# Poisson Distribution

1. Mar 26, 2009

### cse63146

1. The problem statement, all variables and given/known data

Let N,X1, X2, .... be independant random bariables where ?N has a poission Distribution with mean 3 while X1, X2.... each has a poisson distribution with mean 7

Determine $$E[N \sum^N_{i=1} X_i]$$

2. Relevant equations

3. The attempt at a solution

$$E[N \sum^N_{i=1} X_i] = E[N] * E[\sum^N_{i=1} X_i] = (3)(7) = 21$$

but that can't be right.

2. Mar 27, 2009

### lanedance

Hi cse63146

I would first look at for constant sum of m Xi's, with the Xi's representing the same distribution X independently, the sum gives:
$$E[ \sum^m_{i=1} X_i] = *E[X_1 + X_2 + ... + X_m ] = m*E[X]$$

3. Mar 27, 2009

### cse63146

so the expectation would be E[N2]*E[X]

E[N2] = Var(x) + E[N]2 - 3 + 32 = 12

E[N2]*E[X] = (12)(7) = 84

4. Mar 28, 2009

### Focus

You may want to use the tower law
$$\mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X|Y]]$$

You need to condition on the N to pull out the sum.

5. Mar 28, 2009

### cse63146

Let T = X1 + X2 + XN. E[T|N] = E[X1 + X2 + XN | N ]

E[T|N] = E[X1 + X2 + XN] = N*E[X]

E[T] = E[E[T|N]] = E[NE[X]] = E[N]*E[X]

$$E[N \sum^N_{i=1} X_i] = E[N^2]E[X]$$

E[N2] = Var(x) + E[N2] = 3 + 32 = 12

E[N2]*E[X] = (12)(7) = 84.

There's a second part of the question - Determine the variance of

$$\sum^N_{i=1} X_i$$

so Var (T|N) = Var(X1+ ... + Var(XN). Since X1 X2 XN are independant: Var(T|N) = N*Var(X).

Is that correct?