# Thread: Poisson and Binomial Variables

1. ## Poisson and Binomial Variables

See Attachment

2. 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)

3. Originally Posted by lord12
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. $\sum_{x = 0}^1 \sum_{y=0}^{1} p_{XY}(x,y) = 1$. (Some clues to doing this are given in the solution to (b))

2. $p_{XY}(x,y) \geq 0$ for x = 0, 1, 2, ..... and y = 0, 1.

(b) $p_X(x) = \sum_{y=0}^{1} p_{XY}(x,y)$ $=\frac{\lambda^x(1 - e^{-\lambda})}{e^{\lambda} x!} + \frac{\lambda^x}{e^{2 \lambda} x!}$ $= \frac{\lambda^x e^{-\lambda}}{x!}$.

$p_Y(y) = \sum_{x=0}^{\infty} p_{XY}(x,y)$ $= \left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} =$ $\left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] e^{\lambda}$ $= (1 - e^{-y})^{1-y} (e^{-\lambda})^y$

which is clearly recognised as the pdf for a binomial random variable with n = 1 and $p = e^{-\lambda}$.

(c) Does $p_{XY}(x, y) = p_{X}(x) p_{Y}(y)$ .... ?

4. Originally Posted by lord12
I need help
This is the fourth infraction you have received for bumping, you are banned for a week.

CB