Integration of the product of pdf & cdf of normal distribution

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!

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 Is $a>0$? In this case, write the integral, do a substitution on the integral which defines $\Phi((x-b)/a)$ in order to get a bound $x$ in this integral. Then integrate by parts, and complete the squares. – Davide Giraudo Jul 6 at 14:07 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? – unknown (google) Jul 8 at 7:10 In fact I made a miscomputation. I used below a different approach. – Davide Giraudo Jul 9 at 9:12

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

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 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! – unknown (google) Jul 10 at 1:48 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. – unknown (google) Jul 10 at 2:22 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 $\phi$. – Davide Giraudo Jul 10 at 9:35 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! – unknown (google) Jul 12 at 6:10 The only fixed parameters are $a$ and $b$, we integrate with respect to $x$. Hence it's normal that $x$ doesn't appears in the final result. – Davide Giraudo Jul 12 at 9:53

shouldn't we divide - rather than multiplying - by the square root of 1+a^2 in ∫+∞ba2+1√ϕ(t)dt? thanks for your answer.

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Thanks for your answer Davide. Is there a way to calculate the integral given by I(a,b) on a specific interval let say [u,v]? Following Davide's method there would probably be a problem when integrating I(a,b) with regard to b as there will appear a cdf term depending on b. Any help would be greatly appreciated.

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Yes, I think that the integral should divide instead of multiplying by the square root term.

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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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