In my Probability book I have come across the following passage: "Since the PMF is symmetric around 3.5, we conclude that E[X] = 3.5". The PMF in question is one of a binomial random variable uniformely distributed around 3.5. My doubt is: is it always true that if a distribution is symmetric around a value its expectation will be equal to that value? If so, how can one prove it without recourse to calculus? I have seen similar proofs on the internet, but all calculus-based and/or with zero as the point of symmetry. Thanks in advance.
Proving that expectation equals point of symmetry without calculus
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symmetry
1 Answers
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Let me answer for discrete random variable, since you are talking about pmf. Any discrete r.v. can be described by $(x_{i},p_{i})_{i=1}^{n}$, where $x_{i}$ is realization and $p_{i}$ its probability.
Suppose $n$ odd. Denote $m=\frac{n+1}{2}$. Then, $\forall i\in\{1,\ldots,m-1\}$, $x_{m}-x_{m-i}=x_{m+i}-x_{m}$ and $p_{m-i}=p_{m+i}$. The former equality rewrites $x_{m+i}+x_{m-i}=2x_{m}$.
For $(x_{i},p_{i})_{i=1}^{n}$, $\mathbb{E}[X]=\sum_{i=1}^{n}x_{i}p_{i}$. With the equalities above you can rewrite it as $\left(\sum_{i=1}^{m-1}p_{i}2x_{m}\right)+p_{m}x_{m}=x_{m}$.
If $n$ is even, then denoting $m=\frac{n}{2}$, $\mathbb{E}[X]=\frac{x_{m}+x_{m+1}}{2}$ is easy to show for symmetric discrete r.v..