Definite integral of a product of normal pdf and cdf

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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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.

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.

$$\int_{-\infty}^y \phi(x) \Phi(\frac{x-b}{a})dx = BvN\left[\frac{-b}{\sqrt{a^2+1}}, y; \rho= \frac{-1}{\sqrt{a^2+1}}\right]$$

where $BvN(w, z; \rho)$ is the bivariate normal cumulative with upper bounds $w$ and $z$, and correlation $\rho$.

For reference, see equation (10,010.1) in Owen (Comm. in Stat., 1980).

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