Question

For Example $2i$, show that the variance of the number of coupons needed to a mass a full set is equal to$\sum _{i=1}^{N-1} \frac{iN}{(N-i{)}^2}$

When $N$is large, this can be shown to be approximately equal (in the sense that their ratio approaches 1 as $N\to \infty$) to $N^2{\pi }^2/6$.

Short answer

It has been shown that the variance of the number of coupons needed to amass a full set is equal to$N\to \infty .$

Step-by-step solution

Given Information

The variance of the number of coupons needed to a mass a full set $=\sum _{i=1}^{N-1}\frac{iN}{(N-i{)}^2}$

Ratio approaches 1 as$N\to \mathrm{\infty }$

Explanation

The variables in the summation are independent. Hence:

$X=X_0+X_1+\dots +X_{N-1}$

$Var\left(X\right)=\sum _{i=1}^{N-1}Var\left(X_i\right)$

Now, to compute the required variance, one needs to use the following result:

$E\left[X_i\right]=\frac{N}{N-i}$

$P\left\{X_i=k\right\}=\left(\frac{N-i}{N}\right){\left(\frac{i}{N}\right)}^{k-1}$

Using this result, one has that:

$\mathrm{Var}\left(X_i\right)=\sum _{k=1}^{\mathrm{\infty }} \left(\frac{N-i}{N}\right){\left(\frac{i}{N}\right)}^{k-1}{k}^2-\frac{N^2}{(N-i{)}^2}$

$=\left(\frac{N-i}{N}\right)\sum _{k=1}^{\infty }{\left(\frac{i}{N}\right)}^{k-1}{k}^2-\frac{N^2}{(N-i{)}^2}$

$=\left(\frac{N-i}{N}\right)\frac{N^2(N+i)}{(N-i{)}^3}-\frac{N^2}{(N-i{)}^2}$

Explanation

Simplifying the expression, one obtains that:

$\mathrm{Var}\left(X_i\right)=\left(\frac{N-i}{N}\right)\frac{N^2(N+i)}{(N-i{)}^3}-\frac{N^2}{(N-i{)}^2}$

$=\frac{N(N+i)}{(N-i{)}^2}-\frac{N^2}{(N-i{)}^2}$

$=\frac{iN}{(N-i{)}^2}$

$=\sum _{i=1}^{N-1}Var\left(X_i\right)$

$=\sum _{i=1}^{N-1}\frac{iN}{(N-i{)}^2}$

Final Answer

Therefore, the variance of the number of coupons needed to amass a full set is equal to$Var\left(X_i\right)=\sum _{i=1}^{N-1}\frac{iN}{(N-i{)}^2}.$