1. P(x<2y)

Pretty simple question:

f(x,y) = {2 if 0 <= x <= y <= 1 , 0 elsewhere}

How do I compute P(X<2Y) ?

Similarly, if I'm given the following CDF,

F(x,y) = (1-exp(-x^2))(1-exp(-y^2)) on {x,y>0}

How would I compute P(X<2Y) here?

2. Originally Posted by rockyraccoon
Pretty simple question:

f(x,y) = {2 if 0 <= x <= y <= 1 , 0 elsewhere}

How do I compute P(X<2Y) ?

Similarly, if I'm given the following CDF,

F(x,y) = (1-exp(-x^2))(1-exp(-y^2)) on {x,y>0}

How would I compute P(X<2Y) here?

In both cases draw a picture and the integral terminals become obvious.

1. $\displaystyle \int_{x = 0}^{1/2} \int_{y = x/2}^{1} f(x, y) \, dy \, dx$.

2. $\displaystyle \int_{x = 0}^{+ \infty} \int_{y = x/2}^{+ \infty} f(x, y) \, dy \, dx$.

3. That's what I was doing as well, although the answers in the back of the book don't match up.

For the first question, the answer is 1/2.

For the second, the answer is 1/5.

Thoughts?

4. Originally Posted by rockyraccoon
That's what I was doing as well, although the answers in the back of the book don't match up.

For the first question, the answer is 1/2.

For the second, the answer is 1/5.

Thoughts?
My reply to question 1 is wrong - I misread the support - but my advice remains: Draw a picture and the integral terminals become obvious (provide the question is read properly):

1. $\displaystyle \int_{x = 0}^{1} \int_{y = x/2}^{x} f(x, y) \, dy \, dx$.

My reply to question 2 is also wrong - I misread the equation as the pdf when it is in fact the cdf. You have a number of options for answering the question, including using the given cdf to get the pdf and then integrating the pdf over the region I gave.