I am trying to prove that whenever you have a binomial distribution $X \sim Bin(n, p)$, that it is symmetric whenever the probability is $p = 0.5$. I am not sure if there is a more precise way to prove it but my reasoning. I know the binomial distribution formula is
\begin{equation*}
P(X = k) = \dfrac{n!}{k!(n – k)!}p^k(1 – p)^{n – k}
\end{equation*}
My attempt to prove:
Case 1: If $p = 0.5$, then $X \sim Bin(n, 0.5)$ is symmetric.
When $p = 0.5$, then for any $k$, the binomial distribution becomes:
\begin{equation*}
P(X = k) = \dfrac{n!}{k!(n – k)!}(0.5)^k(1 – 0.5)^{n – k} = \dfrac{n!}{k!(n – k)!}(0.5)^n
\end{equation*}
The probability 0.5 in this case would mean that it is non-skewed to the left or right, and would be considered to be even probability. Otherwise, if $p < 0.5$, then the probability will be skewed to the left, or if $p > 0.5$, then the probability will be skewed to the right, in both which do not have any type of symmetry in the distribution. So to prevent any skewness, $p = 0.5$, and therefore, symmetric.
(Edit: Also, whenever $p = 0.5$, then we would have two mutually exclusive outcomes. For example, flipping a fair coin for $n$ amount of times)
Case 2: If $X \sim Bin(n, p)$ is symmetric, then $p = 0.5$.
I am not sure how to proceed with this case of the proof if it is possible to prove this. Also I am not sure if my first case is the right reasoning or if I am missing some details. Would appreciate some help.
Best Answer
Assume $n>0,p\in(0,1)$ (the extreme cases are trivial since the distribution is then degenerate). You want to show
$$P(X=k)=P(X=n-k)\forall k\in \{0,...,n\}\iff p=0.5,$$
which follows since
$$P(X=k)={n \choose k} p^k (1-p)^{n-k}={n \choose n-k} p^k (1-p)^{n-k}$$
is equal to
$$P(X=n-k)={n \choose n-k} p^{n-k} (1-p)^k$$
for all $k$ if and only if $$p^k (1-p)^{n-k} =p^{n-k} (1-p)^k\iff (1-p)^{n-2k}=p^{n-2k}\iff 1-p=p\iff p=0.5.$$