Suppose $F$ is the cumulative distribution function of the normal distribution with mean $a$ and standard deviation $b$, and suppose $f$ is the probability density function of the normal distribution with mean $w$ and standard deviation $z$. How can I calculate this integral in closed form?

$$\int F(x \mid a,b)f(x \mid w,z) {}dx$$

It seems like my question is very similar to this one, but crucially I'm not using that standard versions of the cdf or pdf. I tried converting my functions to their standard forms, but ended up with this:

$$\int \Phi\left(\frac{x-a}{b}\right)\frac{\phi\left(\frac{x-w}{z}\right)}{z} {}dx$$

And now I'm really confused!

P.S.: If the above does have a closed-form simplification, my true problem is actually closer to this:

$$\int F(2a - \xi - x \mid a,b)f(x \mid w,z) {}dx$$


2 Answers 2


$$\int F(x \mid a,b)f(x \mid w,z) {}dx=E[F(Y \mid a,b)]=\Pr(X<Y)$$ where $Y \sim F(\cdot \mid w,z)$ is independent of $X \sim F(\cdot \mid a,b).$

Now, $X-Y \sim {\cal N}(a-w, z^2+b^2)$. The result is $\boxed{F(0 \mid a-w, \sqrt{z^2+b^2})}$.

Just to check:

> a <- 1; b <- 1
> w <- 2; z <- 2
> integrate(function(x) pnorm(x,a,b)*dnorm(x,w,z), lower=-Inf, upper=Inf)
0.6726396 with absolute error < 2.2e-09
> pnorm(0, a-w, sqrt(b^2+z^2))
[1] 0.6726396

Your last question directly follows from the first one by noting that $$ F(2a - \xi - x \mid a,b) = 1 - F(x \mid 2a-\xi-a,b), $$ therefore the result for your last integral is $\boxed{1-F(0 \mid a-\xi-w, \sqrt{z^2+b^2})}$.

> ksi <- 3
> integrate(function(x) pnorm(2*a-ksi-x,a,b)*dnorm(x,w,z), lower=-Inf, upper=Inf)
0.03681913 with absolute error < 0.00012
> 1- pnorm(0, a-ksi-w, sqrt(b^2+z^2))
[1] 0.03681914
  • $\begingroup$ Wow, this is awesome. Do you know how I would solve my P.S. problem similarly? $\endgroup$
    – sammosummo
    Aug 8, 2015 at 0:07
  • $\begingroup$ @user1637894 that's done now $\endgroup$ Aug 8, 2015 at 7:05

I'll gibe you the special case, which may point you to the general case, if it exists.

If the parameters of your distributions were the same, then we have

$$\int F(x|\mu,\sigma)f(x|\mu,\sigma)dx=\int F(x|\mu,\sigma)dF(x|\mu,\sigma)=F^2(x|\mu,\sigma)/2+Const$$

When parameters are not the same, then it gets hairy. There's got to be a way to convert them to standard normal, then it's be the same parameters (0,1). I think it'll give you solution with constants and scaling

So, for different parameters, something like this should be true: $$F(x|\mu_2,\sigma_2)= F(\frac{x-\mu_2}{\sigma_2}\sigma+\mu|\mu,\sigma)$$

Then the integral becomes: $$\int F(x|\mu_2,\sigma_2)f(x|\mu,\sigma)dx=\int F(\frac{x-\mu_2}{\sigma_2}\sigma+\mu|\mu,\sigma)dF(x|\mu,\sigma)$$


$$=\int y(a x +b) dy(x)=?$$ This got to have some beautiful solution, Stiltjes integral or something

UPDATE: I vaguely remember that this type of integrals was used a lot on game theory. The stochastic dominance of the second kind and stuff like that is all based on these integrals

  • 1
    $\begingroup$ Gaussians have nice Fourier transform properties, so thats somewhere to look. $\endgroup$ Aug 7, 2015 at 19:52
  • $\begingroup$ Thanks for your reply! Can you give any more details to go on re: the game theory stuff? $\endgroup$
    – sammosummo
    Aug 7, 2015 at 23:16
  • $\begingroup$ Look at this book, there's pdf of the early version published by rand Corp somewhere in Internet amazon.com/exec/obidos/ASIN/048664216X/ref=nosim/weisstein-20 $\endgroup$
    – Aksakal
    Aug 7, 2015 at 23:22

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