Question

Let $X$be the smallest value obtained when $k$numbers are randomly chosen from the set $1,\dots ,n$. Find $E\left[X\right]$by interpreting $X$as a negative hypergeometric random variable.

Short answer

The required mean is equal to$\frac{n+1}{k+1}$.

Step-by-step solution

Given Information

The smallest value from the random variable is$1,\dots ,n$.

Explanation

Observe that $X\ge j$means that we have not chosen any of value less than $j$, i.e., all values are greater or equal to $j$. The probability of that event is

$P(X\ge j)=\frac{\left(\begin{array}{c}n-j+1\\ k\end{array}\right)}{\left(\begin{array}{l}n\\ k\end{array}\right)}=\frac{\left(\begin{array}{c}n-k\\ j-1\end{array}\right)}{\left(\begin{array}{c}n\\ j-1\end{array}\right)}$

Explanation

So, we see that $X$has the same distribution as the random variable in Example $3e$, but here we have that the total number of balls is equal to $n$and the number of special balls is equal to $k$. Therefore, we have that $E\left(X\right)$is equal to

$E\left(X\right)=1+\frac{n-k}{k+1}=\frac{n+1}{k+1}$

Final answer

The required mean is equal to $\frac{n+1}{k+1}$.