Expectation of a Joint Distribution

cborse
The below gives all the information I was given. I'm pretty sure my answer is right, but a part of me isn't, and that's why I'm asking here.

Homework Statement


Let (X,Y) have the joint pdf:

f_{XY}(x,y)=e^{-y}, 0 < x < y < \infty

Find E(XY).

Homework Equations



E(XY)=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}xyf_{XY}(x,y)dxdy


The Attempt at a Solution


Using the limits 0 < x < y and 0 < y < ∞,

E(XY)=\int^{\infty}_{0}\int^{y}_{0}xye^{-y}dxdy=\frac{1}{2}\int^{\infty}_{0}y^{3}e^{-y}dy=3

Also, are my bounds correct?
 
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Looks good, and I can confirm the result of the integrations.
 
Thank you very much. Now if I was trying to find f_{X}(x) and f_{Y}(y), which bounds would I use? I tried this (same info as above):

Since f_{X}(x)=\int^{\infty}_{-\infty}f_{XY}(x,y)dy,

f_{X}(x)=\int^{\infty}_{x}e^{-y}dy=e^{-x}

and

f_{Y}(y)=\int^{y}_{0}e^{-y}dx=ye^{-y}

Assuming those are correct, which bounds do I use for E(X) and E(Y)? Because if I use the same bounds that I used for the marginal functions, I'll get the variable x in Y's mean, and the variable y in X's mean.
 
Because if I use the same bounds that I used for the marginal functions, I'll get the variable x in Y's mean, and the variable y in X's mean.
Why? You can calculate E(X) and E(Y) with the original distribution and get a real number, or calculate it with your new distributions (in post 3) and get a real number.
 
If I use 0 < x < y for E(X) and x < y < ∞ for E(Y), they'll each contain the other variable. For example,

E(X)=\int^{y}_{0}xe^{-x}=1-e^{-y}(y+1)

I have a very strong feeling I'm using the wrong bounds...

EDIT: If I use 0 < x < ∞ and 0 < y < ∞,

E(X)=\int^{∞}_{0}xe^{-x}=1

E(Y)=\int^{∞}_{0}y^2e^{-y}=2

Is that correct?
 
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Your edit looks right.
 
cborse said:
Thank you very much. Now if I was trying to find f_{X}(x) and f_{Y}(y), which bounds would I use? I tried this (same info as above):

Since f_{X}(x)=\int^{\infty}_{-\infty}f_{XY}(x,y)dy,

f_{X}(x)=\int^{\infty}_{x}e^{-y}dy=e^{-x}

and

f_{Y}(y)=\int^{y}_{0}e^{-y}dx=ye^{-y}

Assuming those are correct, which bounds do I use for E(X) and E(Y)? Because if I use the same bounds that I used for the marginal functions, I'll get the variable x in Y's mean, and the variable y in X's mean.

No. You seem to be plugging into formulas without understanding what you are doing. The marginal ##f_X(x)## has no ##y## in it, and the marginal ##f_Y(y)## has no ##x## in it. Here is how it works:
f_X(x) \, \Delta x = P\{ x &lt; X &lt; x+\Delta x \} = \int_{y=x}^{\infty} f(x,y)\, \Delta x \, dy<br /> = \Delta x \times \int_{y=x}^{\infty} f(x,y)\, dy,
so ##y## is gone: it has been "integrated out".
 
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