Poisson Distribution: Find E[N ∑Nᵢ₁Xᵢ]

[cse63146]

Homework Statement

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

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

Homework Equations

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.

 

[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:
$$E[ \sum^m_{i=1} X_i] = *E[X_1 + X_2 + ... + X_m ] = m*E[X]$$

 

[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

 

[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.

 

[cse63146]

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

E[T|N] = E[X₁ + X₂ + X_N] = 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(X_N). Since X₁ X₂ X_N are independent: Var(T|N) = N*Var(X).

Is that correct?