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!


closed as off-topic by Did, Yemon Choi, David White, David Roberts, Andrés E. Caicedo Jul 12 '13 at 5:52

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  • $\begingroup$ 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 $u:x\mapsto\Phi((x-b)/a)$ and assuming furthermore that $X$ and $Y$ are independent, one sees that the integral to be computed is $$ (\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a). $$ Thus, $$ (\ast)=P(Z\geqslant b), $$ where $Z=X-aY$ (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 $a\ne0$, $(\ast)=\frac12$.


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 $$ I(a,b):=\int\phi(x)\Phi\left(\frac{x-b}a\right)dx.$$ Using the dominated convergence theorem, we are allowed to take the derivative with respect to $b$ inside the integral. We have $$\partial_bI(a,b)=\int\phi(x)\left(-\frac 1a\right)\phi\left(\frac{x-b}a\right)dx$$ and \begin{align} 2\pi\phi(x)\phi\left(\frac{x-b}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^2-2\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+1-1}{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 double-check 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
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    $\begingroup$ 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
  1. The last equation should be integral from $b/\sqrt{a^2+1}$
  2. In I(a, b), a is supposed to be positive. When $a<0$, the answer will be $\int_{-\infty}^{b/\sqrt{a^2+1}} \phi(t) dt$

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