# Statistics/Probability: Joint Densities

1. Dec 5, 2005

### ak416

Ok Whats the general way to solve problems such as these?
-Let X and Y have the joint pdf f_X,Y(x,y) = 2e^-(x+y) , 0 < x < y, 0 < y
Find P(Y < 3X).
-Find P(X < 2Y) if f_X,Y(x,y) = x + y for X and Y defined over the unit interval (meaning from 0 to 1).
I know you have to set up the double integral, like for the first one the answer is:

but P(Y < 3X) => P(x > 1/3Y) so why cant you take the integral from 1/3Y to Y wrt X (because x has to be smaller than y) and the integral from 0 to infinity with respect to y (because y has to be greater than 0)?
Note:I am currently taking a multivariable calculus course and I havent gotten to multi-integration yet, so i would just like to know the general method for determining the boundaries of integration.
I also have other questions (about Transforming random variables) but ill save those for later, I would like to understand this first.
Thanks.

2. Dec 5, 2005

### HallsofIvy

You can. And you would get the same answer.

In the xy-plane, 0< x< y, 0< y is infinite triangle between the y-axis and the line y= x. In order to be a probability density over that region, the integral of 2e^-(x+y) over that region must be 1.
In subset of that region such that y< 3x is the infinite triangle between the lines y= 3x and y= x. P(Y< 3X) is the integeral of 2e^-(x+y) over that region.
There are always two ways of setting up a double integral over a region like that. We can take the outer integral in y taking y going from 0 to infinity. For each such y, a horizontal line in that region must go from y= 3x to y=x which is equivalent to x ranging from y/3 up to y. The probability is
$$2\int_{y= 0}^\infty\int_{x= y/3}^y e^{-(x+y)}dxdy$$

But you could also take the outer integral to be in terms of x, with x running form 0 to infinity. In that case, for each x, a vertical line covering the region would have y running from x to 3x. The probability is
$$2\int_{x=0}^\infty\int_{y= x}^{3x} e^{-(x+y)}dydx$$

3. Dec 5, 2005

### ak416

Thanks, that really helped me visualize what's going on. But what about about the second problem, find P(X < 2y)... either my computations are wrong or integral from 0 to 1 wrt y and integral from 0 to 2y wrt x (with the pdf in between) is wrong. (Gives me a probability of 4/3). and Why?