# Bivariate probability

• forty
In summary: Since X and Y are both continuous (uniformly distributed), the probability of them being exactly equal is zero.

#### 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.

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

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.

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

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

Yep.

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

Correct.

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.

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?

Thanks for the help..

I can see by looking at the graph that the answer should be 1/2 but i don't 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)?

Thanks for the help..

I can see by looking at the graph that the answer should be 1/2 but i don't know how to set up the integrals, i know it involves the f(x,y) and a double integral over dx and dy.

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

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

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.