Given $X$ is a continuous random variable whose density is symmetric about a point $a$.
Show that $V=X-a$ and $U=a-X$ have same distribution.
$$F_U(u) = P(U \leq u) = P(X-a \leq u) = F_X(a+u)$$ and similarly $$F_W(w) = 1 - F_X(a-w) \longrightarrow f_U(w) = f_X(a+w) = f_X(a-w)$$ by symmetry. Therefore, $f_U(u) = f_X(a+u)$ by changing variable $w$ to $u$, which shows $f_U(u)=f_W(u)$. Is this solution right? Thanks!