Denote the pdf of normal distribution as $\phi(x)$ and cdf as $\Phi(x)$. Does anyone know how to calculate $\int \phi(x) \Phi(\frac{x b}{a}) dx$? Notice that when $a = 1$ and $b = 0$ the answer is $1/2\Phi(x)^2$. Thank you!

1$\begingroup$ Thank you very much for the answer. Yes, a>0, but I am not clear about what you mean.. Can you explain that in details? $\endgroup$ – user9836 Jul 8 '12 at 7:10

$\begingroup$ In fact I made a miscomputation. I used below a different approach. $\endgroup$ – Davide Giraudo Jul 9 '12 at 9:12

$\begingroup$ This question is available on mathoverflow.net at mathoverflow.net/questions/127086/… $\endgroup$ – Hugh Perkins Oct 2 '17 at 11:42
This might be a setting where relying on the probabilistic meaning of the functions $\phi$ and $\Phi$ saves ink and tedious computations.
Let $X$ and $Y$ denote standard normal random variables. Then $\int\limits_{\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ for every suitable function $u$ and $\Phi(x)=P(Y\leqslant x)$ for every real number $x$. Using this for the function $u:x\mapsto\Phi((xb)/a)$ and assuming furthermore that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((Xa)/b))=P(Y\leqslant(Xb)/a). $$ Thus, $$ (\ast)=P(Z\geqslant b), $$ where $Z=XaY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $1+a^2$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus, $$ (\ast)=P(T\geqslant b/\sqrt{a^2+1})=1\Phi\left(b/\sqrt{a^2+1}\right). $$ Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$
In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.

1
We have $\phi(x)=\frac 1{\sqrt{2\pi}}\exp\left(\frac{†^2}2\right)$ and $\Phi(x)=\int_{\infty}^x\phi(t)dt$. We try to compute $$ I(a,b):=\int\phi(x)\Phi\left(\frac{xb}a\right)dx.$$ Using the dominated convergence theorem, we are allowed to take the derivative with respect to $b$ inside the integral. We have $$\partial_bI(a,b)=\int\phi(x)\left(\frac 1a\right)\phi\left(\frac{xb}a\right)dx$$ and \begin{align} 2\pi\phi(x)\phi\left(\frac{xb}a\right)&=\exp\left(\frac 12\left(x^2+\frac{x^2}{a^2}2\frac{bx}{a^2}+\frac{b^2}{a^2}\right)\right)\\\ &=\exp\left(\frac 12\frac{a^2+1}{a^2}\left(x^22\frac b{a^2+1}x+\frac{b^2}{a^2+1}\right)\right)\\\ &=\exp\left(\frac 12\frac{a^2+1}{a^2}\left(x\frac b{a^2+1}\right)^2\frac 12\frac{a^2+1}{a^2}\left(\frac{b^2}{a^2+1}\frac{b^2}{(a^2+1)^2}\right)\right)\\\ &=\exp\left(\frac 12\frac{a^2+1}{a^2}\left(x\frac b{a^2+1}\right)^2\right)\exp\left(\frac{b^2}{2a^2}\frac{a^2+11}{a^2+1}\right)\\\ &=\exp\left(\frac 12\frac{a^2+1}{a^2}\left(x\frac b{a^2+1}\right)^2\right)\exp\left(\frac{b^2}{2(a^2+1)}\right). \end{align} Integrating with respect to $x$, we get that $$\partial_b I(a,b)=\frac 1{\sqrt{a^2+1}}\phi\left(\frac b{\sqrt{a^2+1}}\right).$$ Since $\lim_{b\to +\infty}I(a,b)=0$, we have \begin{align}I(a,b)&=\int_b^{+\infty}\frac 1{\sqrt{a^2+1}}\phi\left(\frac s{\sqrt{a^2+1}}\right)ds\\\ &=\int_{b/\sqrt{a^2+1}}^{+\infty}\phi(t)dt = 1  \Phi(b/\sqrt{a^2+1}). \end{align} This can be expressed with the traditional erf function.

$\begingroup$ Thank you very much for the answer but it seems that the "Integrating with respect to x" part is not correct. Can you doublecheck or add some more explanation on that part? Thank you! $\endgroup$ – user9836 Jul 10 '12 at 1:48

$\begingroup$ The result of the integration step should contains some erf function of x instead of $\phi$ in the result, and the exp(b^2/2/(a^2+1)) part should not be omitted. $\endgroup$ – user9836 Jul 10 '12 at 2:22

$\begingroup$ I integrate on the whole real line with respect to $x$, and I do the substitution $\frac{a^2+1}{a^2}\left(x\frac b{a^2+1}\right)^2=t^2$. I didn't omit the exponential term, since I wrote it using $\phi$. $\endgroup$ – Davide Giraudo Jul 10 '12 at 9:35

$\begingroup$ Thank you but I still think that you miss something. I think the result is not correct as $\int_{b\sqrt{a^2+1}}^{+\infty}\phi(t)dt$ is a constant that does not depends on $x$ or $t$. Would you like to double check your result? Thank you very much! $\endgroup$ – user9836 Jul 12 '12 at 6:10

1$\begingroup$ After the last integral, instead of saying "This can be expressed with the traditional erf function.", why not just say "$=1\Phi\left(b\sqrt{a^2+1}\right)$"? $\endgroup$ – Michael Hardy Jun 5 '13 at 19:54
 The last equation should be integral from $b/\sqrt{a^2+1}$
 In
I(a, b)
,a
is supposed to be positive. When $a<0$, the answer will be $\int_{\infty}^{b/\sqrt{a^2+1}} \phi(t) dt$