Is P(X<Y) Equal to 1/3 for f(x,y) = e^{-x-2y}?

In summary, the conversation discusses finding the probability of X being less than Y, given the function f(x,y) = e^{-x-2y}. The domain for the function is 0 < x < \infty and 0 < y < \infty, and the probability should add up to 1. However, there is a discrepancy in the lower bound, with a suggested lower bound of -Log[2]/3 or defining a cumulative distribution function F*(x,y) = F(x,y) + 1/2. Overall, the formula for finding the probability is correct, but there may be some issues with the bounds.
  • #1
tronter
185
1
If [tex] f(x,y) = e^{-x-2y} [/tex] find [tex] P(X<Y) [/tex].

So is this equaled to [tex] 1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?[/tex]
 
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  • #2
What is the domain? [itex]\int_0^\infty\int_0^\infty f(x,y)dydx[/itex]=1/2, not 1.
 
  • #3
the domain is [tex] 0 < x < \infty [/tex], [tex] 0 < y < \infty [/tex]. It is equaled to [tex] 1 [/tex], so its a valid pdf.
 
  • #4
On that domain the probability doesn't add up to 1.
 
  • #5
whoops, I meant [tex] -\infty < x < \infty [/tex], [tex] -\infty < y< \infty [/tex].
 
  • #6
It looks like the lower bound should be -Log[2]/3.

For a "suitable" lower bound that equates the double integral to 1, your formula is correct.

Alternatively, since integration over [0, +infinity) gives 1/2, you might define the cumulative distribution F*(x,y) = F(x,y) + 1/2, where F(x,y) is the double integral of f(s,t) over s from 0 to x and t from 0 to y. In this case f is a valid pdf.
 
Last edited:
  • #7
forgetting about the bounds for the moment, is the general set up correct?
 
  • #8
See my previous post (edited).
 

Related to Is P(X<Y) Equal to 1/3 for f(x,y) = e^{-x-2y}?

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