# Bivariate probability

1. May 9, 2009

### forty

Let the joint probability density function of random variables X and Y be given by

f(x,y) = 2 if 0 <= y <= x <= 1
and 0 otherwise

a) calculate the marginal probability density functions.

f(x) = 2x
f(y) = 2-2y

b) find E(X) and E(Y).

E(X) = 2/3
E(Y) = 1/3

c) calculate P(X<1/2) , P(X<2Y) and P(X=Y).

P(X<1/2) = integral from 0 to 1/2 of 2x dx = 1/4

but as for the other 2 I have no idea on how to do them. do they also involve the marginal distribution functions? Any help as always is greatly appreciated.

Thanks.

2. May 9, 2009

### jbunniii

These are correct, except remember that the equations are valid only for $$0 \leq x \leq 1$$ and $$0 \leq y \leq 1$$. Both f(x) and f(y) are zero for x and y outside those intervals.

Yep.

Correct.

Try calculating P(X<2Y) directly from the joint probability density function. All you need to do is work out what the region "X < 2Y" looks like and how to perform a double integral over that region.

For P(X=Y) my suggestion is similar: what does the region "X=Y" look like in the X-Y plane, and can you immediately identify the probability without even integrating?

3. May 10, 2009

### forty

Thanks for the help..

I can see by looking at the graph that the answer should be 1/2 but i dont know how to set up the integrals, i know it involves the f(x,y) and a double integral over dx and dy.

as for P(X=Y) is that 0 (does this have anything to do with it being continuous)?

4. May 10, 2009

### jbunniii

Notice that the original joint density function is zero unless Y <= X. What does this say about the probability that X < 2Y?

Yes and yes. The line X = Y is a region with zero area (it's just a line), so in order for the probability to be nonzero, there would have to be a concentration of probability on that line, which requires at least one non-continuous random variable.