How to prove that the expectation and variance of a Poisson random variable = lambda?

mh03

New Member
#1
This probably has an easy solution, but I was wondering if anybody could help me solve and explain these two questions:

(a) Show that the expectation of the random variable is E(X) = lambda

(b) Show that the variance of the random variable is Var(X) = lambda

Any help would be very much appreciated.
 

Dason

Ambassador to the humans
#2
Re: How to prove that the expectation and variance of a Poisson random variable = lam

Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.

Specifically what have you tried so far?
 

mh03

New Member
#3
Re: How to prove that the expectation and variance of a Poisson random variable = lam

Thank you for posting and letting me know about the protocol.
For question (a) I had this:

E(X)=

n
∑ x * p(x)
x=0

n
∑ x * λ^x * e^(-λ)/ x!
x=0

n
∑ x * λ^x * e^(-λ)/ (x-1)! * x
x=1

n
∑ λ^x * e^(-λ)/ (x-1)!
x=1

it was here that i had trouble and i got shown the next step, without explaination. I was wondering how the λ was factored across leaving λ^(x-1)?

n
λ ∑ λ^(x-1) * e^(-λ)/ (x-1)! then let y= x-1
x=1

n
λ ∑ λ^y * e^(-λ)/ y!
y=0


= λ

For Question (b) i was having more trouble. I started with:

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

where we know E(X^2 - X) = E(X^2) - E(X)

Therefore: Var(X)= E[X(X-1)] + E(X) - E(X)^2 so we first need to solve E[X(X-1)]


E[X(X-1)]=

n
∑ x * (x-1) * λ^x * e^(-λ)/ x! this is where i got stuck
x=0


Any help with explanations and help as to where to go with these, or if i have made a mistake will be very much appreciated
 
Last edited:

Dason

Ambassador to the humans
#4
Re: How to prove that the expectation and variance of a Poisson random variable = lam

n
∑ λ^x * e^(-λ)/ (x-1)!
x=1

it was here that i had trouble and i got shown the next step, without explaination. I was wondering how the λ was factored across leaving λ^(x-1)?

n
λ ∑ λ^(x-1) * e^(-λ)/ (x-1)! then let y= x-1
x=1
Note that a^b = a*(a^(b-1)) which is why you can factor the single \(\lambda\) out. If you don't see that just plug in some actual numbers for a and b and you should see that it's just simple exponentiation rules being applied.

E[X(X-1)]=

n
∑ x * (x-1) * λ^x * e^(-λ)/ x! this is where i got stuck
x=0


Any help with explanations and help as to where to go with these, or if i have made a mistake will be very much appreciated
Try doing something analogous to what you did to find E[X]. Cancel out the x(x-1) with some of the factors in the x! in the denominator and then pull some stuff out so that what you're left in the summation is essentially a Poisson distribution again (so that it sums to 1).