# Maximum Likelihood

1. Apr 7, 2005

### Thinkmarble

How do I estimate the standart deviation for the mean average of an poisson-distribution ?
The mean average was estimated with the maximum-likelihood method by graphing the likelihood in dependence of the mean average, then just reading off the value for which the likelihood became maximal.
Up to this point I had not problem.
But I also have to determine the standard deviation for my estimation of the mean average.
And that is where I run into problems.
I'm told that "by an deviation from (...) the mean average of the standart deviation the -2*ln(L) function increases by one unit compared with the minimum".
If I understand that correctly:
Minimum of the log-likelihood is 100 at an mean average of a.
At an mean average of b my log-likelihood has the value 101(99).
So my standart deviation is a-b.

Is this interpretation correct ?

2. Apr 8, 2005

### xanthym

Not exactly clear what you're trying to say. For traditional approaches to Maximum Likelihood Estimators, the Likelihood Function "L(θ)" is obtained, and the Variance of the Maximum Likelihood Estimator $$\mathbf{\hat{\theta}}$$ is given by:

$$1: \ \ \ \ \ Var(\mathbf{\hat{\theta}}) \ = \ \left ( - \, \frac{d^{2}L(\theta) } {d \theta^{2}} \right )^{\mathbf{-1}}_{\mathbf{ \theta = \hat{\theta}} }$$

Thus, for the Poisson Distribution with Log Likelihood Function "logL(λ)" (and using the fact that the Maximum Likelihood Estimator of "$\lambda$" is {$$\hat{\lambda} = \overline{x}$$}:

$$2: \ \ \ \ \ \ Var(\hat{\lambda}) \ = \ Var(\overline{x}}) \ = \ \frac {(\hat{\lambda})^{2}} {\sum_{i=1}^{n} \, x_{i} } \ = \ \frac {(\bar{x})^{2}} {n \bar{x} } \ = \ \frac { \overline x} {n}$$

Thus, for the Poisson Distribution, the Variance of "($\overline{x}$)" is estimated by dividing the value of "($\overline{x}$)" by the sample size "n". Remember that the Standard Deviation will be the Sqrt of the Variance.

(Note: FYI, recall for the Poisson Distribution, we have {λ = μ = σ2}).

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Last edited: Apr 8, 2005