Finding the MLE of a Poisson Distribution

[laura_a]

Homework Statement

Suppose that X has a poisson distribution with parameter $$\lambda$$. Given a random sample of n observations, find the MLE of $$\lambda$$, $$\hat{\lambda}$$.

Homework Equations

The MLE can be found by
$$\Sigma^{n}_{i=1} \frac{e^{- \lambda} \lambda^{x_{i}}}{x_{i}!}$$
= $$e^{- \lambda} \times ( \frac{\lambda^{x_1}}{x_{1}!} + \frac{\lambda^{x_2}}{x_{2}!} + . . . )$$
=e^{- \lambda} \Sigma \lambda^{x_i} \div x_{i}! (couldn't get the latex to work from this line)
=e^{- \lambda} \Sigma \lambda^{x_i} \times \frac{1}{x_{i}!}

Now this is the bit that I think I've stuffed up - if not already

= - \lambda log(y) \times x_i \times log(\frac{1}{x_{i}}

I know what the answer is meant to be, I was given it in the text

\hat{\lambda} = \bar{x}

Can anyone tell me where I went wrong, or if I'm even close to being on the right track. I'm a correspondance student so pretty much learning it from reading the notes and the textbook which has no examples!

Thanks heaps

The Attempt at a Solution

 

[mighty2000]

Your approach seems to be on the right track, but there are a few errors in your equations. First, the MLE for the Poisson distribution is actually just the sample mean, as given by \hat{\lambda} = \bar{x}. This is a well-known property of the Poisson distribution and can be derived using calculus.

Secondly, your equations for the MLE are a bit confusing and not entirely correct. The correct equation for the MLE is:

L(\lambda) = \prod_{i=1}^{n} \frac{e^{-\lambda} \lambda^{x_i}}{x_i!}

Taking the natural logarithm of both sides and using the log property log(ab) = log(a) + log(b), we get:

ln(L(\lambda)) = \sum_{i=1}^{n} \left[ -\lambda + x_i ln(\lambda) - ln(x_i!) \right]

Next, we take the derivative of ln(L(\lambda)) with respect to \lambda and set it equal to 0 to find the maximum:

\frac{d}{d\lambda} ln(L(\lambda)) = -n + \sum_{i=1}^{n} \frac{x_i}{\lambda} = 0

Solving for \lambda, we get \hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} x_i = \bar{x}, which is the sample mean.

So in summary, the MLE for the Poisson distribution with parameter \lambda is simply the sample mean \bar{x}. I hope this helps clarify the process for you. Good luck with your studies!