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Denote the pdf of the standard normal distribution as $\phi(x)$ and cdf as $\Phi(x)$. Does anyone know how to calculate $\int_{-\infty}^y \phi(x)\Phi(\frac{x−b}{a})dx$?

Notice that this question is similar to an existing one,

the only difference being that I'm computing the integral over $(-\infty, y)$ for some real $y$, rather than over the entire real line.

Thank you!

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migrated from Jun 30 '13 at 20:37

This question came from our site for professional mathematicians.

1) not a research level question 2) what does "ba" mean??? – Alexey Jun 30 '13 at 18:52
Where does the problem come from? Why do you need this result? – Davide Giraudo Jun 30 '13 at 19:20
Alexey: 1) It's difficult for me to judge the level and complexity of this question. It is likely that you are in a better position to do so, so I'll take your word for it. 2) This was a typo, thank you very much for pointing it out. I have corrected the question. The argument of $\Phi$ should read as $(\frac{x - b}{a})$, not $(x - ba)$ as it was. Here $a, b$ are real constants, $a \neq 0$. – user36247 Jul 1 '13 at 10:40
Davide: This integral has appeared in several contexts in my investigation of Brownian motion over compact intervals, hence the need to integrate over $(-\infty, y)$. (In fact, I need to find this integral over $(y, z)$ for some real $y$ and $z$.) I don't know whether it has an analytic solution. Hopefully it does. I would very much appreciate your help. – user36247 Jul 1 '13 at 10:43
Are $a$ and $b$ the mean and standard deviation of the normal density in question? Or are they independent of the parameters of the normal distribution? – rajb245 Jul 1 '13 at 14:08

As already explained, when $a\gt0$ the full integral is $1-\Phi\left(b/\sqrt{a^2+1}\right)$. The same approach shows that the integral considered here is $$ I(y)=P(Y\leqslant(X-b)/a,X\leqslant y), $$ where $(X,Y)$ are i.i.d. standard normal, that is, $$ I(y)=P(aY+b\leqslant X\leqslant y). $$ I see no reason to expect more explicit formulas.

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Hi, Did, I doubt there is a more explicit formula. Someone told me this integral may appear in an old paper by Owen. I'll look it up and let you know. – user36247 Jul 3 '13 at 16:06
By all means, please do. :-) – Did Jul 3 '13 at 19:03

If you check out the integral tables in section 4.2 and 4.3 of, you will find what you need to get this done. I used equations 4.3.2 and 4.2.1 (latter is same as eqn 7.4.36 in Abranovitz and Stegun). Just change variables on the error function and complete the square on the exponential. You will end up with one 4.3.2 integral, one 4.2.1, and one simple erf(x) integral of the exponential square.

I am a bioinformatician. This problem occured for me in the context of statistics. I was trying to compute conditional probabilities to input in my factor graph model.

I verified equation 4.2.1 from the source litterature. I will have to doublecheck if 4.3.2 is correct (via numerical integration), since this solution is original to this work.

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It's quite easy to calculate the indefinite integral by integrating by parts, using $\Phi (x)^{'} = \phi (x)$. The result is then straightforward.

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Alexey: My first instinct would also be to try integration by parts. Indeed, if the integral in question were $\int_{-\infty}^y \phi(x) \Phi(x)$, this would work nicely. Here, however, we are dealing with $\int_{-\infty}^y \phi(x) \Phi(\frac{x - b}{a})$. And when I tried integrating by parts as you are proposing, I ended up with a rather cumbersome mess. Can you please clarify how you would deal with $\frac{x - b}{a}$ in the argument of $\Phi$? Many thanks. – user36247 Jul 1 '13 at 13:59
P.S. There is already a question here where the argument of $\Phi$ is simpler (no linear transformation) and integration by parts does work:… Whereas if integration by parts were so easy in this case, all of Davide Giraudo's working in the solution to… would be quite unnecessary. – user36247 Jul 1 '13 at 14:02
...To conclude, I don't think this answer works, I'm sorry, Alexey. Please do correct me if I'm wrong. – user36247 Jul 1 '13 at 14:38

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