Hi Guys,(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]P_{X,Y}[/tex] which is

[tex]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.[/tex]

where [tex]\lambda > 0[/tex] and c > 0

(a) The support supp P_{X,Y} = {-2,-1,0,1, \ldots}

(b)

The Probability functions [tex]P_X[/tex] and [tex]P_Y[/tex] for X and Y are

[tex]P(X=x) = \left\{ \begin{array}{ll}c & \textrm{where} \ x \in (-2,-1,0,1) \\0 & \textrm{other.}&\\\end{array} \right.[/tex]

[tex]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.[/tex]

This is done by showing that [tex]Y \ \~{} pol(\lambda)[/tex],

[tex]\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[/tex]

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 [tex]\lambda = 1[/tex], then [tex]P(X=Y) = 1/2 e^ -1[/tex]. Any hits on how I show that ?

Sincerely and God bless You

Fred

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Probability Problem(Urgend)

**Physics Forums | Science Articles, Homework Help, Discussion**