# finding the probability from joint distribution of x and y

Below is an examination problem. Please consider the question in image

This is what I tried:

$$\max(X,-X)= \frac{x-x}{2} -{|x+x|}= |x|$$

using the identity $\max(X,Y)= \frac{x+y}{2} -{|x-y|}$

So that now we need to find $1 - P(-x \le y \le x)$. Is this approach correct? If so, I'm now stuck choosing limits for $x$. Any suggestions?

• If you look at max(X,Y) suppose X >Y so X is the max. Then the RHS is (X+Y)/2 -|X-Y|/2 = X/2 +Y/2 -(X-Y)/2 =X/2 -X/2 +Y/2 -Y/2 = Y a contradiction when X>Y. So the same mistake will carry over when you replace Y with -X. So at a minimum you have to fix those "identities". – Michael Chernick Jan 10 '17 at 12:45
• That warning bells don't go off in your mind when you find that $\max(x,-x)$ equals $-|x|$ (instead of $|x|$) is worrisome. Surely, $\max(3,-3) = \max(-3,3)$ should be $3$ and not $-3$, no? It isn't what you know that will kill you, it is what you know that just a'int so. – Dilip Sarwate Jan 10 '17 at 14:42
• So now the question is, Why does $\frac{x-x}{2} -{|x+x|}$ equal $|x|$ rather than $-2|x|$? You really need to work on your elementary algebra skills.... – Dilip Sarwate Jan 10 '17 at 17:41

Approach the question in a graphical manner. See that the PDF is non-zero for a circle on the x-y plane. Now visualise the area where Y > max(X, -X). This gives you a sector on the above circle. You should now be able to see that the answer is 1/4

• +1. You should add, "Look, Ma! No integration needed!" – Dilip Sarwate Jan 10 '17 at 14:46
• @ujjwal thanks .,., but if we use integration what will be the limits for x ? – ANUJ NAIN Jan 10 '17 at 16:42
• If you must solve using integration, you'd have to transform you space from coordinate to spherical system. Since your PDF is constant, for r <1, you now just have to integrate areas formed with r and theta varying from pi/4 to 3*pi/4 – Ujjwal Kumar Jan 10 '17 at 17:13
• But integration is certainly an overkill – Ujjwal Kumar Jan 10 '17 at 17:14