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The (old qualifying exam) question is this: if $X,Y$ are independent standard normals, what is the distribution of $Z=X+Y$ given that $X>0,Y>0$?

We must find $P(Z\le z, X>0,Y>0)$ (and then divide this by $P(X>0,Y>0)=\frac14$), which means integrating $\frac1{2\pi}e^{-(x^2+y^2)/2}$ over the triangle $x+y\le z$ in the first quadrant. I figured using polar coordinates was the right idea. Converting $x+y=z$ to polar, we get $r(\cos\theta+\sin\theta)=r\sqrt{2}\cos(\theta-\pi/4)=z$, so \begin{align*} P(Z\le z,X>0,Y>0) &=\frac1{2\pi}\int_0^{\pi/2}\int_0^{\frac{z\sec(\theta-\frac\pi4)}{\sqrt2}}e^{-r^2/2}\cdot r\,dr\,d\theta\\ &=\frac1{2\pi}\int_0^{\pi/2}(-e^{-r^2/2})\bigg|^{\frac{z\sec(\theta-\frac\pi4)}{\sqrt2}}_0\,d\theta\\ &=\frac1{2\pi}\int_0^{\pi/2}1-\exp\left(-\frac{z^2\sec^2(\theta-\pi/4)}{4}\right)\,d\theta\\ &=\frac14-\frac{e^{-z^2/4}}{2\pi}\int_0^{\pi/2}\exp\left(-\frac{\tan^2(\theta-\pi/4)}{4}\right)\,d\theta \end{align*} However, neither I nor Mathematica know what to do with the last integral. Am I missing any clever tricks, or is the best we can do to write $P(Z\le z|X>0,Y>0)=1-\frac{2}\pi\cdot e^{-z^2/4}\cdot\tau$, with $\tau$ being the value of the horrible integral?

Mike Earnest
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    If it's really true that $P(Z\le z|X>0,Y>0)=1-\frac{2}\pi\cdot e^{-z^2/4}\cdot\tau$ (apologies, I haven't checked your work), then set $z=0$. Then $0 = 1 - {2\over\pi}\tau$, and you've got the value of $\tau$. – grand_chat Feb 24 '15 at 00:53
  • Your comment has made me realize my work was wrong. – Mike Earnest Feb 24 '15 at 01:28
  • For $z>0$, isn't $P(Z\le z,X>0,Y>0)=\int_0^\infty P(0<X\le z-y)\phi(y),dy=\int_0^\infty \Phi(z-y)\phi(y),dy-1/4$ ? Then to find the density one would have to find $\int_0^\infty \phi(z-y)\phi(y),dy$, which could be possibly done. – StubbornAtom Apr 12 '18 at 08:37
  • Could you tell me what the final answer should turn out to be for the density of $Z\mid X>0,Y>0$ ? – StubbornAtom Apr 12 '18 at 12:00

1 Answers1

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Because of the circular symmetry of the joint density of two independent standard normal random variables,

  • $P\{(X,Y) \in \text{triangle with vertices at $(0,0), (z,0), (0,z)$}\}$ is one-fourth the probability that the random point lies in the square with vertices $(z,0), (0,z), (-z,0), (0,-z)$.

  • Rotating the square about the origin till the sides are parallel to the axes does not change $P\{(X,Y) \in \text{rotated square}\}$

  • $P\{(X,Y) \in \text{rotated square}\}$, whose vertices are at $\left(\pm \frac{z}{\sqrt{2}}, \pm \frac{z}{\sqrt{2}}, \right)$, is left to you as an exercise.

Can you take it from here?

Dilip Sarwate
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  • Yes I can, very nice answer! – Mike Earnest Feb 24 '15 at 01:27
  • Apologies for reviving this old post but I was looking for a solution to this very problem and found this. So, if I am not wrong (correct me please!) the conditional distribution turns out to be $P(X\in[-z/\sqrt{2},z/\sqrt{2}],Y\in[-z/\sqrt{2},z/\sqrt{2}])=(2\Phi\big(\dfrac{z}{\sqrt{2}})-1)^2$ from which the conditional density is $f(z)=2(2\Phi(\dfrac{z}{\sqrt{2}}-1)\sqrt{2}\phi(\dfrac{z}{\sqrt{2}})$. Now, if I want to take $E[X+Y|X>0,Y>0]$ then this form will not really help me, will it? Can $\int_0^\infty zf(z)dz$ be evaluated from this? Thanks!! – Landon Carter Apr 19 '15 at 16:22
  • @yedaynara Don't ask a different question in a comment; write up a separate question. – Dilip Sarwate Apr 19 '15 at 19:13
  • @Dilip Sarwate please look at http://math.stackexchange.com/questions/1242305/clever-way-of-finding-int-0-infty-x-phix-phixdx Here is the part I am stuck at, while finding the expectation. – Landon Carter Apr 19 '15 at 19:14