Marginal PDF Problem: Accidental Edit Fixed

[Ninty64]

Oops, I accidentally edited over this post =(
Thankfully it's quoted in the next one.

 

[LCKurtz]

This is my first post, please excuse my inability to make my posts look nice.

Homework Statement

Let f(x,y) = 3/2, x²[tex]\leq[/tex]y[tex]\leq[/tex]1, 0[tex]\leq[/tex]x[tex]\leq[/tex]1, be the joint pdf of X and Y.
(a) Find P(0 [tex]\leq[/tex] x [tex]\leq[/tex] 1/2)
(b) Find P(1/2 [tex]\leq[/tex] y [tex]\leq[/tex] 1)
(c) Find P(1/2 [tex]\leq[/tex] x [tex]\leq1[/tex] 1, 1/2 [tex]\leq[/tex] y [tex]\leq[/tex] 1)
(d) Find P(X [tex]\geq[/tex] 1/2, Y [tex]\geq[/tex] 1/2)
(e) Are X and Y independent? Why or why not?

Homework Equations

f₁(x) = integral(f(x,y)dy)
f₂(y) = integral(f(x,y)dx)

They are independent iff f(x,y) = f₁(x) * f₂(y)

The Attempt at a Solution

(a) I did integral(3/2dy) from y = x² to y = 1, which gave me a marginal pdf of x to be:
f₁(x) = (3/2) - (3/2)x²
Then I integrated from 0 to 1/2 to get the answer 11/16.

Good so far.

(b) I have a problem with this question. I tried to get the marginal pdf of y:
f₂(y) = integral(3/2dx) from 0 to 1
which gives me:
f₂(y) = 3/2

If you draw a picture of your region you will see that the limits on x depend on y. x only goes from 0 to 1 if y = 1. Otherwise the upper limit depends on y. You need to redo f₂.

 

[Ninty64]

If you draw a picture of your region you will see that the limits on x depend on y. x only goes from 0 to 1 if y = 1. Otherwise the upper limit depends on y. You need to redo f₂.

Oh! I get it! x goes from 0 to sqrt(y)
So...
f₂(y) = integral(3/2dx) from 0 to sqrt(y)
which becomes:
f₂(y) = (3/2)sqrt(y)
Then when I integrate from 1/2 to 1, I get that the probability is 1 - sqrt(2)/4

Well that also answers (e). They are definitely dependent since
(3/2)sqrt(y) * (3/2 - (3/2)x^) =/= 3/2

Thank you! =)

That leaves me with one last question. (Assuming that my other work was right at least). Does this mean that (c) is asking me for the double integral of x(1/2 to 1) and y(1/2 to 1) while (d) is asking me for the double integral of x(1/2 to sqrt(y)) and y(1/2 to 1)?

 

[LCKurtz]

Oh! I get it! x goes from 0 to sqrt(y)
So...
f₂(y) = integral(3/2dx) from 0 to sqrt(y)
which becomes:
f₂(y) = (3/2)sqrt(y)
Then when I integrate from 1/2 to 1, I get that the probability is 1 - sqrt(2)/4

Well that also answers (e). They are definitely dependent since
(3/2)sqrt(y) * (3/2 - (3/2)x^) =/= 3/2

Thank you! =)

That leaves me with one last question. (Assuming that my other work was right at least). Does this mean that (c) is asking me for the double integral of x(1/2 to 1) and y(1/2 to 1) while (d) is asking me for the double integral of x(1/2 to sqrt(y)) and y(1/2 to 1)?

You could have seen intuitively that X and Y were not independent because, for example, if you know Y < 1 you would know X couldn't be 1.

As to (c) and (d) it looks to me as if they are asking the same question. And integrating f(x,y) with x from 1/2 to 1 and y from 1/2 to 1 is the same thing as integrating f(x,y) with x from 1/2 to sqrt(y) and y from 1/2 to 1. Remember f(x,y) is zero everywhere except where it is given to be 3/2.

 

[Ninty64]

You could have seen intuitively that X and Y were not independent because, for example, if you know Y < 1 you would know X couldn't be 1.

As to (c) and (d) it looks to me as if they are asking the same question. And integrating f(x,y) with x from 1/2 to 1 and y from 1/2 to 1 is the same thing as integrating f(x,y) with x from 1/2 to sqrt(y) and y from 1/2 to 1. Remember f(x,y) is zero everywhere except where it is given to be 3/2.

Thanks! That makes a lot of sense. =)