Find $E[B 1_{\{ B\ge \frac{n}{2} \} }]$ where $B$ is Binomial $(n,\frac{1}{2})$.

Question

How to find the following expectation $$E[B 1_{\{ B\ge \frac{n}{2} \} }]$$ where $B$ is Binomial random variable with $(n,\frac{1}{2})$.

Here is what I did. The pmf in this case is given by \begin{align} E[B 1_{\{ B\ge \frac{n}{2} \} }]= \sum_{k \ge \lceil n/2 \rceil }^n k {n \choose k} \left( \frac{1}{2}\right)^n \end{align}

However, I am not sure how to compute this sum. An upper bound would be fine too.

Answer 1

Note that we have $$ E[B 1_{\{ B\ge \frac{n}{2} \} }]=\frac n2 - E[B1_{\{B< \frac n2\}}]. $$ Then we evaluate $E[B1_{\{B< \frac n2\}}]$. We have $$ E[B1_{\{B< \frac n2\}}]=\sum_{k<\frac n2} k\binom nk \frac 1{2^n}=\sum_{k<\frac n2} n \binom{n-1}{k-1} \frac1{2^n}. $$ The second equality is due to $k\binom nk = n \binom{n-1}{k-1}$.

Now, the index $k$ of the sum is up to $k\leq \frac n2 -1$ if $n$ is even, and $k\leq \frac{n-1}2$ if $n$ is odd. Then $k-1\leq \frac n2-2$ if $n$ is even, and $k-1\leq \frac{n-3}2$ if $n$ is odd.

We apply the symmetry of binomial coefficients $\binom{n-1}k=\binom{n-1}{n-1-k}$. Then the last sum is $$ \begin{cases} \frac n4 - \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\ \frac n4- \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\ \end{cases}. $$ Therefore, $$ E[B 1_{\{ B\ge \frac{n}{2} \} }]=\begin{cases} \frac n4 + \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\ \frac n4+ \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\ \end{cases}. $$

Answer 2

If $n$ is odd, it's $1/2$ due to symmetry.

If $n$ is even, it's $$ \frac{1}{2}(1 - P(B = n/2)) + P(B = n/2) $$ again due to symmetry.

Here, by symmetry I mean $P(B = k) = P( B = n-k)$ for any $k= 0, \dots, n$.

[EDIT] Seems that I misread the question; my apologies.