If $X$ satisfies $B (2n, \frac 1 2)$, then what is the mode?

Question

A mode is the value of $X$ for which $p(x)$ is the maximum.

The maximum value of $p(x)$ in the distribution $B(2n,\frac12)$ is $n$.

Is my approach is correct?

Answer 1

Your solution is correct. In order to prove it you can write down $p(k)={2n\choose k} 2^{-2n}$ and therefore try to find the max of $g(k)={2n\choose k}$.

Claim : $g(k)$ is increasing for $0\leq k\leq n$ and decreasing for $n\leq k\leq 2n$.

This is true since for any $0\leq k\leq n-1$, $g(k+1)=g(k)\cdot\frac{(2n-k)}{(k+1)}\geq g(k)$. Furthermore $g(k)=g(2n-k)$ and so for $n+1\leq k\leq 2n$, $g(k-1)\geq g(k)$.

Now it should be clear that this claim actually shows that $n$ is the maximum of $p(x)$.