Distribution of sums of symmetric independent random variables If random variables $X$ and $Y$ are independent, then the density function of $X+Y$ is $$f(z)=\int_Rf_X(x)f_Y(z-x)ds.$$ Moreover, if $X$ and $Y$ are symmetric, then $$f(z)=f(-z).$$For $z>0$, how to prove the following distribution function:
$$P(X+Y>z)=P(X>z)+\int_0^\infty P(Y>y)\big(f_X(|z-y|)-f_X(z+y)\big)dy?$$
Thanks for helping!
 A: $$P(X+Y>z)=P(X+Y>z, X>z)+ P(X+Y>z, X<z) $$
Add and subtract $P(X+Y<z, X>z)$:
$$P(X+Y>z)= \underbrace{P(X+Y>z, X>z)+ P(X+Y<z, X>z)}_{P(X>z)} + $$ 
$$+\,P(X+Y>z, X<z)-P(X+Y<z, X>z)=P(X>z)+\Delta.$$
Let us consider $\Delta=P(X+Y>z,X<z)−P(X+Y<z,X>z)$:
$$
\Delta = \int\limits_{-\infty}^z f_X(x) P(Y>z-x) dx - \int\limits_z^\infty f_X(x) P(Y<z-x)\, dx.
$$
Change variables in both integrals: $y=z-x$, $dy=-dx$, $x=z \mapsto y=0$, $x=\pm\infty\mapsto y=\mp\infty$. Then
$$
\Delta = \int\limits_0^{\infty} f_X(z-y) P(Y>y)\, dy - \int\limits_{-\infty}^0 f_X(z-y) P(Y<y)\, dy.
$$
Use that the distribution of $Y$ is symmetric and $P(Y<y)=P(Y>-y)$ and change $y$ by $-y$ in the second integral:
$$
\Delta = \int\limits_0^{\infty} f_X(z-y)\, P(Y>y)\, dy - \int\limits_0^{\infty} f_X(z+y)\, P(Y>y)\, dy = $$
$$=\int\limits_0^{\infty} P(Y>y)\,\bigl(\,f_X(|z-y|)-f_X(z+y)\bigr)\,dy.
$$
Here $f_X(z-y)=f_X(|z-y|)$ by symmetry. We get
$$P(X+Y>z)= P(X>z)+\Delta=P(X>z)+\int\limits_0^\infty P(Y>y)\,\big(\,f_X(|z-y|)-f_X(z+y)\big)\,dy.$$
