probability - Expectation value of symmetric random variable

Can someone please help me answer this question?

From the definition of symmetry it follows that: $P(X≥b+x)=P(X≤b-x)$. What I did:
$E[X]$$=$$∑$$xi$$P(X=xi)=$$∑$$(b+(xi-b))$$P(X=xi)$$=$$∑$$b$$P(X=xi)$$+$$∑$$(xi-b)$$P(X=xi)$$=$$b⋅1$ $+$ $(a-b)$$P(X=a)$ $+$ $(b-a)$$P(X=2b-a)$ $=$ $b$ due to the symmetry of the random varible $X$ since it follows that $P(X=2b-a)$$=$$P(X=a)$. Is this correct?

Thanks in advance!

Answer

We have $a<0. Thus $$\mathbb{P}(X=a)=\mathbb{P}(Xand
$$\mathbb{P}(X=2b-a)=\mathbb{P}(X>b)$$

By symmetry of $X$ about $b$:

$$\mathbb{P}(X\geq b)=\mathbb{P}(X\leq b)=1-q$$

for some $q\in(0,1)$

Thus $$\begin{align*}\mathbb{P}(Xb)&=q\\\mathbb{P}(X=b)&=1-2q\end{align*}$$

It follows that

$$\mathbb{E}[X]=qa+q(2b-a)+(1-2q)b=b.$$