# Clever way of finding $\int_0^\infty x\Phi(x)\phi(x)dx$

Suppose that $\Phi$ and $\phi$ are the Standard Normal c.d.f and p.d.f. respectively. Then, evaluate $$\int_0^\infty x\Phi(x)\phi(x)dx$$

There is no use of my trying to show my approach because none of the techniques I used could initiate the solving process. I have, however, obtained the value of $$\int_{-\infty}^\infty x\Phi(x)\phi(x)dx=\dfrac{1}{\sqrt{2\pi}}$$Any help is appreciated.

The way I proceeded:

1. Orthogonalization of the unit square in which the points $(X,Y)$ are situated. This did not work.

2. Supposing that the centroid of the triangle with vertices at $(0,0),(0,z),(z,0)$ will be the answer due to the symmetry between $X$ and $Y$.

3. Integration by parts repeatedly taking $\Phi$ as the second function, which yielded nothing practically.

4. Using the conditional m.g.f.

The question arose in an attempt to find $E[X+Y|X>0,Y>0]$ where one term contained this integral. The question came in a semester exam. I will not add "self-study" to this because the question precisely stated that this is an immensely important problem in Monte Carlo Methods, something which I have not read.

• You could show how you arrived at the integral you did obtain. Then we could see why you might have had trouble with the other integral. – robjohn Apr 19 '15 at 19:41
• Well I did not think that finding this integral was in any way similar to the problem. So, I did not post it. Okay, if you insist, let me post the solution to this integral. – Landon Carter Apr 20 '15 at 3:20
• By the way, why has this been voted as off-topic? – Landon Carter Apr 20 '15 at 3:22
• robjohn, Actually while I was uploading the solution to find this new integral, I realized there has been a major error. So the integral that I wrote is incorrect. I want to delete that part. But then, will it not seem as if no research effort has been shown, while I had spent literally one whole day trying in vain everything I knew? – Landon Carter Apr 20 '15 at 3:28
• And you've gotten one reopen vote already. I would vote to reopen myself, but since I am a moderator, my vote would immediately reopen the question. It is best to let the community decide. – robjohn Apr 20 '15 at 9:52

Note that $$-xe^{-x^2/2}\,\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}x}e^{-x^2/2}$$ Since $\Phi(0)=\frac12$, we can integrate by parts to get \begin{align} \int_0^\infty x\Phi(x)\phi(x)\,\mathrm{d}x &=\frac1{\sqrt{2\pi}}\int_0^\infty x\Phi(x)e^{-x^2/2}\,\mathrm{d}x\\ &=-\frac1{\sqrt{2\pi}}\int_0^\infty \Phi(x)\,\mathrm{d}e^{-x^2/2}\\ &=\frac1{2\sqrt{2\pi}}+\frac1{\sqrt{2\pi}}\int_0^\infty e^{-x^2/2}\left(\frac1{\sqrt{2\pi}}e^{-x^2/2}\right)\,\mathrm{d}x\\ &=\frac1{2\sqrt{2\pi}}+\frac1{2\pi}\int_0^\infty e^{-x^2}\,\mathrm{d}x\\ &=\frac1{2\sqrt{2\pi}}+\frac1{2\pi}\frac{\sqrt\pi}2\\ &=\frac{1+\sqrt2}{4\sqrt\pi} \end{align}
• Using the same procedure as above, $$\int_{-\infty}^\infty x\Phi(x)\phi(x)\,\mathrm{d}x=\frac1{2\sqrt\pi}$$ – robjohn Apr 19 '15 at 19:48
Have you tried integration by parts together with the knowledge that the antiderivative of $x\phi(x)$ is $-\phi(x)$? The answer is readily computable via this approach.