Calculating E(XY) for Jointly Distributed Discrete Random Variables | Homework

[kingwinner]

Homework Statement

X and Y are joointly distributed discrete random variables with probability mass function
p_X,Y(x,y)=(c/e^c)(1-c)^xc^y/y!, x,y=0,1,2,..., 0<c<1
Find E(X^Y)

Homework Equations

The Attempt at a Solution

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/e^c, but what's next?

Thanks for any help!

 

[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?

 

[kingwinner]

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?

The x^y 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?

 

[Pere Callahan]

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

 

[kingwinner]

OK, so
E(X^Y)
∞
= ∑ [e^c (1-c)]^x (c/e^c)
x=0
={1/[1-e^c (1-c)]} (c/e^c)

Did I get it right?


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


Thank you!

 

[kingwinner]

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

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

 

[Pere Callahan]

OK, so
E(X^Y)
∞
= ∑ [e^c (1-c)]^x (c/e^c)
x=0
={1/[1-e^c (1-c)]} (c/e^c)

Did I get it right?

Yes.

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

Try and find the maximum of e^c (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.

 

[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?

 

[kingwinner]

Is it possible to prove it without using calculus?

Still wondering...