0

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.

Yuan Gao
  • 597
  • Be careful. You need restrictions on the parameters to get a finite $I$. – herb steinberg Aug 16 '21 at 21:54
  • @herbsteinberg Can you explain a bit more about your insight? For finite $a, b, c$ the integral can be infinity? – Yuan Gao Aug 16 '21 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 Aug 16 '21 at 23:33
  • @Yuan Gao My error. I forgot the $e^{-x^2}$ term. – herb steinberg Aug 17 '21 at 00:02

1 Answers1

0

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.

Kurt G.
  • 14,198