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      1 March, 2006
      March 2006

      A collector's piece: solution

      In last issue's outer space we worked out how many pruchases you would expect to have to make in order to obtain every one of a set of N distinct cards. The puzzle was to show that the standard deviation of this result is close to 1.3N for large N. To work out the expectation, we considered N random variables X0, X1, etc, up to XN-1. Each random variable Xi counts how many purchases you have to make to get a new card if you already have i distinct cards. The overall expectation is then just the sum of the individual expectations of the N variables.

      You can use the same technique to work out the standard deviation (which is of course the square root of the variance): since the N random variables are independent of each other, the overall variance is simply the sum of the individual values.

      For each individual variable Xi, the probability that you have to make j purchases to get a new card is N−iN×(1−N−iN)j−1. So each individual Xi has the well-known geometric probability distribution with parameter p=N−iN.

      This distribution has variance 1−pp2=iN(N−i)2. Now if X is the sum of the N random variables X0, X1, ... , XN-1, then the variance of X is variance(X)=∑i=0N−1iN(N−i)2=N∑i=0N−1i(N−i)2. A little thought shows that this sum can be expressed as N∑i=1NN−ii2=N∑i=1NNi2−N∑i=1Nii2, which is equal to N2∑i=1N1i2−N∑i=1N1i.

      Dividing through by N2 gives variance(X)N2=∑i=1N1i2−1/N∑i=1N1i. As N gets large, the first term tends to π2/6, as was stated in the hint, while the second term tends to zero. This proves that the standard deviation for large N is close to π2N2/6, which is approximately 1.3 N — QED.


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