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June 21st, 2016, 02:41 PM  #1 
Member Joined: Nov 2013 Posts: 34 Thanks: 1  How to calculate P(X>=3Y) having density function f(x,y) =
How can I calculate P(X>=3Y) ? Thanks 
June 21st, 2016, 05:03 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493 
$\displaystyle \int_0^2 \int_0^{\frac{x}{3}}dydx=\frac{2}{3}$

June 22nd, 2016, 11:08 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,426 Thanks: 614 
Notice that, for x< 2, y= x/2< 1 so you really don't need "y< 1". You are integrating of the triangle bounded by y= 0, y= x/2, and x= 1. The line y= x/3 lies between y= 0 and y= x/2 for all x between 0 and 1 so "3y> x", which is the same as "y= x/3", we take x from 0 to 2 and, for each x, y from x/3 up to x/2. Mathman has a slight error he takes the integral, with respect to y, from 0 to x/3. That would give $\displaystyle P(X\le 3Y)$. Instead, to get $\displaystyle P(X\ge 3Y)$ you want to take the integral, with respect to y, from x/3 upto x/2: $\displaystyle \int_0^2\int_{x/3}^{x/2} 1 dy dx= \int_0^2 \frac{x}{2} \frac{x}{3}dx= \int_0^2\frac{1}{6} dx= \left[\frac{x}{6}\right]_0^2= \frac{2}{6}= \frac{1}{3}$. 
June 22nd, 2016, 01:37 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493  Quote:
 
June 22nd, 2016, 01:43 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493 
The problem can be solved without calculus, using elementary geometry. The density function can be visualized as points on a triangle (0,0), (2,0), (2,1). The condition X>3Y restricts to a triangle (0,0),(2,0),(2,2/3) which has area=2/3.


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