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

• tronter
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.
tronter
If $$f(x,y) = e^{-x-2y}$$ find $$P(X<Y)$$.

So is this equaled to $$1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?$$

What is the domain? $\int_0^\infty\int_0^\infty f(x,y)dydx$=1/2, not 1.

the domain is $$0 < x < \infty$$, $$0 < y < \infty$$. It is equaled to $$1$$, so its a valid pdf.

On that domain the probability doesn't add up to 1.

whoops, I meant $$-\infty < x < \infty$$, $$-\infty < y< \infty$$.

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:
forgetting about the bounds for the moment, is the general set up correct?

See my previous post (edited).

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