expectation of a standard normal within the CDF I was wondering if there is a way to compute the following expectation analytically?
$
\mathbb{E}\left[e^{cZ}\Phi\left(a+bZ\right)\right]
$
where $Z$ is  a standard normal random variable and $\Phi$ is the CDF of the standard normal.
Thanks in advance.
 A: If $Z\sim N\left(0,1\right)$ and $f$ is a 'neat' function then:
$$\mathbb{E}\left[e^{cZ}f\left(Z\right)\right]=e^{\frac{1}{2}c^{2}}\mathbb{E}f\left(Z+c\right)$$
We start with:
$\mathbb{E}\left[e^{cZ}f\left(Z\right)\right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{cz-\frac{1}{2}z^{2}}f\left(z\right)dz$. 
Applying $v=z-c$ and we find:
$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{cz-\frac{1}{2}z^{2}}f\left(z\right)dz=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{cv-\frac{1}{2}\left(v+c\right)^{2}}f\left(v+c\right)dv=\frac{e^{-\frac{1}{2}c^{2}}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}v^{2}}f\left(v+c\right)dv=e^{-\frac{1}{2}c^{2}}\mathbb{E}\left[f\left(Z+c\right)\right]$
Next to that we have: $$\mathbb{E}\Phi\left(a+bZ\right)=\Phi\left(\frac{a}{\sqrt{1+b^{2}}}\right)$$
$\mathbb{E}\Phi\left(a+bZ\right)$can be recognized as $P\left(V\leq a+bZ\right)$
where $Z,V\sim N\left(0,1\right)$ are independent. 
So $\mathbb{E}\Phi\left(a+bZ\right)=P\left(W\leq a\right)$ where
$W=V-bZ$. 
Here $W\sim N\left(0,1+b^{2}\right)$ so $W'=\frac{W}{\sqrt{1+b^{2}}}\sim N\left(0,1\right)$.
This leads to $\mathbb{E}\Phi\left(a+bZ\right)=P\left(W\leq a\right)=P\left(W'\leq\frac{a}{\sqrt{1+b^{2}}}\right)=\Phi\left(\frac{a}{\sqrt{1+b^{2}}}\right)$.
Applying these facts we find: 

$$\mathbb{E}e^{cZ}\Phi\left(a+bZ\right)=e^{\frac{1}{2}c^{2}}\mathbb{E}\Phi\left(a+bc+bZ\right)=e^{\frac{1}{2}c^{2}}\Phi\left(\frac{a+bc}{\sqrt{1+b^{2}}}\right)$$

