Expected value of a discrete random variable whose PMF satisfies the symmetry property

Question

Let $X$ be a discrete random variable with support $\{-n, -n+1, \cdots, 0, \cdots, n-1, n\}$ for some positive integer $n$. Further suppose the PMF of $X$ satisfies the symmetry property such that $P(X=-k)=P(X=k)$ for all integers $k$. What is $E(X)$?

Intuitively, I want to say $E(X) = 0$ because the distribution is symmetric around 0. But how would I prove this? I was thinking:

$$E(X)=\sum_{k=-n}^n kP(X=k)$$ $$E(X)=\sum_{k=-n}^{-1} kP(X=k) + 0P(X=0) + \sum_{k=1}^n kP(X=k)$$ (the one above the first sigma should be negative, I am sorry I cannot seem to make the formatting do that).

Then the summation of the probabilities of the negative values for $k$ should give a negative mean that cancels out the positive mean of the second summation, but I have no idea how to prove that.

Does this make any sense? Is it the right approach at all?

Answer 1

\begin{align}E(X)&=\sum_{k=-n}^{-1} kP(X=k) + 0P(X=0) + \sum_{k=1}^n kP(X=k)\\&= \sum_{k=1}^n (-k)P(X=-k) + \sum_{k=1}^n kP(X=k) \\&= \sum_{k=1}^n (-k)P(X=k) + \sum_{k=1}^n kP(X=k) \\&= \sum_{k=1}^n (-k+k)P(X=k) \\&= \sum_{k=1}^n (0)P(X=k)\\&=0\end{align}

Answer 2

Then the summation of the probabilities of the negative values for $k$ should give a negative mean that cancels out the positive mean of the second summation, but I have no idea how to prove that.

Change of bound variables theorem.

$$\begin{split}\sum\limits_{k=-n}^{-1} k~\mathsf P(X{=}k)&=\sum\limits_{(-k)=1}^n -(-k)~\mathsf P(X{=}{-}(-k))\\ &=\sum_{h=1}^n -h~\mathsf P(X{=}{-}h)\\ &=\sum_{k=1}^n -k~\mathsf P(X{=}{-}k)\\ &=\sum_{k=1}^n -k~\mathsf P(X{=}k)\end{split}$$