
<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/visual_icon.png" alt="" width="98" height="97" /></div>
<p><a href="/content/seeing-pythagoras">Seeing Pythagoras</a> — Here are three geometric proofs of Pythagoras' theorem. No algebra necessary!</p></div><br clear="all" />
<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/visual_icon_0.png" alt="" width="98" height="97" /></div>
<p><a href="/content/seeing-proof">Seeing proof</a> — Having trouble with algebra? Then try these visual proofs of two well-known algebraic identities.</p></div>
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/nishi_icon.png" alt="" width="98" height="97" /></div>
<p><a href="/content/some-lovely-proofs-picture">Some lovely proofs by picture</a> — Don't like trigonometry? Don't worry, here are three beautiful proofs of a well-known geometric result that make do without it.
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/5/9_jun_2011_-_1543/diagram.jpg" alt="" width="98" height="97" /></div>
<p><a href="/content/outer-space-series">Outer space: The rule of two</a> — The idea that an infinite sum can converge can be a little hard to grasp. Luckily, when it comes to a well-known geometric series, there's a neat visual demonstration.
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/issue43/features/korner/icon.jpg" alt="" width="98" height="97" /></div>
<p><a href="/content/os/issue43/features/korner/index">What is the area of a circle?</a> — You might know the formula for the area of a circle, but why does this formula work? Our visual explanation really is a piece of cake, served up with a hefty estimate of pi.
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/5/20_jun_2014_-_1209/icon.jpg" alt="" width="98" height="97" /></div>
<p><a href="/content/art-gallery-problem">The art gallery problem</a> — A beautiful result where a clever shift in viewpoint coupled with some geometric intuition delivers the proof (almost) on a plate.
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/5/31_oct_2014_-_1120/origami_icon.png" alt="" width="98" height="97" /></div>
<p><a href="/content/trisecting-angle-origami">Trisecting the angle with origami</a> — You can't divide an angle into three equal parts using traditional geometric methods. But you can do it with origami! The best way to do this is to try it and see. Here's how.
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<div class="pkgblock"><div class="leftimage" style="width: 98px;"><img src="/content/sites/plus.maths.org/files/abstractpics/5/5_nov_2014_-_1736/axioms_icon.jpg" alt="" width="98" height="97" /></div>
<p><a href="/content/maths-minute-euclids-axioms">Maths in a minute: Euclid's axioms</a> — Finally, no talk of geometric proofs would be complete without a nod to Euclid's axioms of geometry and a couple of examples of what they allow you to do.
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<br clear="all" /></div></div></div>Fri, 21 Jan 2022 12:50:41 +0000Marianne7574 at https://plus.maths.org/contenthttps://plus.maths.org/content/proof-picture#commentsUnderstanding waning immunity
https://plus.maths.org/content/so-whats-waning
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/vaccines_icon_0.jpg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-field-author field-type-text field-label-hidden"><div class="field-items"><div class="field-item even">Marianne Freiberger</div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>The Spanish flu pandemic came to an end because people who caught the disease (and didn't die) acquired long-lasting immunity. The pandemic killed tens of millions of people between 1918 and 1920. After an estimated 500 million people around the world had become infected there simply weren't enough susceptible people left, and so the pandemic petered out. The world had acquired herd immunity, though at an enormous cost.</p>
<div class="rightimage" style="max-width: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/oxford_astrazeneca_and_pfizer_biontech_covid-19_vaccine.jpg" alt="SIR model" width="350" height="221" /><p>Boosters vaccinations are given because the immunity gained through vaccination wanes. Photo: <a href="http://www.arne-mueseler.com" target="blank">Arne Müseler</a>, <a href="www.arne-mueseler.com target="blank">CC BY-SA 3.0 DE</a>.
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<p>When it comes to the virus which causes COVID-19, we don't yet know for how long a person who's been infected or vaccinated retains their immunity. Indeed, there is strong evidence that natural immunity, as well as the immunity provided by the vaccines, wanes over time. </p>
<p>That's the reason why boosters have dominated Christmas, yet very little is currently known about the exact nature of how immunity wanes. While we are continually learning more about what is happening within an infected or vaccinated person's body, for example what kind of antibodies tend to be produced and so on, it's hard to know what exactly these processes mean for the dynamics of the disease as it passes between people. </p>
<p>"The link between what is happening at the within-host level and the actual risk of getting infected or the probability of transmitting [the disease] — that link is still a mystery," says <a href="https://www.research.manchester.ac.uk/portal/lorenzo.pellis.html" target="blank">Lorenzo Pellis</a>, an epidemiologist at the University of Manchester and member of the <a href="https://maths.org/juniper/" target="blank">JUNIPER modelling consortium</a>.</p>
<div class="rightshoutout"><p>See <a href="/content/tags/covid-19">here</a> for all our coverage of the COVID-19 pandemic.</div><p>
<p> Another way of finding out about waning immunity is to forget about the biology and instead look at statistics: observe how many people become re-infected, or infected after vaccination, and when this happens. But this too is hard to do. You'd need linked-up data that tells you if and when a person who has tested positive was previously infected or vaccinated. And this kind of data, when available, is often incomplete and difficult to analyse because many infections go unreported. </p>
<p>So can we say anything at all about what we might expect from an infectious disease for which immunity wanes?</p>
<h3>A toy model</h3>
<p>The answer is yes, in broad terms we can. If immunity were to last for a long time, then we would expect the epidemiological curve to consist of a single bump. In the beginning, when everyone is susceptible, the number of infections rises exponentially (assuming there's no intervention such as a lockdown). After a while, though, the pool of susceptibles will have been sufficiently diminished for the dynamics to turn around and the number of infections to fall, eventually diminishing to zero. This is the curve we were told should be flattened at the beginning of the COVID-19 pandemic.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/sir.png" alt="SIR model" width="400" height="300" /><p style="max-width: 400px;">The typical output of an SIR model: The red curve shows the number of infected individuals over time, the blue curve the number of recovered and immune people, and the yellow curve the number of susceptible people.
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</div><!- Image from wikipedia, in public domain -->
<p>The one-bumped curve chimes with intuition, and it's also what you get from a simple mathematical model, called the SIR model, which describes the course of the disease. It divides a hypothetical population into three classes — susceptible (S), infected (I) and recovered (and therefore immune, R). People pass from one class into the next at a given rate described by a mathematical equation. The parameters in that equation will be specific to the particular disease and societal factors, and scientists estimate them from available data. (Find out more about the SIR model in <a href="https://plus.maths.org/content/mathematics-diseases" target="blank">this article</a>.)</p>
<p>If the immunity of a disease wanes, then this means that the pool of susceptible people is eventually refilled, providing further "fuel" for the epidemic. A simple way of modelling this set-up is to use an SIS model: people pass from the susceptible class (S) to the infected class (I), and then back to the susceptible class (S) as they loose immunity. (The infected class, I, can also include people who are recovered and still immune: the fact that these people are no longer infectious to others can be captured by letting the parameter that describes their infectiousness drop to zero after they have spent a while in the infected class.)</p>
<p><a href="https://sites.google.com/view/scarabelfrancesca/home" target="blank">Francesca Scarabel</a>, a research associate at the University of Manchester and member of <a href="https://maths.org/juniper/" target="blank">JUNIPER</a>, has used such a SIS model to explore waning immunity. "It's a toy model which is not realistic [enough to decide on] COVID response just now," says Scarabel. Instead, the model gives us a broad indication of what might happen with a disease with waning immunity and what factors may be important. </p>
<p>If you run the model, simulating how the disease will spread through your hypothetical population in the absence of any interventions, you'll find that you either get recurring peaks of infection or eventually settle down to a more or less constant number of infections within the population — a stable <em>endemic state</em>.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/graph1.png" alt="SIS model" width="800" height="434" /><p style="max-width: 800px;">One of the outputs of Scarabel’s model, showing the number of daily new infections (in jargon, the incidence) over time under the assumption that the mean duration of time a person is immune after infection is 9 months (further assumptions are explained further down in the article). Different colours correspond to different values of the basic reproduction number, as indicated in the box at the top right of the figure. Figure: Francesca Scarabel.
