Estimate Standard Deviation of Mean Average w/ Maximum Likelihood

[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 ?

 

[xanthym]

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 ?

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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[thepatientmental]

Yes, your interpretation is correct. The standard deviation for the mean average of a Poisson distribution can be estimated using the maximum likelihood method. This is done by graphing the likelihood function in terms of the mean average and finding the value that maximizes the likelihood. This value will be the estimated mean average. To determine the standard deviation, you can use the fact that a deviation of one unit from the mean average will result in a change of -2*ln(L) in the likelihood function. This means that the difference between the estimated mean average and the true mean average is equal to the standard deviation. In your example, if the minimum of the log-likelihood is at a mean average of a=100 and at a mean average of b=101, the standard deviation would be 1. This approach is a valid way to estimate the standard deviation for the mean average of a Poisson distribution.