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June 21st, 2016, 02:41 PM   #1
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How to calculate P(X>=3Y) having density function

f(x,y) =
  • 1 if 0<= x <= 2 ; 0<=y<=1 ; 2y<=x
  • 0 (another combination)

How can I calculate P(X>=3Y) ?
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June 21st, 2016, 05:03 PM   #2
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$\displaystyle \int_0^2 \int_0^{\frac{x}{3}}dydx=\frac{2}{3}$
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June 22nd, 2016, 11:08 AM   #3
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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:
$\displaystyle \int_0^2\int_{x/3}^{x/2} 1 dy dx= \int_0^2 \frac{x}{2}- \frac{x}{3}dx= \int_0^2\frac{1}{6} dx= \left[\frac{x}{6}\right]_0^2= \frac{2}{6}= \frac{1}{3}$.
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June 22nd, 2016, 01:37 PM   #4
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Quote:
Originally Posted by Country Boy View Post
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:
$\displaystyle \int_0^2\int_{x/3}^{x/2} 1 dy dx= \int_0^2 \frac{x}{2}- \frac{x}{3}dx= \int_0^2\frac{1}{6} dx= \left[\frac{x}{6}\right]_0^2= \frac{2}{6}= \frac{1}{3}$.
I have it right. Your assertion has the reverse. x/3<y<x/2 or P(2Y<X<3Y)
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June 22nd, 2016, 01:43 PM   #5
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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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