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<p>According to the model, one of the factors that decides which of the two situations we are in, and how high the peaks will get, is the <a href="/content/maths-minute-r0-and-herd-immunity" target="blank">basic reproduction number</a> of the disease. This is the average number of people an infected person goes on to infect, in the absence of any interventions or vaccination, usually denoted by <em>R</em><sub>0</sub>. A relatively low value for this number gives a stable endemic state, and the higher this number, the higher the peaks. This is shown in the figure above.</p>
<p>Thus, the mantra of "flattening the curve" still applies in situations described by the model. "From a public health perspective the aim is always to contain the waves, to lower the peaks, or even push the endemic equilibrium as low as possible," says Scarabel. "Lowering transmission by social distancing, masks, or even lockdowns is one mechanism for pushing the waves down and stabilising the system."</p>
<h3>Don't underestimate the boosters</h3>
<p>Vaccination is of course another mechanism for lowering transmission, and Scarabel has been able to extend her model to reflect a situation in which a proportion of the population is vaccinated and therefore less likely to become infected or transmit the disease. She assumed that vaccine induced immunity wanes over time and saw what would happen. In this case, the model suggests, in tune with intuition, that a lower vaccination rate leads to higher peaks, with a high vaccination rate leading to an endemic equilibrium that has the number of infections stable over time and close to zero.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/graph2.png" alt="SIS model with vaccination" width="800" height="380" /><p style="max-width: 800px;">One of the outputs of Scarabel’s model, showing incidence over time under various assumptions, some of which are explained further down in the article. Different colours correspond to different levels of vaccination coverage, as indicated in the box at the top right of the figure. Figure: Francesca Scarabel.
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<p>The intuitive explanation for this is very simple: if many people lose their immunity all at once, or within a short window of time, the population will suddenly become largely susceptible, allowing for a new large outbreak to take off. And this is where the strategic use of boosters comes in. "The results suggest that we want to make sure that there is some level of immunity in the population at all times," says Scarabel. "This is why we should care about vaccine boosters and not under-estimate their impact."</p>
<h3>The modelling mosaic</h3>
<p>The reality we actually live in is of course more complex than Scarabel assumed in her modelling, but making accurate predictions isn't her aim at this stage. Policy decisions in the UK are based on complex, large-scale models, such as the Warwick model we <a href="https://plus.maths.org/content/winter-coming-where-are-we-going" target="blank">reported on recently</a>. These do reflect that immunity can wane by making some simple assumptions about the nature of the waning</p>
<p>Given our incomplete knowledge about the precise mechanisms of waning immunity, one of Scarabel's aims is to figure out what kinds of assumptions are safe to make, or where we need to be more discerning. "When you make a model you always make simplifications," she explains. "We have to make choices as to what to simplify and what assumptions to make. You have to be careful with these choices because they might affect the predictions you get [from the model]."</p>
<p>The "shape" of the waning of immunity is one of the aspects Scarabel explored in her work. She ran simulations based on the assumption that people lose their immunity suddenly, instantly returning to being fully susceptible to catching the disease. This is assumed to happen at some random point in time after a person was infected, chosen according to a specific <a href="/content/maths-minute-probability-distributions" target="blank">probability distribution</a>: there's a mean duration of immunity that applies to the population as a whole, but single individuals may have longer or shorter periods of immunity.
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<p>Scarabel compared this <em>all-or-nothing</em> assumption with one where people lose immunity continuously over a given period, in a <em>leaky</em> way described by a particular mathematical curve. Her results suggest that qualitatively there is no substantial difference between the two assumptions: in both cases you either get recurring peaks or a stable endemic state, depending on the value of the <em>R</em> number and the mean duration of immunity. This is shown in the figures below, where the solid line indicates the all-or-nothing assumption and the dashed line the leaky assumption. You can see that dashed and solid lines are very close.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/graph3.jpg" alt="SIS model different assumptions" width="800" height="467" /><p style="max-width: 800px;">
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<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/graph4.jpg" alt="SIS model different assumptions" width="800" height="467" /><p style="max-width: 800px;">Top plot: Model predictions for incidence over time, with different colours representing different values of the basic reproduction number. Bottom plot: Model predictions for incidence over time, with different colours representing different mean duration of immunity. The solid lines reflect the output of the model when assuming all-or-nothing waning of immunity and the dashed lines represent the output when assuming leaky immunity.
Figures: Francesca Scarabel.
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<p>The results suggest that if you are mostly interested in the general shape of the future epidemic — for example if there will be peaks or if it will settle down — it doesn't really matter which of the two assumptions you choose in your modelling. The figures above also show, however, that quantitative differences may be substantial, in particular concerning the exact height of the peaks or the endemic state, especially in the long run. So if you are after more precise numbers, you might need to be careful in your choice of assumption.</p>
<h3>Windows of time</h3>
<p>Scarabel also explored what happens if you vary the probability distribution that controls the loss of immunity in the all-or-nothing assumption, and the shape of the curves that describes the loss in the leaky assumption. Her results indicate that peaks become higher if the population loses immunity within a short window of time. Again this makes sense: if many people lose immunity in a short period of time, many can promptly be re-infected.</p><p>
The same mechanism also explains a curious feature of the bottom figure above: the figure suggests that if the period of time people are immune for is shorter then the peaks will be lower, albeit more frequent (compare, for example, the green and the blue lines). A short duration of immunity means that some individuals can become susceptible again and be reinfected during the same epidemic wave. This avoids a situation where many people in the population are susceptible at the same time because people's immunity is no longer in sync. Thus, recurrent epidemic peaks can be avoided, at the cost of a high stable prevalence.</p>
<p>As with all research on COVID-19, Scarabel's work will be continually updated and refined as more scientific information and statistical data becomes available. Waning immunity is an important issue, so the research will play an essential role in the vast modelling mosaic researchers are building to understand the pandemic as we move into the future. "It's only when you sit down and try to model [waning immunity] properly that you discover that you have many choices in how you can render it in the model," says Pellis. "What you really want to do is make sure that your results are as robust as possible."</p>
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<h3>About this article</h3>
<p>This article is based on an interview with <a href="https://sites.google.com/view/scarabelfrancesca/home" target="blank">Francesca Scarabel</a> and <a href="https://www.research.manchester.ac.uk/portal/lorenzo.pellis.html" target="blank">Lorenzo Pellis</a>, both of the University of Manchester and members of the <a href="https://maths.org/juniper/" target="blank">JUNIPER modelling consortium</a>, and a talk given by Scarabel at a <a href="https://warwick.ac.uk/fac/sci/maths/research/miraw/days/mathematicalmodellingofcovid19/" target="blank">research meeting organised by the University of Warwick</a> in October 2021.</p>
<p><a href="https://plus.maths.org/content/people/index.html#marianne">Marianne Freiberger</a> is Editor of <em>Plus</em>.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2022/waning/francesca_lorenzo.jpg" alt="Francesca Scarabel and Lorenzo Pellis" width="420" height="200" /><p>Francesca Scarabel and Lorenzo Pellis
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<p><em>This article is part of our <a href="/content/joining-forces-covid19" target="blank">collaboration</a> with <a href="https://maths.org/juniper/" target="blank">JUNIPER</a>, the Joint UNIversity Pandemic and Epidemic Response modelling consortium. JUNIPER comprises academics from the universities of Cambridge, Warwick, Bristol, Exeter, Oxford, Manchester, and Lancaster, who are using a range of mathematical and statistical techniques to address pressing questions about the control of COVID-19. You can see more content produced with JUNIPER
<a href="/content/juniper" target="blank">here</a>.
