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

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# Probability Problem(Urgend)

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