Suppose $X$ has a density function $f$ that is symmetric about $a$. Let $Y = X + a$. Show that the density function $g$ of $Y$ is symmetric about $0$.
Setting $f(x) = g(x-a)$ gives you the result through basic algebra. But I'm having an inordinate amount of difficulty justifying the equality to myself. Could someone explain to me very clearly why/ if the equality is allowed?
You have that $P(X\leq x) = \int_{-\infty}^x f(t)dt$. Now using variable change $t\mapsto t+a$ gives $$ P(Y\leq y) = P(X+a\leq y) = P(X\leq y-a) = \int_{-\infty}^{y-a}f(t)dt = \int_{-\infty}^y f(t+a)dt $$ so $Y$ has density $g(t) = f(t+a)$. But by symmetry of $f$ about $a$ we have $g(-t) = f(a-t) = f(a+t) = g(t)$ for each $t$, so $g$ is symmetric about $a$.