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</p><div class="centreimage"><img src="/content/sites/plus.maths.org/files/packages/2021/Juniper-logos/juniper-light-bg.png" alt="Juniper logo" width="400" height="83" />
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</div></div></div>Mon, 10 Jan 2022 14:33:39 +0000Marianne7568 at https://plus.maths.org/contenthttps://plus.maths.org/content/so-whats-waning#commentsMaths in a minute: The normal distribution
https://plus.maths.org/content/normal-distribution
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/normal_icon.png" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>What's the chance that a random woman you meet on the street is exactly 170cm tall? </p>
<p>The question seems impossible to answer, but luckily maths can help. It tells us that the heights of people follow a <a href="/content/maths-minute-probability-distributions" target="blank">probability distribution</a> known as the <em>normal distribution</em> (also sometimes called a <em>Gaussian distribution</em>) represented by a bell-shaped curve.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/normal_height.png" alt="The normal distribution" width="600" height="306" />
<p style="max-width: 600px;">The normal distribution with mean 1.647 and standard deviation 7.07. This curve represents the distribution of heights of women based on a large study of twenty countries across North America, Europe, East Asia and Australia. Source: <a href="https://ourworldindata.org/human-height#height-is-normally-distributed" target="blank">Our world in data</a></p>
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<p>Here's how to interpret the curve. The top of the curve represents the mean (or average) height, which is 164.7cm. The probability that a random woman is between 165cm and 175cm is given by the area under the bell curve and on top of the interval from 165cm to 175cm, which is 0.17. Similarly, the probability that the height lies in any other range is given by the area under the curve and above that range.</p>
<p>Mathematically, the curve shown above is the <em>probability density function</em> of the normal distribution (find out more <a href="/content/maths-minute-probability-distributions" target="blank">here</a>). It's given mathematically by the formula</p>
<p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/faa04b666515b96b6e718cd06cdb9e94/images/img-0001.png" alt="\[ f(x)=\frac{1}{s\sqrt{2\pi }}e^{-\frac{1}{2}\left(\frac{x-m}{s}\right)^2}, \]" style="width:179px;
height:38px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table></p><p> where <img src="/MI/a81d4f19317d04058871e2c129446642/images/img-0001.png" alt="$m$" style="vertical-align:0px;
width:14px;
height:8px" class="math gen" /> is the mean we already mentioned above and <img src="/MI/a81d4f19317d04058871e2c129446642/images/img-0002.png" alt="$s$" style="vertical-align:0px;
width:6px;
height:8px" class="math gen" /> is the so-called <em>standard deviation</em>, which measures how fat or thin the bell curve is, in other words, how spread out the probabilities are. In our example of heights, the mean is <img src="/MI/a81d4f19317d04058871e2c129446642/images/img-0003.png" alt="$m=164.7cm$" style="vertical-align:0px;
width:94px;
height:12px" class="math gen" /> and the standard deviation is <img src="/MI/a81d4f19317d04058871e2c129446642/images/img-0004.png" alt="$s=7.07cm$" style="vertical-align:0px;
width:79px;
height:12px" class="math gen" />.</p>
<p> As the figure above shows, the probability that a woman's height lies within one standard deviation to the left of the mean is 0.34. By the symmetry of the curve, the probability the height lies within one standard deviation to the right of the mean is also 0.34. This means that the probability that a woman's height lies within one standard deviation of the mean no matter what side is 0.68. This is true for all normal distributions, no matter what their mean or standard deviation are: you always know that the chance that your random variable takes a value that's within one standard deviation of the mean is 0.68. (The standard deviation is the square root of the <em>variance</em>, which you can read about <a href="/content/maths-minute-variance target="blank">here</a>.)</p>
<p>If, instead of women, we were interested in the height of sausage dogs, we'd also get a normal distribution, but its exact shape would be different (see below for normal distributions of different shapes). The mean would be much smaller, so the entire bell-curve would be less tall. Presumably the heights also wouldn't be as spread out, so the standard deviation would be smaller and therefore the curve would be thinner than it is for humans (this isn't based on real data, we couldn't find any on sausage dogs, but surely there can't be that much variation in their heights).</p>
<p>How do we know that the heights of people and dogs follow a normal distribution? And how do we know that many, many other things — people's shoe sizes, blood pressure, measurement errors — also follow normal distributions? This is due in part to a mathematical result called the <em>central limit theorem</em>, which says that if you have lots of independent random variables (quantities that can take on a range of values) then as long as certain conditions are met their sum will follow a normal distribution. So if, in nature or the human-made world, something can be thought of as the sum of many independent factors, there's a chance that this something will follow a normal distribution. To see the central limit theorem in action, read <a href="/content/maths-three-minutes-central-limit-theorem" target="blank">this article</a>.</p>
<p>The figure below shows the normal distribution for various values of the mean and standard deviation.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/normal_many3.png" alt="Several normal distributions" width="600" height="517" />
<p style="max-width: 600px;">The normal distribution for various values of the mean and standard deviation.</p>
</div>
<!-- Image made by MF -->
</div></div></div>Fri, 07 Jan 2022 19:12:52 +0000Marianne7565 at https://plus.maths.org/contenthttps://plus.maths.org/content/normal-distribution#commentsWhat are probability distributions?
