Add new comment

Proving the marginal value theorem

The marginal value theorem holds true under three fairly mild conditions:

  1. The fixed cost $T$ is larger than zero.

  2. The reward function $E(t)$ increases with $t$.

  3. The slope $dE(t)/dt$ of the reward function decreases with $t$ (i.e. $E(t)$ is a diminishing returns function).

We wish to prove that the instantaneous reward rate $r(t)$ equals the average reward rate $R(t)$ when $R(t)$ is maximal. To achieve this, we need to find the value of $r(t)$ when $R(t)$ is maximal. To find the maximal average reward rate, we make use of the fact that its slope is zero at a maximum.

The average reward rate is defined as

  $\displaystyle  R(t)  $ $\displaystyle  =  $ $\displaystyle  \frac{E(t)}{T+t},  $   (1)

and its derivative is

  $\displaystyle  \frac{dR}{dt}  $ $\displaystyle  =  $ $\displaystyle  \frac{1}{T+t} \frac{dE}{dt} + E(t) \frac{d(T+t)^{-1}}{dt}, \label{eqaa}  $   (2)

where (by definition)

  $\displaystyle  \frac{dE}{dt}  $ $\displaystyle  =  $ $\displaystyle  r(t), \label{eqa}  $   (3)

is the instantaneous reward rate, and where

  $\displaystyle  \frac{d((T+t)^{-1}) }{ dt}  $ $\displaystyle  =  $ $\displaystyle  \frac{-1}{(T+t)^{2}}. \label{eqb}  $   (4)

Substituting Equations 3 and 4 into Equation 2,

  $\displaystyle  \frac{dR}{dt}  $ $\displaystyle  =  $ $\displaystyle  r(t) \frac{1}{T+t} - \frac{E(t)}{(T+t)^{2}} , \label{eqaab}  $   (5)

where

  $\displaystyle  \frac{ E(t) }{T+t}  $ $\displaystyle  =  $ $\displaystyle  R(t),  $   (6)

is the average reward rate, so that Equation 5 becomes

  $\displaystyle  \frac{dR}{dt}  $ $\displaystyle  =  $ $\displaystyle  \frac{r(t) }{T+t} - \frac{R(t)}{T+t}.  $   (7)

At a maximum, this is equal to zero,

  $\displaystyle  \frac{r(t) }{T+t} - \frac{R(t)}{T+t}  $ $\displaystyle  =  $ $\displaystyle  0.  $   (8)

Finally, multiplying both sides by $(T+t)$, and re-arranging yields

  $\displaystyle  r(t)  $ $\displaystyle  =  $ $\displaystyle  R(t).  $   (9)

This proves that the average reward rate is maximal when the instantaneous reward rate equals the average reward rate.

Back to the main article

Filtered HTML

  • Web page addresses and email addresses turn into links automatically.
  • Allowed HTML tags: <a href hreflang> <em> <strong> <cite> <code> <ul type> <ol start type> <li> <dl> <dt> <dd>
  • Lines and paragraphs break automatically.
  • Want facts and want them fast? Our Maths in a minute series explores key mathematical concepts in just a few words.

  • What do chocolate and mayonnaise have in common? It's maths! Find out how in this podcast featuring engineer Valerie Pinfield.

  • Is it possible to write unique music with the limited quantity of notes and chords available? We ask musician Oli Freke!

  • How can maths help to understand the Southern Ocean, a vital component of the Earth's climate system?

  • Was the mathematical modelling projecting the course of the pandemic too pessimistic, or were the projections justified? Matt Keeling tells our colleagues from SBIDER about the COVID models that fed into public policy.

  • PhD student Daniel Kreuter tells us about his work on the BloodCounts! project, which uses maths to make optimal use of the billions of blood tests performed every year around the globe.