joint pmfs

[taz]

Can someone help with these questions...? I have no clue where to start...Thank you

Let X and Y be independent random variables taking values in the positive integers
and having the same PMF f(x) = 2−x for x = 1, 2, .... Find:
• (a) P(min{X, Y }  x)
• (b) P(X > Y )
• (c) P(X = Y )
• (d) P(X  kY ) for a given positive integer k
• (e) P(X divides Y )
• (f) P(X = rY ) for a given positive rational r

 

[pka]

I have no clue where to start
Let X and Y be independent random variables taking values in the positive integers
and having the same PMF f(x) = 2−x for x = 1, 2, .... Find:

I do not doubt that! Because as stated te question totally meaningless.
The PMF cannot be negative!

 

[taz]

yeah i just realized...sorry this actually meant f(x)=2^-x

 

[taz]

I still have no idea...help please...

 

[pka]

You have said nothing about Y. Have you?

 

[taz]

they are both positive integers and have the same pmf

 

[pka]

If Prof Haas stops by he may have seen such a problem before. He has far more experience with mathematical statistics than most of us here. I admit that I do not understand the problem.

 

[mark07]

Independence gives

\(\displaystyle \L f_{X,Y} (x,y) = \frac{1}{2^x 2^y}\)

With discrete random variables, integration means summation over sets (since the measure is the counting measure, for those interested in technical details).

You forgot to type the relation in some parts of your problem. So I'll do just one. Others are similar.

b) \(\displaystyle \L P(X>Y) = \sum_{x>y} f_{X,Y} (x,y)
= \sum_{i=1}^\infty \sum_{j>i}^\infty \frac{1}{2^i 2^j}\)

\(\displaystyle \L = \sum_{i=1}^\infty \left( \sum_{j=i+1}^\infty \frac{1}{2^j} \right) \frac{1}{2^i}\)

These are geometric series. I hope you can finish now...

 

[taz]

Thank you so much!!

 

[taz]

Is the answer for (a) 1?
the question was P(min{X,Y} <= x)
I got everything else, but the first one is a little comfusing

 

[mark07]

Well, it can't be 1. It has to be a function of x right?

Here's the trick:

\(\displaystyle \L P(\min(X,Y) \leq x) = 1 - P(\min(X,Y) > x)\)

\(\displaystyle \L = 1- P(X > x \text{ and } Y>x)
= 1 - \sum_{i=x+1}^\infty \sum_{j=x+1}^\infty \frac{1}{2^i 2^j}\)

I get

\(\displaystyle \L 1 - \frac{1}{2^{2x}}\)

using geometric sum formula again.