https://plus.maths.org/content/what-are-probability-distributions
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/dic_green.jpg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>Probability distributions turn up in all areas of science (and in many <em>Plus</em> articles) so we've decided to have a closer look at them. The short explainers below explore what a probability distribution actually is, visit some of the most commonly used distributions, and a few important concepts in the probability context. So next time you're not sure what is going to happen next, don't despair: one of these distributions may well be able to help you deal with the uncertainty.</p>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/dice_icon.jpg" alt="" width="100" height="100" /> </div>
<p><a href="/content/maths-minute-probability-distributions">Maths in a Minute: Probability distributions</a> — Here's a short and sweet introduction to probability distributions with a couple of easy examples.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/dice_icon_1.jpeg" alt="" width="100" height="100" /> </div>
<p><a href="/content/maths-minute-binomial-distribution">Maths in a Minute: The binomial distribution</a> — When you're repeating the same process over and over and want to know the chance you get a given sequence of outcomes, then binomial is the way to go. </p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/poisson_icon.png" alt="" width="100" height="100" /> </div>
<p><a href="/content/poisson-distribution">Maths in a Minute: The Poisson distribution</a> — If you know that an event happens three times an hour <em>on average</em>, then this doesn't mean it happens exactly three times every hour. So how many times is it going to happen in the next hour? That's the kind of situation where the Poisson distribution can help.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/exp_cul_many_icon.png" alt="" width="100" height="100" /> </div>
<p><a href="/content/exponential-distribution">Maths in a Minute: The exponential distribution</a> — If you know that an event happens three times an hour <em>on average</em>, then this doesn't mean it happens exactly every twenty minutes. So how long will you have to wait for it to happen? Here's where the exponential distribution comes in.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/gamma_icon.png" alt="" width="100" height="100" /> </div>
<p><a href="/content/gamma-distribution">Maths in a Minute: The gamma distribution</a> — This is related to the exponential distribution, only rather than asking "how long do I have to wait for one event to happen" you ask "how long do I have to wait until the event has happened two, three, four, or any other number of times?" </p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/normal_icon.png" alt="" width="100" height="100" /> </div>
<p><a href="/content/normal-distribution">Maths in a Minute: The normal distribution</a> — This is a heavy weight among probability distributions because it turns up in many, many different situations. Here's a quick introduction.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/5/15_apr_2016_-_1508/crowd_icon.jpg" alt="" width="100" height="100" /> </div>
<p><a href="/content/maths-three-minutes-central-limit-theorem">Maths in a Minute: The central limit theorem</a> — The central limit theorem is a reason why the normal distribution is so ubiquitous, so we've included it here.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/dic_blue.jpg" alt="" width="100" height="100" /> </div>
<p><a href="/content/maths-minute-expectation">Maths in a Minute: Expectation</a> — The expectation, also called the mean, of a probability distribution gives you a sense of what you'll see on average. This article explains how.</p></div>
<div class="pkgblock">
<div class="leftimage" style="width: 100px;"><img src="/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/variance_icon.jpg" alt="" width="100" height="100" /> </div>
<p><a href="/content/maths-minute-variance">Maths in a Minute: Variance</a> — The variance of a probability distribution gives you a sense of how spread out the probabilities captured by a distribution are. This article explains how.</p></div>
</div></div></div>Fri, 07 Jan 2022 18:21:07 +0000Marianne7569 at https://plus.maths.org/contenthttps://plus.maths.org/content/what-are-probability-distributions#commentsMaths in a minute: The gamma distribution
https://plus.maths.org/content/gamma-distribution
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/gamma_icon.png" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><p>Suppose that during a given time period an event happens on average <img src="/MI/8ccae9a106610c141c41b6af2bb40a8e/images/img-0001.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> times. For example, you might know that on average you'll see three new posts on your social media feed per minute. This doesn't mean that the event will occur at regular intervals: seeing three posts a minute on average doesn't mean you'll see one exactly every twenty seconds (which is a third of a minute). </p>
<p>One question you might ask yourself is, "when I switch on my feed, what's the chance I have to wait no more than <img src="/MI/0150a12da5a41decdfec5da10a3ff148/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes until I see the first new post?" As we explain <a href="/content/exponential-distribution" target="blank">in this article</a>, the answer is given by the <em>exponential distribution</em>.</p>
<p>Here we go one step further and ask, "what's the chance I have to wait no more than <img src="/MI/bff9c5aaf2c97d4ec640b114ddefa668/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes until I have seen two new posts, or three new posts, or any number of new posts?"</p>
<p>In this case the answer is given by a <a href="/content/maths-minute-probability-distributions" target="blank">probability distribution</a> called the <em>gamma distribution</em>. The probability <img src="/MI/f6a468524778adfd41d57979657099d5/images/img-0001.png" alt="$F(t,k)$" style="vertical-align:-4px;
width:47px;
height:17px" class="math gen" /> you have to wait at most <img src="/MI/f6a468524778adfd41d57979657099d5/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes to see <img src="/MI/f6a468524778adfd41d57979657099d5/images/img-0003.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> events is given by </p><p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0001.png" alt="\[ F(t,k) =1-\left(\frac{(at)^0e^{-at}}{0!}+\frac{(at)^1e^{-at}}{1!}+\frac{(at)^2e^{-at}}{2!}+...+\frac{(at)^{k-1}e^{-at}}{(k-1)!}\right). \]" style="width:523px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> Here <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> could be any positive integer and <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0003.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> any amount of time measured in minutes. </p><p>So for <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0004.png" alt="$a=3$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" /> the probability you have to wait at most a minute to see <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0005.png" alt="$k=3$" style="vertical-align:0px;
width:38px;
height:12px" class="math gen" /> posts is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0006.png" alt="\[ F(1,3) = 1-\left(\frac{(3)^0e^{-3}}{0!}+\frac{(3)^1e^{-3}}{1!}+\frac{(3)^2e^{-3}}{2!}\right)=0.577, \]" style="width:403px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> rounded to 3 decimal places. The plot below shows the distribution for <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0007.png" alt="$k=1,$" style="vertical-align:-4px;
width:42px;
height:16px" class="math gen" /> <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0008.png" alt="$k=2$" style="vertical-align:0px;
width:38px;
height:12px" class="math gen" />, and <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0005.png" alt="$k=3$" style="vertical-align:0px;
width:38px;
height:12px" class="math gen" />. The dashed lines correspond to our example of <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0009.png" alt="$t=1$" style="vertical-align:0px;
width:35px;
height:12px" class="math gen" /> and <img src="/MI/bfded44a58d27eb8720bc7210048d2d8/images/img-0005.png" alt="$k=3$" style="vertical-align:0px;
width:38px;
height:12px" class="math gen" />. </p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/gamma_cul3.png" alt="The cumulative function for the gamma distribution with <em>a</em>=3 and <em>k</em>=1, <em>k</em>=2, and <em>k</em>=3" width="600" height="374" />
<p style="max-width: 600px;">The cumulative function for the gamma distribution with <em>a</em>=3 and <em>k</em>=1, <em>k</em>=2, and <em>k</em>=3.</p>
</div>
<!-- Image made by MF -->
<p>Time is of course a continuous quantity, that is, it doesn't vary in discrete steps but instead flows along. As we explained in our <a href="/content/maths-minute-probability-distributions" target="blank">brief introduction to probability distributions</a>, when a continuous random variable is involved, a probability distribution comes with a <em>probability density function</em>. The density function in this case is </p>
