# Expected Value of Exponential of Normal Random Variable times its CDF

Let $$\Phi$$ and $$\varphi$$ denote the cumulative density function and the density function of a standard normal random variable. I'd like to calculate

$$I=\int_{-\infty}^{\infty}e^{cx}\varphi(x)\Phi(a+bx)dx.$$

This is similar to Expected Value of Normal Random Variable times its CDF where the term $$e^{ax}$$ is replaced with a linear term. Tried to apply similar tricks here but didn't work out.

• Be careful. You need restrictions on the parameters to get a finite $I$. Commented Aug 16, 2021 at 21:54
• @herbsteinberg Can you explain a bit more about your insight? For finite $a, b, c$ the integral can be infinity? Commented Aug 16, 2021 at 22:08
• Write $e^{ax} \psi=e^{ax} e^{-x^2}$ (with constants). Complete the square in the exponent. Change variable $u=x+\text{const}$ so that the ($u$ dependant) exponential term looks like $e^{-Cu^2}$. Now you have an integral of Gaussian and error function of the form $e^{-Cu^2}\operatorname{erf}(A+Bu)$. $A$, $B$, and $C$ are related to your original constants. I'm not sure how the integral could diverge as $e^{ax}$ is swamped by the Gaussian for large $|x|$
– Sal
Commented Aug 16, 2021 at 23:33
• @Yuan Gao My error. I forgot the $e^{-x^2}$ term. Commented Aug 17, 2021 at 0:02

This integral evaluates as $$e^{c^2/2}\Phi\left(\frac{bc+a}{\sqrt{1+b^2}}\right)\,.$$ I do not see a restriction on the parameters. I suggest to verify this with a simple python program before we get into the details of the proof.