Maximum likelihood of Poisson distribution

[dim&dimmer]

Homework Statement

Suppose X has a Poisson distribution with parameter lambda. Given a random sample of n observations,
Find the MLE of lambda, and hat lambda.
Find the expected value and variance of hat lambda.
Show that hat lambda is a consistent estimator of lambda.

Homework Equations

PX(x) = e^-lambda.lambda^x/x!
E(X) = lambda ? or mean
var (X) = lambda ?? or mean

The Attempt at a Solution

I am really struggling with stats, hope someone can help.
I try to find likelihood function
L(x1,x2,...,xn|lambda) = e^-lambda.lambda^x/x!
= capital pi from i=1 to n, e^-lambda.lambda^xi/xi!
then I'm stuck.
Differentiating with respect to lambda,
-e^-lambda.xlambda^(x-1)/x!
is that the answer?
sorry about the notation also, is there somewhere on this site to get the symbols?

Any help greatly appreciated.

 

[dim&dimmer]

OK, I think I've got the first part but if my work could be checked it would be grate.(at least I'll get a reply for that if not for anything else).
for the sake of notation I'll use $ for lambda

L($|x1, x2, ... , xn) =e^-$.$^xi/xi!...e^-$.$^xn/xn!
=e^-n$.$^(sigma xi)/(xi!...xn!)

log L = -n$ + (ln$)sigma xi -ln(product xi!)

(d log L($))/d$ = -n + sigma xi/$ =0 gives
hat$ = sigma xi/n , the MLE

For part b, poisson distributions have lambda = mean = variance, so the mean and variance equal the result above.
Part c , the sample mean is a consistent estimator for lambda when the Xi are distributed Poisson, and the sample mean is = to the MLE, therefore the MLE is a consistent estimator.

Corrections are most welcome.

 

[statdad]

You are correct in saying

<br /> \hat \lambda = \frac 1 n \sum_{i=1}^n x_i<br />

is the MLE of \lambda. to obtain the mean and variance of \hat \lambda, think of the general properties that (since the x_i are independent)

<br /> E \hat \lambda = \frac 1 n \sum_{i=1}^n Ex_i<br />

and

<br /> \text{Var} \left( \hat \lambda \right) = \frac 1 {n^2} \sum_{i=1}^n \text{Var}(x_i)<br />

simplifying the variance of \hat \lambda should also help you show that it is a consistent estimator of \lambda

 

[dim&dimmer]

Thank you for your reply, but I'm still stuck. By using general properties do you mean that
E(xi) = sum i=1 n, (xi e^-lambda lambda^xi)/xi! that is E(x) = sum x pX(x) for discrete rv.
If so, I just seem to go round in circles

 

[statdad]

No, rather these two notions.
First,

<br /> \text{E} \left(\sum_{i=1}^n \frac{X_i} n \right) = \sum_{i=1}^n \frac{\text{E}X_i}{n}<br />

and

<br /> \tex{Var} \left(\sum_{i=1}^n \frac{X_i} n\right) = \sum_{i=1}^n \frac{\text{Var}(X_i)}{n^2}<br />

You already know the mean and variance of a Poisson random variable; calculating the mean and variance of the MLE for \lambda will allow you to conclude that the variance goes to zero, so ...

 

[dim&dimmer]

Please excuse my ignorance, but is EX(i) = lambda or lambda(i).
If the latter then sample mean is sum 1 to n lambda(i)/n, and variance has n^2 as denominator.
Variance approaches 0 as n approaches infinity, and 0 variance means that the estimator is consistent.
Have I got it?

 

[statdad]

Since all of the X_i values come from the same distribution, they all have the same mean (\lambda) and they all have the same variance. You should be able to show these points (\hat \lambda is the MLE)

1. \text{E}\left(\hat \lambda\right) = \lambda

2. \text{Var}\left(\hat \lambda\right) converges to zero as n \to \infty

With these things done you've shown (by an appeal to the result you seem to mention in your most recent post) that the MLE is consistent.

 

[dim&dimmer]

You are patient!
So summing EXi gives n lambda as each EX = lambda, then dividing by n gives Ehatlambda = lambda.
similarly Var(hat lambda) = lambda/n which converges to 0.
Please tell me I'm done.

 

[statdad]

Okay: You are done. :-)
Write it up neatly.

 

[dim&dimmer]

Thank you very much for your help

 

[Owumi]

Thanks for your help and patience with this, statdad! I googled a similar question and found this thread extremely helpful.