This is my following problem:
$$ CDF_A:F_A(x)=\Phi(x)^2 $$ $$ CDF_B:F_B(x)=1-\Phi(-x)^3 $$ $$ Defining: X=A+(-B) $$ I have those two CDF's and I want to calculate the probability that X is smaller than some positive value: $$ \mathbb{P}(X\le\mu) ; 0\le\mu $$ I know I have to use the convolution integral, but I don't know how to start doing it. *no need a closed form(don't think there is one) , I'm intending to calculate it numerically. thank's in advance.
Assuming that $A$ and $B$ are independent, $$ P(X\leqslant x)=P(A\leqslant B+x)=E(F_A(B+x))=\int_\mathbb Rf_B(t)F_A(x+t)\mathrm dt, $$ where $F_A$ is the CDF of $A$ and $f_B$ is the PDF of $B$, thus, $$ P(X\leqslant x)=\int_\mathbb R 3\varphi(t)\Phi(-t)^2\Phi(x+t)^2\mathrm dt. $$ If this can help, note that $$ X=\max(Z_1,Z_2)+\max(Z_3,Z_4,Z_5), $$ where the random variables $Z_n$ are i.i.d. standard normal.