See Attachment
For the Poisson random variables I get ((lamda)^x)/((e^2lambda)*x!) when y =1. Is this a Poisson Random variable? Also, when y = 0, I get ((lamda)^x)/((e^lambda)*x!) - ((lamda)^x)/((e^2lambda)*x!) I cannot get the binomial. I don't know how to show that this is Binomial as x is in the range from 1...infinity.
For part 1 I used the fact that P(X=x and Y=y) = P(X|Y)P(Y)
and P(Y=y) = sum(1 to x)P(X=x and Y=y) = 1, so P(X=x|Y=y)(P(y) = P(X=x and Y=y)
(a) You need to show that:
1. $\displaystyle \sum_{x = 0}^1 \sum_{y=0}^{1} p_{XY}(x,y) = 1$. (Some clues to doing this are given in the solution to (b))
2. $\displaystyle p_{XY}(x,y) \geq 0$ for x = 0, 1, 2, ..... and y = 0, 1.
(b) $\displaystyle p_X(x) = \sum_{y=0}^{1} p_{XY}(x,y)$$\displaystyle =\frac{\lambda^x(1 - e^{-\lambda})}{e^{\lambda} x!} + \frac{\lambda^x}{e^{2 \lambda} x!}$ $\displaystyle = \frac{\lambda^x e^{-\lambda}}{x!}$.
$\displaystyle p_Y(y) = \sum_{x=0}^{\infty} p_{XY}(x,y)$$\displaystyle = \left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} =$$\displaystyle \left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] e^{\lambda} $$\displaystyle = (1 - e^{-y})^{1-y} (e^{-\lambda})^y$
which is clearly recognised as the pdf for a binomial random variable with n = 1 and $\displaystyle p = e^{-\lambda}$.
(c) Does $\displaystyle p_{XY}(x, y) = p_{X}(x) p_{Y}(y)$ .... ?