<p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0001.png" alt="\[ f(t,k)=\frac{a^ k t^{k-1}e^{-at}}{(k-1)\times (k-2) \times ... \times 2 \times 1)}, \]" style="width:294px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>where <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> is a positive real number and <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0003.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> a positive integer. The function <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0004.png" alt="$F(t,k)$" style="vertical-align:-4px;
width:47px;
height:17px" class="math gen" /> above is the corresponding <em>cumulative function</em>. (To be absolutely precise, because we are assuming <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0003.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> is an integer, this is actually a special case of the gamma distribution, called the Erlang distribution. For a general gamma distribution <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0003.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> can be a continuous quantity.) </p><p>The plot below shows the density function for the gamma distribution for <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0005.png" alt="$a=3$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" /> and <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0006.png" alt="$k=1,$" style="vertical-align:-4px;
width:42px;
height:16px" class="math gen" /> <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0007.png" alt="$k=2,$" style="vertical-align:-4px;
width:42px;
height:16px" class="math gen" /> and <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0008.png" alt="$k=3.$" style="vertical-align:0px;
width:41px;
height:12px" class="math gen" /> The purple curve corresponds to the example <img src="/MI/e1744bde2e73b6dc89e5d111770bb0bd/images/img-0005.png" alt="$a=3$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" /> and the shaded area gives the probability that you have to wait no more than one minute to see three posts. </p></p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/gamma_cul2.png" alt="The density function for the gamma distribution with <em>a</em>=3 and <em>k</em>=1, <em>k</em>=2, and <em>k</em>=3" width="600" height="356" />
<p style="max-width: 600px;">The density function for the gamma distribution with <em>a</em>=3 and <em>k</em>=1, <em>k</em>=2, and <em>k</em>=3.</p>
</div>
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<p>The <em>mean</em> of the exponential distribution, also known as the <a href="/content/maths-minute-expectation" target="blank"><em>expectation</em></a> is <img src="/MI/889bcb733412e638efb01253a4f519b4/images/img-0001.png" alt="$k/a.$" style="vertical-align:-4px;
width:28px;
height:17px" class="math gen" /> Loosely speaking, this means that if we switched our feed on lots and lots of time and each time counted how long we waited to see the first <img src="/MI/889bcb733412e638efb01253a4f519b4/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> post, the average of wait times would be a <img src="/MI/889bcb733412e638efb01253a4f519b4/images/img-0003.png" alt="$k/a$" style="vertical-align:-4px;
width:25px;
height:17px" class="math gen" /> minutes. </p>
<p>The <a href="/content/maths-minute-variance" target="blank"><em>variance </em></a> of the exponential distribution, which measures how the individual probabilities are spread around the mean is <img src="/MI/9f293038659198635f5cef9c2490cb89/images/img-0001.png" alt="$k/a^2.$" style="vertical-align:-4px;
width:35px;
height:18px" class="math gen" /> </p></div></div></div>Fri, 07 Jan 2022 16:48:40 +0000Marianne7564 at https://plus.maths.org/contenthttps://plus.maths.org/content/gamma-distribution#commentsMaths in a minute: Probability distributions
https://plus.maths.org/content/maths-minute-probability-distributions
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/dice_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>Probability distributions describe processes that can have more than one outcome which you can't be sure about in advance. An example is rolling a die. There are six possible outcomes, namely the numbers 1 up to 6. As long as the die is fair, these are all equally likely to happen — you're just as likely to roll a 1, say, as you are to roll a 6. This means you have a <img src="/MI/cd844657e1b2a643169bb58f66ef8cf4/images/img-0001.png" alt="$100/6\approx 16.6\% $" style="vertical-align:-4px;
width:104px;
height:17px" class="math gen" /> chance of rolling any one number, which translates to a probability of <img src="/MI/cd844657e1b2a643169bb58f66ef8cf4/images/img-0002.png" alt="$0.166.$" style="vertical-align:0px;
width:40px;
height:11px" class="math gen" /> And that's your probability distribution: it tells you the probability of each of the possible outcomes of the process. You can visualise this using a histogram:</p>
<div class="centreimage"><img src="content/sites/plus.maths.org/files/articles/2021/Prob_dist/histo1.png" alt="Histogram" width="400" height="298" />
<p style="max-width: 400px;">A histogram visualising the distribution associated to rolling a die. The probability of rolling each of the six possible numbers is 1/6=0.166.</p>
</div>
<!-- Image made by MF -->
<p>What we've just considered is an example of a <em>uniform distribution</em>, one for which each outcome is equally likely so that the histogram is flat. Of course there are many processes for which this isn't the case. If you roll two dice and add the numbers they show, then you've got 11 possible outcomes (the numbers 2 up to 12) and not all are equally likely. Because a lot of different combinations can give you a final result of 7 (1 and 6, 2 and 5, etc), you're much more likely to end up with a 7 than you are to end up with a 2, which only happens when both dice show a 1. </p>
<p>Here is the histogram visualising this distribution. Probability histograms can be drawn in such a way that the area of the rectangle above each possible outcome is equal to the probability of this outcome. This is useful: to get a sense of the probability of either one of two outcomes occurring (say 2 or 3) you simply look at the combined area of the corresponding two rectangles.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/histo2.png" alt="Histogram" width="600" height="316" />
<p style="max-width: 600px;">A histogram visualising the distribution associated to rolling two dice and adding up the results. </p>
</div>
<!-- Image made by MF -->
<p>In these two examples, the <em>random variable</em> we are interested in — which stands for the different possible outcomes of our rolls of the die or dice — can only take on integer values. It's a <em>discrete variable</em>. This isn't always the case. If, for example, you would like to know the chance that a person you randomly picked from the population has a certain height, then your variable is continuous: the height can be any real number within the range of possible human heights.</p>
<p>
In this case, you can imagine the distribution of the probabilities being visualised by a curve, such as the one shown below (this is actually the famous <em>normal distribution</em>, which you can read more about <a href="/content/normal-distribution" target="blank">here</a>). In keeping with our histograms above, where the area of the rectangles represent probabilities, the probability of our random variable lying in a particular interval is represented by the area under the curve above that interval. A curve representing probabilities in this way (or rather the mathematical formula that describes it) is called a <em>probability density function</em>.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/normal_simple.png" alt="Normal distribution" width="600" height="332" />
<p style="max-width: 600px;">A normal distribution representing the women's height (using information from <a href="https://ourworldindata.org/human-height#height-is-normally-distributed" target="blank">Our world in data</a>). The mean of this distribution is 1.647m, and also happens to be the median. The probability of a random woman having a height in a given interval is given by the area under the curve exactly above the interval. </p>
</div>
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<p>
One thing that can be a little puzzling about this set-up is that the probability of your random variable taking on one particular value — for example that a random person is exactly 1.7m tall — is zero. That's because the area that sits exactly above a single value on the horizontal axis is just a line, which has zero area. The probability being zero can be explained in terms of the way infinity lurks within the continuum that is the real line. And intuitively it's ok too: in practice we'll never be able to distinguish someone with a height of 1.7m from someone with a height of 1.7001m, so all we're ever going to be interested in is the probability that someone's height falls within a given measurable range. And for that our probability density function is perfectly suitable. </p><p>The height <em>y</em> of the bell curve above a given value <em>x</em> on the horizontal axis, such as <em>x</em>=1.7m, in this case isn't the probability the random variable will take on the value <em>x</em>. But it does still give you information on how likely it is for values of the random variable to cluster around <em>x</em>: the larger <em>y</em>, the larger the area of the region under the curve sitting above an interval centred on <em>x</em>. Therefore, the larger <em>y</em>, the larger the chance that the random variable takes on values clustering near <em>x</em>.
