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 offtopic by Did, Yemon Choi, David White, David Roberts, Andres Caicedo Jul 12 '13 at 5:52This question appears to be offtopic. The users who voted to close gave this specific reason:



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. \end{align} This can be expressed with the traditional erf function. 


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 $a\ne0$, $(\ast)=\frac12$. 





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