Point estimation with MLE and MoM

Hi,

My question to answer is: Estimate Lambda in a Poisson distribution if X1, X2, etc. are iid, with the:
a) Method of Moments.
b) Maximum Likelihood Estimator.
Are the estimators unbiased? Are they the best estimators for Lambda?

I'm really having trouble understanding how these two methods work. I just don't see what's going on. I've read so much about them though that I think I have the answers to the questions at least. For Method of moments & MLE, estimator is 1/n * (Sum from i=1 to n of: x[sub:bzzisawh]i[/sub:bzzisawh]) Both unbiased estimators, and they are efficient estimators.

But I really want to understand what's going on here, so I was wondering if someone could explain how each of these is derived? From what i've gathered both can be found from the pmf of poisson: P(X=x)=[Lambda^x * e^(-Lambda)]/x!
I'm not sure how to derive them though.

MOM solves for \(\displaystyle \lambda\) in terms of the mean of the distribution. MLE solves for the value of \(\displaystyle \lambda\) which maximizes the log-likelihood of the (log of) the pmf of a sample. It just so happens they are equal in the Poisson case. This is not always what happens

Not so sure about the method of moments, but I can show you how MLE works.

The idea is : "Given this data, what set of parameters assign this data the highest probability?"

So, if you have X1, X2, X3... Xn drawn independently from a normal distribution N(mu, sig^2), with KNOWN sig and UNKNOWN mu, the probability of getting (within dX1, dX2, dX3.... of) this data is

constant * exp(-(X1-mu)^2/(2*sig^2)) * exp(-(X2-mu)^2/(2*sig^2)) * ... * exp(-(Xn-mu)^2/(2*sig^2)) * dX1 * dX2 * ... * dXn

Now we need to find mu to maximise this, ie, to maximise

-(X1-mu)^2/(2*sig^2)-(X2-mu)^2/(2*sig^2)-...-(Xn-mu)^2/(2*sig^2), that is, to minimise the sum of (Xi-mu)^2. Differentiate, solve, and mu=sum(Xi)/n

If we didn't know sig either, the 'constant' wouldn't be constant any more, and we'd have to minimise a function of two variables mu and sig.