</p>
<p>Now we've got that sorted out, here's a big question: given a process that's more complex than just rolling a couple of dice, how are you ever going to know the associated probability distribution? Luckily, there are various families of distributions that apply to certain types of set-ups. They are clearly defined by mathematics and can be calibrated to fit a whole range of situation you might want to consider in practice. Here are a few famous examples.</p>
<ul><li><a href="/content/maths-minute-binomial-distribution" target="blank">The binomial distribution</a></li>
<li><a href="/content/poisson-distribution" target="blank">The Poisson distribution</a></li><li><a href="/content/exponential-distribution" target="blank">The exponential distribution</a></li><li><a href="/content/gamma-distribution" target="blank">The gamma distribution</a></li><li><a href="/content/normal-distribution" target="blank">The normal distribution.</li></ul>
</div></div></div>Fri, 07 Jan 2022 16:46:54 +0000Marianne7560 at https://plus.maths.org/contenthttps://plus.maths.org/content/maths-minute-probability-distributions#commentsMaths in a minute: The exponential distribution
https://plus.maths.org/content/exponential-distribution
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/exp_cul_many_icon.png" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>Suppose that during a given time period an event happens on average <img src="/MI/fa8847d19ed36fdab383ee022d375798/images/img-0001.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> times. For example, you might know that on average you'll see three new posts on your social media feed per minute. This doesn't mean that the event will occur at regular intervals: seeing three posts a minute on average doesn't mean you'll see one every twenty seconds (which is a third of a minute). So you're justified in asking, "if I switch my feed on now what's the probability I will have to wait at most <img src="/MI/fa8847d19ed36fdab383ee022d375798/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes before I see the first new post?" Here <img src="/MI/fa8847d19ed36fdab383ee022d375798/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> could be 0 minutes, half a minute, five minutes, or any other positive real number of minutes.</p>
<p><p> It turns out that, assuming the events are independent, the probability <img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0001.png" alt="$F(Wait \leq t)$" style="vertical-align:-4px;
width:91px;
height:17px" class="math gen" /> that you’ll have to wait at most <img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes is </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0003.png" alt="\[ F(Wait \leq t)=1-e^{-a t}, \]" style="width:179px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> where <img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0004.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> is the average number of events per unit time. Below is the plot of this function for our social media example where <img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0005.png" alt="$a=3.$" style="vertical-align:0px;
width:42px;
height:11px" class="math gen" /> The probability of waiting at most one minute for the first post is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/940c0db7e4b39fd9a1d4e56cbefd3807/images/img-0006.png" alt="\[ F(Wait \leq 1) = 1-e^{-3}=0.95. \]" style="width:226px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> This makes sense: since there are three posts per minute on average, it’s very likely you’ll see at least one in the first minute. </p></p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/generation_time/exp_cul3.png" alt="The cumulative function of the exponential distribution" width="400" height="344" />
<p style="max-width: 600px;">The cumulative function for the exponential distribution with <em>a</em>=3.</p>
</div>
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<p>
What we have here is an example of a <em>exponential distribution</em>. Generally it is used to describe the time you have to wait for an event to occur for the first time, when you know the average number of events occurring per unit time and the events are independent.</p>
<p>Time is of course a continuous quantity, that is, it doesn't vary in discrete steps but instead flows along. As we explained in our <a href="/content/maths-minute-probability-distributions" target="blank">brief introduction to probability distributions</a>, when a continuous random variable is involved, a probability distribution comes with a <em>probability density function</em>. The density function in this case is </p>
<p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/a861a783175d558a52cc9d55e37a9600/images/img-0001.png" alt="\[ f(t) = ae^{-at}, \]" style="width:94px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> where <img src="/MI/a861a783175d558a52cc9d55e37a9600/images/img-0002.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> is a positive real number. The probability density function tells you the probability that the time you have to wait lies within a given interval: that probability is given by the area under the curve sitting on top of that interval. The function <img src="/MI/a861a783175d558a52cc9d55e37a9600/images/img-0003.png" alt="$F$" style="vertical-align:0px;
width:13px;
height:12px" class="math gen" /> above is the corresponding <em>cumulative function</em>, which tells you the probability you have to wait at most a given amount of time. </p></p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/exp_density.png" alt="The cumulative function of the exponential distribution" width="400" height="280" />
<p style="max-width: 400px;">The probability density function for the exponential distribution with <em>a</em>=3. The probability of the wait time being at most one minute is given by the shaded region.</p>
</div>
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<p>The <em>mean</em> of the exponential distribution, also known as the <a href="/content/maths-minute-expectation" target="blank"><em>expectation</em></a> is <img src="/MI/a2b0f434bd9423bca6890d7eda511112/images/img-0001.png" alt="$1/a,$" style="vertical-align:-4px;
width:28px;
height:17px" class="math gen" /> which in our example is equal to <img src="/MI/a2b0f434bd9423bca6890d7eda511112/images/img-0002.png" alt="$1/3$" style="vertical-align:-4px;
width:23px;
height:17px" class="math gen" />. Loosely speaking, this means that if we switched our feed on lots and lots of times and each time counted how long we waited to see the first post, the average of wait times would be a third of a minute. </p>
<p>The <a href="/content/maths-minute-variance" target="blank"><em>variance </em></a> of the exponential distribution, which measures how the individual probabilities are spread out, is <img src="/MI/4d754a333f6ecc3ef4629b9ba0252415/images/img-0001.png" alt="$1/a^2.$" style="vertical-align:-4px;
width:34px;
height:18px" class="math gen" /> Below are the curves of the cumulative (top) and probability density (bottom) functions for <img src="/MI/4d754a333f6ecc3ef4629b9ba0252415/images/img-0002.png" alt="$a=0.5$" style="vertical-align:0px;
width:52px;
height:12px" class="math gen" />, <img src="/MI/4d754a333f6ecc3ef4629b9ba0252415/images/img-0003.png" alt="$a=1$" style="vertical-align:0px;
width:38px;
height:11px" class="math gen" />, and <img src="/MI/4d754a333f6ecc3ef4629b9ba0252415/images/img-0004.png" alt="$a=2.$" style="vertical-align:0px;
width:42px;
height:11px" class="math gen" /> </p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/exp_cul_many.png" alt="The cumulative function of the exponential distribution for different values of a" width="600" height="481" />
<p style="max-width: 600px;">The cumulative function for the exponential distribution for <em>a</em>=0.5,
<em>a</em>=1, and <em>a</em>=2.</p>
</div>
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<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/exp_density_many.png" alt="The probability density function of the exponential distribution for different values of a" width="600" height="481" />
<p style="max-width: 600px;">The probability density function for the exponential distribution for <em>a</em>=0.5, <em>a</em>=1, and <em>a</em>=2.</p>
</div>
<!-- Image made by MF -->
<p>If you have read our post about the <a href="/content/poisson-distribution" target="blank">Poisson distribution</a>, then you can get a sense of how the exponential distribution comes about. <p> If the probability of waiting at most <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes for an event is <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0002.png" alt="$F(Wait \leq t),$" style="vertical-align:-4px;
width:96px;
height:17px" class="math gen" /> then the probability of waiting longer than <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0003.png" alt="$T$" style="vertical-align:0px;
