How to prove that the expectation and variance of a Poisson random variable = lambda?
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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?
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?
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
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:
Re: How to prove that the expectation and variance of a Poisson random variable = lam
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.
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).
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.
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).