Poisson distribution

[Guest Chris123]

Hey,

 

I came across this question:

 

Suppose X is a Poisson random quantity with parameter lambda>0

i.e suppose P[X=k}= e^-lambda * lambda^k /k! for k=0,1,2 etc

 

a). derive a formula for E (X) in terms of lambda

b). Show that E[X(X-1)] = lambda squared

c). Derive a formula for Var(X) in terms of lambda

 

For part a i'm assuming that the answer is just 'lambda' and also for part c since variance and E(X) is equal in the Poisson dist. and lambda = mean. Is this correct? However Im sttuck on part b. I deducted the following ....

 

E(X(X-1)) = E(X^2 -X) = E(X)E(X) - E(X)

 

but this equates to lambda^2 - lambda. where am i going wrong??

 

Thanks,

 

Chris

[fafalone]

Les poissons, les poissons, how I love les poissons!

[Dave]

I spent a lot of time thinking about this until I realised it was quite easy.

 

(btw, your answers to a and c are right)

 

Now I don't know whether they want you to prove this from scratch, but my method is this:

 

Var(X) = lambda = E(X^2) - [E(X)]^2

 

therefore:

 

lambda + lambda^2 = E(X^2)

 

From part (b), E(X^2 - X) = E(X^2) - E(X)

= lambda^2 (by substitution).

 

There may be another shorter method, I'm not entirely sure.

[Dave]

Originally posted by Chris123

a). derive a formula for E (X) in terms of lambda

b). Show that E[X(X-1)] = lambda squared

c). Derive a formula for Var(X) in terms of lambda

 

okay, after thinking about this I realised that I was being completely stupid. The question itself is completely trivial when you just look at it. The key word here is 'derive' - hence you have to prove it yourself.

 

here's a pdf with the answers in (too much time on my hands ).

scistuff.pdf