Thread: Bounds for integration in joint pdf

Hi,

Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

Find

P(X=1/4, 0<y<3)

So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

Thanks.

Originally Posted by shogunhd

Hi,

Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

Find

P(X=1/4, 0<y<3)

So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

Thanks.

.

But since you have a continuous distribution, ......

So, do you mean find ?

If so:

First draw the region of integration. Then:

Originally Posted by mr fantastic

.

But since you have a continuous distribution, ......

So, do you mean find ?

If so:

First draw the region of integration. Then:

Note:

Originally Posted by mr fantastic

Note:

the question asks for P(x=1/4, 0<y<3). I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?

Originally Posted by shogunhd

the question asks for P(x=1/4, 0<y<3). Mr F says: Two things then:

1. Why did you have "My bounds for x are from 0 to 1/4" then?
2. I have shown why P(x=1/4, 0<y<3) = 0.

I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?

Because I thought you had misposted due to the fact that if it wasn't, then the answer is zero and there's no itegrtaion to do at all ......