0
$\begingroup$

We have $n$ different types of coupons in the box. Probability of drawing coupon of $j$th type from the box does not change and is equal to $p_j$. Find the expected value and variance of number of types in randomly chosen set of $k$ coupons.

I've managed to find the expected value, but I cannot calculate the variance. This is what I've got so far.

Let $X_i = \begin{cases} 0, & \text{if}\ \text{the $i$th coupon was not found in $k$ selections} \\ 1, & \text{otherwise} \end{cases}$

We have $X = \sum_{i=1}^{n} X_i$ and $E[X_i] = 1 \times Pr(X_i = 1) + 0 = 1 - (1 - p_i)^k$. $E[X] = E[\sum_{i=1}^{n}X_i] = \sum_{i=1}^nE[X_i] = n - \sum_{i=1}^{n}(1 - p_i)^k$.

Now let's calculate variance. $\text{Var}\left[X\right] = E[X^2] - (E[X])^2$. Since I've managed to calculate $E[X]$ the only problem is to obtain $E[X^2]$.

So $E[X^2] = E[(\sum_{i=1}^{n}X_i)^2] = \sum_{i=1}^{n}E[X_i^2] + 2\sum_{1 \leq i < j \leq n}E[X_iX_j]$.

The first summand is quite straightforward, but the second one certainly not. Could you suggest me any solution?

  • 1
    Hint: $E((1-X_i)(1-X_j))$ is the probability to avoid the $i$ type and the $j$ type in $n$ draws, which is $(1-p_i-p_j)^n$. Can you finish? – 2017-01-03
  • 0
    Yes. If I'm not mistaken I can use linearity of expected value here which yields $(1 - p_i - p_j)^k = E[(1 - X_i)(1 - X_j)] = 1 - E[X_j] - E[X_i] + E[X_iX_j]$. From which I can obtain $E[X_iX_j]$. Thank you for your answer. – 2017-01-03

0 Answers 0