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Maximum Likelihood

  1. Apr 7, 2005 #1
    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. jcsd
  3. Apr 8, 2005 #2

    xanthym

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    Science Advisor

    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 [tex] \mathbf{\hat{\theta}} [/tex] is given by:

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

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

    [tex] 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} [/tex]

    Thus, for the Poisson Distribution, the Variance of "([itex] \overline{x} [/itex])" is estimated by dividing the value of "([itex] \overline{x} [/itex])" 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
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