1. ## Poisson Distributiom

i just started stats uni this year so stats is completely new to me. can someone please help me with this question which i'm sure must be really easy, but i dont understand a thing!

Suppose that X has a poisson distribution with mean lamda.
a) find by summation the mean of X
b) find also the variance of X

If W is a poisson random variable with mean 2, what is P(W>3 given W>1) ?
[for this question we have to use the tables right? and not the equation?)

If x1 = 3 ; x2 = 2 ; x3 = 4 ; x4 = 2 ; x5 = 5, and all are equally likely values for X, what is E[X(X-1)] ?

thanks so much...

2. Originally Posted by star_Asia
i just started stats uni this year so stats is completely new to me. can someone please help me with this question which i'm sure must be really easy, but i dont understand a thing!

Suppose that X has a poisson distribution with mean lamda.
a) find by summation the mean of X
b) find also the variance of X
If $X$ is Poission it means it takes values for $n=0,1,2,...$ with probability $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$.

To find the mean you need to compute: $E[X] = \sum_{n=0}^{\infty} n\frac{\lambda^n e^{-\lambda}}{n!} = \lambda \sum_{n=1}^{\infty}\frac{\lambda^{n-1} e^{-\lambda}}{(n-1)!}=\lambda \sum_{n=0}^{\infty}\frac{\lambda ^n e^{-\lambda}}{n!} = \lambda e^{\lambda} e^{-\lambda} = \lambda$

To find the variance you need to compute: $\text{Var}(X) = E[X^2] - E[X]^2$

It is helpful to consider: $E[X(X-1)] = \sum_{n=0}^{\infty}n(n-1) \frac{\lambda ^ne^{-\lambda}}{n!} = \lambda^2 \sum_{n=2}^{\infty} \frac{\lambda^{n-2}e^{-\lambda}}{(n-2)!} = \lambda^2 \sum_{n=0}^{\infty} \frac{\lambda^n e^{-\lambda}}{n!} = \lambda^2$.

This means that: $\lambda^2 = E[X(X-1)] = E[X^2] - E[X] \implies E[X^2] = \lambda^2 + E[X] = \lambda^2 + \lambda$.

Thus: $E[X^2] - E[X]^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$.

3. Or use moment generating function: $\phi(t) = \text{exp} \{\lambda(e^{t}-1) \}$.

Then $\phi '(t) = \lambda e^{t} \text{exp} \{\lambda(e^{t}-1) \}$ and $\phi ''(t) = (\lambda e^{t})^{2} \text{exp} \{\lambda(e^{t}-1) \} + \lambda e^{t} \text{exp} \{\lambda(e^{t}-1) \}$.

So $E[X] = \phi '(0) = \lambda$ and $E[X^2] = \phi ''(0) = \lambda^{2} + \lambda$.

And $\text{Var}(X) = E[X^2] - (E[X])^{2} = \lambda$.

You get the mgf as follows: $E[e^{tX}] = \sum_{n=0}^{\infty} \frac{e^{tn} e^{-\lambda} \lambda^{n}}{n!}$ which can be considered as summation.