# joint pdf probability

• Nov 12th 2010, 01:32 PM
wik_chick88
joint pdf probability
suppose X and Y have the joint pdf
$\displaystyle f_{X, Y} = x + y, 0 < x, y < 1$
Find P(X < 2Y)
• Nov 12th 2010, 05:21 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
suppose X and Y have the joint pdf
$\displaystyle f_{X, Y} = x + y, 0 < x, y < 1$
Find P(X < 2Y)

$\displaystyle \displaystyle \Pr(X < 2Y) = \int_{y=0}^{y = 1} \int_{x = 2y}^{x = 2} f(x, y) \, dx \, dy$.
• Nov 13th 2010, 07:50 AM
matheagle
Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square.

I'd switch the order of integration...

$\displaystyle 1-\int_0^1\int_0^{x/2} (x+y)dydx$
• Nov 13th 2010, 12:01 PM
mr fantastic
Quote:

Originally Posted by matheagle
Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square.

I'd switch the order of integration...

$\displaystyle 1-\int_0^1\int_0^{x/2} (x+y)dydx$

Right you are. I misread the support.
• Nov 13th 2010, 02:48 PM
matheagle
thats why I'm the math beagle
• Nov 27th 2010, 11:21 PM
wik_chick88
the final answer i got is $\displaystyle \frac{19}{24}$ is this correct? thanks so much!