Elementary proof that a binomial distribution with $p\ne \frac12$ is almost symmetric?

Asked Feb 07 '19 at 16:39

Active Feb 07 '19 at 19:08

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I am looking for an elementary explanation of why a loaded coin gets more and less than the expected number of heads approximately equiprobably, in other words, mean=median.

Mathematically speaking, if $X\sim B(n,p)$ is a Binomial random variable, then $P(X<np)\approx P(X>np)$ for large $n$.

When $p=\frac12$, this is obvious because of the exact symmetry of the distribution around $\frac{n}{2}$.

For a general $0<p<1$ this follows from convergence of $X$ to the Gaussian Normal distribution.

Is there an elementary intuitive explanation?

edited Feb 07 '19 at 18:59

asked Feb 07 '19 at 16:39

sds

4,533

interesting question. for a general r.v. $X$, the mean and the median are not necessarily the same, i.e. $P(X > E[X]) \neq P(X < E[X])$. i suspect we do have to pull in the CLT to explain why mean = median for $X = B(n,p)$. but i would love to see an alternate "elementary intuitive" explanation. – antkam Feb 07 '19 at 18:54

Elementary proof that a binomial distribution with $p\ne \frac12$ is almost symmetric?

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