# Homework Help: Expected Value: E(X^Y)

1. Dec 6, 2008

### kingwinner

1. The problem statement, all variables and given/known data
X and Y are joointly distributed discrete random variables with probability mass function
pX,Y(x,y)=(c/ec)(1-c)xcy/y!, x,y=0,1,2,..., 0<c<1
Find E(XY)

2. Relevant equations
3. The attempt at a solution
Code (Text):
By definition,
E(X[SUP]Y[/SUP])
∞    ∞
= ∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
x=0  y=0

How can we calculate this double sum? We can pull out the constant c/ec, but what's next?

Thanks for any help!

2. Dec 6, 2008

### Pere Callahan

Do the y-sum first and use
$$\sum_{y=0}^\infty{\frac{Z^y}{y!}}=e^Z$$
What is Z here?
Next do the x-sum using
$$\sum_{x=0}^\infty{a^x}=\frac{1}{1-a}$$
What is a here? Is |a|<1?

3. Dec 6, 2008

### kingwinner

The xy is giving me some trouble doing the y-sum, it depends on both x and y, and I cannot separate it into a product, what should I do?

4. Dec 7, 2008

### Pere Callahan

When doing the y-sum treat x as a constant.

5. Dec 9, 2008

### kingwinner

OK, so
E(XY)

= ∑ [ec (1-c)]x (c/ec)
x=0
={1/[1-ec (1-c)]} (c/ec)

Did I get it right?

Also, why is |ec (1-c)|<1 ? How can we see this? I can only show 0<c<1 => 0<1-c<1 => 0<(1-c)ec<ec < e, but we need (1-c)ec<ec < 1, how can we prove this? This is giving me so much headache...please help...

Thank you!

Last edited: Dec 9, 2008
6. Dec 9, 2008

### kingwinner

Just out of curiousity, in this case, is it possible to sum over x FIRST, and then sum over y?

Code (Text):
∞    ∞
∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
y=0 x=0

7. Dec 9, 2008

### Pere Callahan

Yes.
Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function You will find that the function has its maximum when c=0 and this maximum value is 1.

For your other question: Sure you can but it's a lot more complicated.

8. Dec 9, 2008

### kingwinner

"Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function You will find that the function has its maximum when c=0 and this maximum value is 1."

OK, I got it! Thanks!
Is it possible to prove it without using calculus?

9. Dec 11, 2008

### kingwinner

Still wondering...