width:12px;
height:12px" class="math gen" /> minutes is </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0004.png" alt="\[ 1-F(Wait \leq t). \]" style="width:123px;
height:17px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> Now if you have to wait more than <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes for the event to occur, then this means that no event happens during each of the first <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes. The Poisson distribution tells us that the probability of no event happening during a single minute is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0005.png" alt="\[ P(Event=0)=\frac{a^0 e^{-a}}{0!}=e^{-a}. \]" style="width:222px;
height:36px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>The probability of no post arriving during each of the first <img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0001.png" alt="$t$" style="vertical-align:0px;
width:6px;
height:12px" class="math gen" /> minutes is therefore </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0006.png" alt="\[ (e^{-a})^ t=e^{-a t}. \]" style="width:100px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> This means that </p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/12a07aee38330c420c601490bc61a08b/images/img-0007.png" alt="\[ F(wait \leq t)=1- e^{-a t}, \]" style="width:173px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> as we claimed. </p></p>
</div></div></div>Fri, 07 Jan 2022 16:09:45 +0000Marianne7563 at https://plus.maths.org/contenthttps://plus.maths.org/content/exponential-distribution#commentsMaths in a minute: The Poisson distribution
https://plus.maths.org/content/poisson-distribution
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/poisson_icon.png" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>Suppose that during a given time period an event happens on average <img src="/MI/4f8c9ad2ea1496b258f7792e8aa5c24d/images/img-0001.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> times. For example, you might know that on average you'll see three new posts on your social media feed per minute. This doesn't mean that the event will occur exactly <img src="/MI/4f8c9ad2ea1496b258f7792e8aa5c24d/images/img-0001.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> times during every such time period: sometimes you might see five new posts a minute and sometimes just 1. So you're justified in asking, "what's the probability I will see exactly <img src="/MI/4f8c9ad2ea1496b258f7792e8aa5c24d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> new posts in a single minute?" Here <img src="/MI/4f8c9ad2ea1496b258f7792e8aa5c24d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> could be 0, or 1, or 2, or any other positive integer.</p>
<p>To answer this question you need a <a href="/content/maths-minute-probability-distributions" target="blank">probability distribution</a> — in fact, you need the <em>Poisson distribution</em>. It works as long as the events in question are all independent of each other and says that the probability of seeing the event exactly <img src="/MI/506aa37b2be0537bfb810f1b52281713/images/img-0001.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> times in a unit of time is </p><p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/91ab27dd460562e74ccedb4050781c5a/images/img-0001.png" alt="\[ P(Event=k)=\frac{a^ k e^{-a}}{k \times (k-1) \times ... \times 2}, \]" style="width:276px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table> <p>where <img src="/MI/89d975968ea0d7ad86f43dc87158641b/images/img-0001.png" alt="$e=2.78281...$" style="vertical-align:0px;
width:96px;
height:12px" class="math gen" /> is Euler's number. </p>
<p><p>In our social media example, the average <img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0001.png" alt="$a$" style="vertical-align:0px;
width:9px;
height:8px" class="math gen" /> is equal to <img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0002.png" alt="$3$" style="vertical-align:0px;
width:8px;
height:11px" class="math gen" /> posts per minute. Assuming the posts are all independent (so one post doesn’t spawn or prevent other posts), the probability that you see exactly one new post in a given minute is therefore </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0003.png" alt="\[ P(Event=1) = \frac{3^1 e^{-3}}{1} = \frac{3 \times 0.05}{1} = 0.15. \]" style="width:308px;
height:36px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> The chart below shows the poisson distribution for <img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0004.png" alt="$a=3$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" />. As <img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0005.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> gets larger the probability of seeing <img src="/MI/3482c92afc57844e6e3ff425cebbad08/images/img-0005.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> events per unit time tends to zero, as you’d expect. </p></p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/poisson.png" alt="The Poisson distribution" width="600" height="423" />
<p style="max-width: 600px;">The Poisson distribution for <em>a</em>=3.</p>
</div>
<!-- Image made by MF -->
<p>The <em>mean</em> of this distribution, also known as the <a href="/content/maths-minute-expectation" target="blank"><em>expectation</em></a>, is <img src="/MI/58edc117a653cdd57445a5e335a6512d/images/img-0001.png" alt="$a.$" style="vertical-align:0px;
width:12px;
height:8px" class="math gen" /> In our example <img src="/MI/58edc117a653cdd57445a5e335a6512d/images/img-0002.png" alt="$a=3$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" />. Loosely speaking, this means that if we switched our feed on lots and lots of times and each time counted the number of posts we see in the first minute, the average of those counts would be 3 — which is of course what we expect.</p>
<p>The <a href="/content/maths-minute-variance" target="blank"><em>variance </em></a> of the Poisson distribution, which measures how the individual probabilities are spread out is also <img src="/MI/38b7b85bf4cdac0ffd478b7d1609ddf4/images/img-0001.png" alt="$a.$" style="vertical-align:0px;
width:12px;
height:8px" class="math gen" /> Below are three more examples of the Poisson distribution, for <img src="/MI/38b7b85bf4cdac0ffd478b7d1609ddf4/images/img-0002.png" alt="$a=1$" style="vertical-align:0px;
width:38px;
height:11px" class="math gen" />, <img src="/MI/38b7b85bf4cdac0ffd478b7d1609ddf4/images/img-0003.png" alt="$a=6$" style="vertical-align:0px;
width:39px;
height:11px" class="math gen" />, and <img src="/MI/38b7b85bf4cdac0ffd478b7d1609ddf4/images/img-0004.png" alt="$a=10$" style="vertical-align:0px;
width:47px;
height:11px" class="math gen" />.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/poisson_many2.png" alt="The Poisson distribution" width="600" height="332" />
<p style="max-width: 600px;">The Poisson distribution for <em>a</em>=1 (blue), <em>a</em>=6 (red), and <em>a</em>=10 (green).</p>
</div>
<!-- Image made by MF -->
</div></div></div>Fri, 07 Jan 2022 15:41:16 +0000Marianne7562 at https://plus.maths.org/contenthttps://plus.maths.org/content/poisson-distribution#commentsMaths in a minute: The binomial distribution
https://plus.maths.org/content/maths-minute-binomial-distribution
<div class="field field-name-field-abs-img field-type-image field-label-hidden"><div class="field-items"><div class="field-item even"><img class="img-responsive" loading="lazy" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsite-date%5D/dice_icon_1.jpeg" width="100" height="100" alt="" /></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>In our <a href="/content/maths-minute-probability-distributions" target="blank">brief introduction to probability distributions</a> we talked about rolling dice, so let's stick with that example. Imagine I roll a die three times and each time you try and guess what the outcome will be. What's the probability of you guessing exactly <em>k</em> rolls right, where <em>k</em> is 0, 1, 2 or 3?</p>
<p>More generally, imagine you perform an experiment (eg roll a die) <img src="/MI/86db4e1ab1736428e236311b63515285/images/img-0001.png" alt="$N$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" /> times, and each time the result can be success or failure. What's the probability you get exactly <img src="/MI/86db4e1ab1736428e236311b63515285/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> successes, where <img src="/MI/86db4e1ab1736428e236311b63515285/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> can be any integer from <img src="/MI/86db4e1ab1736428e236311b63515285/images/img-0003.png" alt="$0$" style="vertical-align:0px;
width:8px;
height:11px" class="math gen" /> to <img src="/MI/86db4e1ab1736428e236311b63515285/images/img-0001.png" alt="$N$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" />?
