Another difficult Gaussian integral

**sven svenson** · 02-04-2011 03:12 PM

Hello all,

This time, I'm trying to compute

$\int^{\infty}_{-\infty} \frac{1}{c} \Phi\left(\frac{x}{a}\right)\phi\left(\frac{x-b}{c}\right) dx$

where $\phi,\Phi$ are the standard Gaussian pdf and cdf. The parameters a, b, and c are arbitrary constants.

I noticed some other threads on difficult Gaussian integrals, but this one has the added issue of these parameters. I am trying to apply the technique from the other thread -- could somebody please let me know if I'm on the right track? I write

$\int^{\infty}_{-\infty} P(X \leq x) \frac{1}{c} \phi\left(\frac{x-b}{c}\right) dx = P(X \leq Y)$

where $X \sim N\left(0,a^2\right),Y\sim N\left(b,c^2\right)$ . This is $P(X - Y \leq 0)$ where $X - Y \sim N\left(-b,a^2+c^2\right)$ , yielding an answer of $\Phi\left(\frac{b}{\sqrt{a^2+c^2}}\right)$ .

**Dragan** · 02-04-2011 08:16 PM

Originally Posted by sven svenson

Hello all,

This time, I'm trying to compute

$\int^{\infty}_{-\infty} \frac{1}{c} \Phi\left(\frac{x}{a}\right)\phi\left(\frac{x-b}{c}\right) dx$

where $\phi,\Phi$ are the standard Gaussian pdf and cdf. The parameters a, b, and c are arbitrary constants.

yielding an answer of $\Phi\left(\frac{b}{\sqrt{a^2+c^2}}\right)$ .

Yes, your solution looks good to me.

Note that it is a good idea for you to specify the conditions on what values the constant terms can take (for obvious reasons).

**Tom La Bone** · 04-07-2011 06:46 AM

Can you elaborate a bit on why the integral in the second equation (a probability times a pdf) gives the probability P(X <= Y)? It just is not sinking in why this is so.

**BGM** · 04-07-2011 11:28 AM

It is just law of total probability. (in a continuous version)

Here the random variable $Y$ is independent of $X$

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