Cov(X,Y) Distribution Function

[Shackleford]

For (4), when I calculated the Distribution Function, I had to break it up into two cases whether or not the point was above or below the line.

For (6.c), I got the right answer for the Cov(X,Y), but I didn't break up the double integral into two cases. Is this because since the density function has the condition that y is less than or equal to x, it is inherent to the double integral? That's why I was able to set the limits of integration of y from 0 to x?

http://i111.photobucket.com/albums/n149/camarolt4z28/2-2.png?t=1302966321

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-3.png?t=1302966341

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110416_100322.jpg?t=1302966359

 

[Leptos]

For (4), when I calculated the Distribution Function, I had to break it up into two cases whether or not the point was above or below the line.

I couldn't see the picture for 4.

For (6.c), I got the right answer for the Cov(X,Y), but I didn't break up the double integral into two cases. Is this because since the density function has the condition that y is less than or equal to x, it is inherent to the double integral? That's why I was able to set the limits of integration of y from 0 to x?

Yes, that's correct since the restriction on x is that it must be positive but the restriction on y is that it is bounded by x so we could write that 0 < x < ∞ and 0 < y < x and it turns out from there we get our limits of integration with respect to x and y. I'm assuming you found E(X) = 2 and E(Y) = 1 since you found E(XY) = 3(which was calculated perfectly).

 

[Shackleford]

I couldn't see the picture for 4.

Yes, that's correct since the restriction on x is that it must be positive but the restriction on y is that it is bounded by x so we could write that 0 < x < ∞ and 0 < y < x and it turns out from there we get our limits of integration with respect to x and y. I'm assuming you found E(X) = 2 and E(Y) = 1 since you found E(XY) = 3(which was calculated perfectly).

Yes. I found the means all right.

I'm not sure how to start (a). The problem looks easy enough, though.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-4.png?t=1302983928

I assume in the uv-plane I draw the line v = z - u. I then setup a double integral ∫∫e^-z dudv.

I'm looking for P(Z ≤ z). This would be everything integrated above the line in the shaded region.

 

[Leptos]

Yes. I found the means all right.

I'm not sure how to start (a). The problem looks easy enough, though.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-4.png?t=1302983928

I assume in the uv-plane I draw the line v = z - u. I then setup a double integral ∫∫e^-z dudv.

I'm looking for P(Z ≤ z). This would be everything integrated above the line in the shaded region.

Actually the change of variables would transform the original density f(x,y) = e^-(x+y) to a function of z alone, i.e, f(z) = e^-z whose only restriction is that 0 < z and now all you have to do is apply the definition of the cumulative density function: F(Z) = P(X ≤ x).

 

[Shackleford]

Actually the change of variables would transform the original density f(x,y) = e^-(x+y) to a function of z alone, i.e, f(z) = e^-z whose only restriction is that 0 < z and now all you have to do is apply the definition of the cumulative density function: F(Z) = P(X ≤ x).

I assume I integrate f(z) = e^-z. What the heck are my variables and limits of integration? 0 to infinity? dx and dy? Double or single integral?

If there weren't a change of variables, I would calculate a double integral f_X,Y(x,y) = e^-(u+v) with du and dv from 0 to x and 0 to y respectively.