# Integral involving the CDF of normal distribution

I am doing some research and got stuck in solving the following integral (which I am not sure whether it has a closed form solution or not, I hope it has:))

Here is the integral:

$\int_{-\infty}^{+\infty} e^{-(x-a)^2}N(cx+d)dx$

where $N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x} e^{-\frac{z^2}{2}}dz$

any help would greatly be appreciated. (by the way I am not a grad student in math, so please be specific in answers) Thanks a lot.

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When $e^{-(x-a)^2}$ is the derivative of $N(cx+d)$, i.e. $c=\sqrt{2}$,$d=a\sqrt{2}$, then it has a closed form solution of $\frac{\sqrt{\pi}}{8}$, which you can obtain using integration by parts. For the more general case I am unsure, but the indefinite integral doesn't work in any CAS. – Lucas Jun 8 '13 at 1:33
If $a=0$ and $d=0$, then, for any $c$, the solution is: $\frac{\sqrt{\pi }}{2}$ – wolfies Jun 8 '13 at 1:42

In what follows, I am assuming that $c > 0$.

Your $N(x)$ is commonly denoted $\Phi(x)$ and called the cumulative probability distribution function of the standard normal random variable. So, what you have is the evaluation of $$\int_{-\infty}^\infty e^{-(x-a)^2}\Phi(cx+d)\,\mathrm dx = \sqrt{\pi}\int_{-\infty}^\infty \frac{e^{-(x-a)^2}}{\sqrt{\pi}}\Phi(cx+d)\mathrm dx$$ where the thing multiplying $\Phi(cx+d)$ in the integrand is the probability density function of a normal random variable $X$ with mean $a$ and variance $\frac{1}{2}$. If $Y$ is another normal random variable with mean $\mu$ and variance $\sigma^2$, then $$P\{Y \leq x\} = \Phi\left(\frac{x-\mu}{\sigma}\right) = \Phi(cx+d)$$ if we choose $\sigma = \frac{1}{c}$ and $\mu = -\sigma d= -\frac{d}{c}$. Now, if $X$ and $Y$ are taken to be independent random variables, then the conditional probability $P\{Y \leq X \mid X = x\}$ is just $\Phi(cx+d)$ and the unconditional probability that $Y$ does not exceed $X$ is, by the law of total probability, \begin{align} P\{Y \leq X\} &= \int_{-\infty}^\infty P\{Y \leq X \mid X = x\} f_X(x)\,\mathrm dx\\ &= \int_{-\infty}^\infty \frac{e^{-(x-a)^2}}{\sqrt{\pi}}\Phi(cx+d)\mathrm dx \end{align} which is the integral you want to evaluate (except for the $\sqrt{\pi}$ factor).

But, $P\{Y \leq X\} = P\{Y-X\leq 0\}$ where $Y-X$ is also a normal random variable with mean $\hat{\mu}= -\frac{d}{c}-a$ and variance $\hat{\sigma}^2 = \frac{1}{c}+\frac{1}{2}$ and so $$P\{Y \leq X\} = \Phi\left(\frac{0-\hat{\mu}}{\hat{\sigma}}\right) = \Phi\left(\frac{\frac{d}{c}+a}{\sqrt{\frac{1}{c}+\frac{1}{2}}}\right)$$ giving

$$\int_{-\infty}^\infty e^{-(x-a)^2}\Phi(cx+d)\,\mathrm dx = \sqrt{\pi}\cdot\Phi\left(\frac{\frac{d}{c}+a}{\sqrt{\frac{1}{c}+\frac{1}{2}}}\right).$$

For $a=d=0$, the argument of $\Phi(\cdot)$ is $0$ and so, as pointed out in @wolfie's comment, your integral has value $\frac{\sqrt{\pi}}{2}$ since $\Phi(0) = \frac{1}{2}$. Similarly, the value is $\frac{\sqrt{\pi}}{2}$ as long as the ratio $\frac{d}{c}$ equals $-a$, and in particular when $c = \sqrt{2}$ and $d=-a\sqrt{2}$. Note that the value of $d$ is the negative of what is stated in @Lucas's comment. I am not sure about the value $\sqrt{\pi}/8$ stated by Lucas for the case $c=\sqrt{2}$, $d=+a\sqrt{2}$.

What if $c < 0$? Well, $\Phi(cx+d)=1-\Phi(-cx-d)$ and so a similar result can be worked out.

A closely related question came up on stats.SE today and the accepted answer there uses a different method that you might want to try on your problem.

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