Can any one help me how to solve this pronbelm?
I have a random variable $W$, i.e.,
$$W=\Phi(X)^k\Phi(-X)^m=P(Z\le X)^kP(Z \ge X)^m,$$ $X$ is Normal($\mu$,1), $Z \text{ is Normal(0,1)}$, and $k$ and $m$ are two constant integers.
I want to find the expectation of $W$, which is $EW$. In other words I want to find the integration of $$\int \Phi(x)^k\Phi(-x)^mf(x)dx$$
a) I tried to find first the density function of $W$ like the following way and then $EW$ but could not solve:
$$P(W\le w)=P\left(\Phi(X)^k\Phi(Y)^m\le w\right)=\int P\left(X\le \Phi^{-1} \left( \left( \frac{w}{\Phi(y)^m} \right)^{1/k} \right)\right) f(y)\,dy$$
b) I also tried like $$W=\Phi(X)^k\Phi(-X)^m=P(Z\le X)^kP(Z \ge X)^m=P(Z-X\le 0)^kP(Z-X \ge 0)^m=\Phi \left(\frac{-\mu}{\sqrt(2-2\rho)}\right)^k\Phi \left(\frac{\mu}{\sqrt(2+2\rho)}\right)^m.$$ Since $Z-X \text{ is Normal(-$\mu,2-2\rho$) if correlation between Z and X is $\rho.$ }$ Therfore, $$EW=\Phi \left(\frac{-\mu}{\sqrt(2-2\rho)}\right)^k\Phi \left(\frac{\mu}{\sqrt(2+2\rho)}\right)^m.$$ Is the second procedure right?
Thank you I highly appreciate it.