most recent 30 from http://mathoverflow.net 2013-05-27T02:57:15Z http://mathoverflow.net/feeds/question/101469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution unknown (google) 2012-07-06T08:59:44Z 2013-03-10T16:19:34Z

<p>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!</p>

http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution/101753#101753 Davide Giraudo 2012-07-09T09:11:56Z 2012-07-09T09:11:56Z

<p>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)&amp;=\exp\left(-\frac 12\left(x^2+\frac{x^2}{a^2}-2\frac{bx}{a^2}+\frac{b^2}{a^2}\right)\right)\\ &amp;=\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)\\ &amp;=\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)\\ &amp;=\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)\\ &amp;=\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)&amp;=\int_b^{+\infty}\frac 1{\sqrt{a^2+1}}\phi\left(\frac s{\sqrt{a^2+1}}\right)ds\\ &amp;=\int_{b\sqrt{a^2+1}}^{+\infty}\phi(t)dt. \end{align} This can be expressed with the traditional erf function. </p>

http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution/104899#104899 noname 2012-08-17T08:32:36Z 2012-08-17T08:32:36Z

<p>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.</p>

http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution/105793#105793 Boz 2012-08-29T05:21:29Z 2012-08-29T05:21:29Z

<p>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. </p>

http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution/123917#123917 unknown (google) 2013-03-07T22:09:15Z 2013-03-07T22:09:15Z

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

http://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution/124156#124156 liuminzhao 2013-03-10T16:19:34Z 2013-03-10T16:19:34Z

<ol> <li>The last equation should be integral from $b/\sqrt{a^2+1}$</li> <li>In <code>I(a, b)</code>, <code>a</code> is supposed to be positive. When $a&lt;0$, the answer will be $\int_{-\infty}^{b/\sqrt{a^2+1}} \phi(t) dt$</li> </ol>