<p><p>In our example, as long as the die is fair, you have a probability of <img src="/MI/085c90d1ce63a5c0441b648adbd1bc12/images/img-0001.png" alt="$p=1/6$" style="vertical-align:-4px;
width:55px;
height:17px" class="math gen" /> of guessing right. Since the probability of three independent events (ie guessing correctly) is the product of the individual probabilities, your probability of three correct guesses is </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/085c90d1ce63a5c0441b648adbd1bc12/images/img-0002.png" alt="\[ P(Correct = 3) = 1/6 \times 1/6 \times 1/6 = (1/6)^3\approx 0.005 \]" style="width:379px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>Your probability of guessing wrong is <img src="/MI/085c90d1ce63a5c0441b648adbd1bc12/images/img-0003.png" alt="$1-p=1-1/6=5/6$" style="vertical-align:-4px;
width:157px;
height:17px" class="math gen" />. By the same reasoning as above, the probability of getting no guess right is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/085c90d1ce63a5c0441b648adbd1bc12/images/img-0004.png" alt="\[ P(Correct = 0) = 5/6 \times 5/6 \times 5/6 = (5/6)^3\approx 0.579. \]" style="width:381px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table></p><p>
What about the probability of guessing one roll right, so <img src="/MI/4b953ce9dfb4746fa45e20fd280ad8bc/images/img-0001.png" alt="$k=1$" style="vertical-align:0px;
width:37px;
height:12px" class="math gen" />? There are three ways in which this could happen:</p>
<ul><li>You get the 1st roll right and the other two wrong</li>
<li>You get the 2nd roll right and the other two wrong</li>
<li>You get the 3rd roll right and the other two wrong</li></ul>
<p><p>Since each involves one correct and two incorrect guesses, the probability of each of the three scenarios is <img src="/MI/a89ce9037f81404976481bee9a07f320/images/img-0001.png" alt="$1/6 \times (5/6)^2.$" style="vertical-align:-4px;
width:92px;
height:18px" class="math gen" /> And since the probability of any one of three events occurring is the sum of the individual probabilities, we have that the probability of getting exactly one guess right is </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/a89ce9037f81404976481bee9a07f320/images/img-0002.png" alt="\[ P(Correct=1)=3\times 1/6 \times (5/6)^2 \approx 0.347. \]" style="width:318px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>Finally, we look at the probability of two correct guesses. Again this can happen in three ways (we leave it up to you to work these out). Each individual way has a probability of <img src="/MI/a89ce9037f81404976481bee9a07f320/images/img-0003.png" alt="$(1/6)^2\times 5/6$" style="vertical-align:-4px;
width:88px;
height:18px" class="math gen" />, so the overall probability is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/a89ce9037f81404976481bee9a07f320/images/img-0004.png" alt="\[ P(Correct=2)=3\times (1/6)^2 \times 5/6 \approx 0.069. \]" style="width:318px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table>
<p>Here's the histogram displaying the distribution.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/hiato_bonomial.png" alt="Histogram" width="300" height="483" />
<p style="max-width: 400px;"></p>
</div>
<!-- Image made by MF -->
<p><p>Now let’s look at the general set-up. You’re doing <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0001.png" alt="$N$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" /> experiments that can each end in success or failure, and you’re asking for the probability that there are exactly <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> successes among the <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0001.png" alt="$N$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" /> experiments. Write <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0003.png" alt="$p$" style="vertical-align:-4px;
width:10px;
height:12px" class="math gen" /> for the probability of success so <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0004.png" alt="$1-p$" style="vertical-align:-4px;
width:36px;
height:15px" class="math gen" /> is the probability of failure. </p><p>By the same reasoning as above, a particular sequence of <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> successes and <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0005.png" alt="$N-k$" style="vertical-align:0px;
width:44px;
height:12px" class="math gen" /> failures has probability </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0006.png" alt="\[ p^ k(1-p)^{N-k}. \]" style="width:101px;
height:19px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>But, also as above, such a sequence can occur in several ways, each way defined by how the successes are sprinkled in among the failures. It turns out that the number of ways you can sprinkle <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> objects in among a sequence of <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0001.png" alt="$N$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" /> objects, denoted by <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0007.png" alt="${N\choose k}$" style="vertical-align:-5px;
width:22px;
height:21px" class="math gen" />, is given by </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0008.png" alt="\[ {N\choose k} = \frac{n!}{k!(n-k)!}. \]" style="width:137px;
height:40px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> Here the notation <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0009.png" alt="$i!$" style="vertical-align:0px;
width:10px;
height:12px" class="math gen" />, where <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0010.png" alt="$i$" style="vertical-align:0px;
width:5px;
height:12px" class="math gen" /> is a positive integer, stands for </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0011.png" alt="\[ i!=i\times (i-1)\times (i-2)\times ... \times 2 \times 1. \]" style="width:268px;
height:17px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>(and <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0012.png" alt="$0!$" style="vertical-align:0px;
width:12px;
height:12px" class="math gen" /> is defined to equal 1). We now have a neat way of writing the probability of <img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0002.png" alt="$k$" style="vertical-align:0px;
width:8px;
height:12px" class="math gen" /> successes: </p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/0324022fec2b0fd717614158d1fd979d/images/img-0013.png" alt="\[ P(Correct=k) = {N\choose k}p^ k(1-p)^{N-k}. \]" style="width:276px;
height:40px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> That’s the binomial distribution. </p>
</p>
<p>The <em>mean</em> of this distribution, also known as the <a href="/content/maths-minute-expectation" target="blank"><em>expectation</em></a> is <img src="/MI/236baa034cd3418fd6ec64edd38459e1/images/img-0001.png" alt="$Np.$" style="vertical-align:-4px;
width:26px;
height:16px" class="math gen" /> So in our example above where <img src="/MI/236baa034cd3418fd6ec64edd38459e1/images/img-0002.png" alt="$N=3$" style="vertical-align:0px;
width:45px;
height:12px" class="math gen" /> and <img src="/MI/236baa034cd3418fd6ec64edd38459e1/images/img-0003.png" alt="$p=1/6$" style="vertical-align:-4px;
width:55px;
height:17px" class="math gen" /> the mean is <table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/236baa034cd3418fd6ec64edd38459e1/images/img-0004.png" alt="\[ Np=3/6=1/2. \]" style="width:120px;
height:17px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table> Loosely speaking, this means that if we played our game of guessing three rolls lots and lots of times, then on average you could expect to get half a roll per game right. Or, to phrase it in a way that uses whole numbers on average you could expect to get one roll in two games right.</p>
<p>The <a href="/content/maths-minute-variance" target="blank"><em>variance </em></a> of the binomial distribution, which measures how spread out the probabilities are, is <table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/aadf8938090a88220e7984689b91c8bb/images/img-0001.png" alt="\[ Np(1-p). \]" style="width:76px;
height:17px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> So in our example above it is </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/aadf8938090a88220e7984689b91c8bb/images/img-0002.png" alt="\[ Np(1-p)=\frac{3}{6}\times \frac{5}{6}=\frac{5}{12}. \]" style="width:186px;
height:35px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table></p>
<p>The shape of the binomial distribution depends on the value of the mean and the number of experiments. Here are some more examples:</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2021/Prob_dist/434px-binomial_distribution_pmf.svg.png" alt="Bonomial distributions" width="434" height="289" />
<p style="max-width: 400px;"></p>
</div>
<!-- Image from Wikipedia, in public domain --></div></div></div>Fri, 07 Jan 2022 13:23:48 +0000Marianne7561 at https://plus.maths.org/contenthttps://plus.maths.org/content/maths-minute-binomial-distribution#comments