I have a question regarding a problem and I want to know if I am right:
An urn contains $n$ balls, numbered 1 to $n$ and you remove $k$ of them.
What is the expected value value of their sum ?
My suggestion:
Let $X_i$ = the number of the ball taken.
$\mathbf{E}(X_i)= \frac{n+1}{2}$
Since you are removing $k$ balls, $\mathbf{E}X = \mathbf{E}X_1+…+\mathbf{E}X_n$
So $\mathbf{E}X=k \cdot \frac{n+1}{2}$ ??
Any help would be appreciated !
Regards,
Best Answer
You've got the right answer. Here is the mathematical explanation. Let $S$ be the sum of the k balls chosen without replacement, $S = X_1 + X_2 + ... + X_k$ $(i=1,...,k）$, where $X_i$ = value of $i^{th}$ ball chosen, then
$$E[S]=E[\sum\limits_{i=1}^{k}X_i] \overset{(a)}{=} \sum\limits_{i=1}^{k}E[X_i] = kE[X_i] = k(n+1)/2$$
In above, (a) follows from the Linearity of Expectation.