# Probability Problem(Urgend)

1. Dec 12, 2005

### Mathman23

Hi Guys,
I have this probability Problem where I have become stuck, therefore I pray that there is somebody in here who can give me some advice :)
It goes a something like that:
The two dimensional discrete Stochastic vector (X,Y) has the probability function $$P_{X,Y}$$ which is
$$P(X=x,Y=y) = \left\{ \begin{array}{ll}\frac{{c e^{- \lambda}{\lambda ^{y}}}}{{y!}} & \textrm{where} \ x \in (-2,-1,0,1) \ \textrm{and} \ \ y \in (0,1,\ldots)&\\0 & \textrm{other.}&\\\end{array} \right.$$
where $$\lambda > 0$$ and c > 0
(a) The support supp P_{X,Y} = {-2,-1,0,1, \ldots}
(b)
The Probability functions $$P_X$$ and $$P_Y$$ for X and Y are
$$P(X=x) = \left\{ \begin{array}{ll}c & \textrm{where} \ x \in (-2,-1,0,1) \\0 & \textrm{other.}&\\\end{array} \right.$$
$$P(Y=y) = \left\{ \begin{array}{ll}\frac{{c e^{- \lambda}{\lambda ^{y}}}}{{y!}} & \textrm{and} \ \ y \in (0,1,\ldots)&\\0 & \textrm{other.}&\\\end{array} \right.$$

This is done by showing that $$Y \ \~{} pol(\lambda)$$,

$$\sum_{y=0} ^{\infty} \frac{{e^{- \lambda}{\lambda ^{y}}}}{{y!}} =\sum_{y=1} ^{\infty} \frac{{ e^{- \lambda}{\lambda ^{y}}}}{{y-1!}} = \lambda \sum_{y=1} ^{\infty} \frac{{ e^{- \lambda}{\lambda ^{(y-1)}}}}{{(y-1)!}} = \lambda \sum_{v=0} ^{\infty} \frac{{ e^{- \lambda}{\lambda ^{(v)}}}}{{(v)!}} = \lambda$$
where v = (y-1)
I have two question
(c) I need to find the constant 'c'. How do I go about doing that?
(d)
If $$\lambda = 1$$, then $$P(X=Y) = 1/2 e^ -1$$. Any hits on how I show that ?
Sincerely and God bless You
Fred

Last edited: Dec 12, 2005
2. Dec 12, 2005

### EnumaElish

Is this homework?

3. Dec 12, 2005

### Mathman23

Yes,

I there is somebody out there who culd please give me a hit on how to solve question (c) and (d)??

Many thanks in advance and God bless You :)

/Fred

4. Dec 13, 2005

### EnumaElish

5. Dec 14, 2005