Dear Friends,

I have this there discrete stochastic vector

$\displaystyle P(X=x, Y=y) = \left\{ \begin{array}{cccc} \frac{1}{3} \cdot e^{-\lambda} \frac{\lambda^y}{y!} \mbox{x \in {-1,0,1} $\mathrm{and}$ y \in {0,1,\ldots}} \\ 0 \mbox{elsewhere.} \end{array}$

where $\displaystyle \lambda>0$

My question I know that $\displaystyle \sum_{n=0}^{\infty} e^{-\lambda} \frac{\lambda^n}{n!} = 1$

But does $\displaystyle \sum_{n=0}^{\infty} \frac{1}{3} e^{-\lambda} \frac{\lambda^n}{n!} = 1$ also do this??

If yes this doen't this prove that

$\displaystyle P(X=x) = \left\{\begin{array}{ccx} \frac{1}{3} \mbox{x \in {-1,0,1}} \\ 0 & \mathrm{elsewhere.} \end{array}$

and

$\displaystyle P(Y=y) = \left\{ \begin{array}{cccc} e^{-\lambda} \frac{\lambda^y}{y!} \mbox{y \in {0,1,\ldots}} \\ 0 \mbox{elsewhere.} \end{array}$

are the probability functions P_x and P_y for X and Y respectively??

Sincerely Yours and God bless You all

Billy

p.s. Is latex broken here? I heard that one had to use , but if I do that I get a post then build arrays error! Therefore I use . Hope thats allright.