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Math Help - Proving Maximum Likelihood Estimator (MLE) of Theta is the sample mean

  1. #1
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    Question Proving Maximum Likelihood Estimator (MLE) of Theta is the sample mean

    p(x) = [ ((Theta)^x) * (e^(-Theta)) ] / x! X = 0,1,2,...

    You may assume that E(X)= Theta and V(X) = Theta

    A random sample were examined....

    Show that the maximum likelihood estimator (MLE) of (Theta) is the sample mean.



    Any help, or a point in the right direction would be great. Any worked solutions are also welcome!

    Thankyou!
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  2. #2
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    Re: Proving Maximum Likelihood Estimator (MLE) of Theta is the sample mean

    The likelihood of a sample of independent x values is: (I assume there are n values in total)

    L(\theta) = \prod_{i=1}^{n} \left[ \frac{\left( \theta^{x_i}  \right) e^{-\theta}}{{x_i}!} \right]

    use log likelihood...

    Log L(\theta) =  \ln \prod_{i}  \left(\frac{\left( \theta^x  \right) e^{-\theta}}{x!} \right)

    simplify using log rules
    Log L(\theta) =  \sum_i \ln   \left[\frac{\left( \theta^x  \right) e^{-\theta}}{x!} \right]

    Log L(\theta) = \sum_i \left[\ln(\theta^x) + \ln(e^{-\theta}) +  \ln(x!) \right]
    Log L(\theta) = \sum_i \left[x\ln(\theta) -\theta +  \ln(x!) \right]

    Simplify the sums
    Log L(\theta) = \sum_i x\ln(\theta) -\sum_i \theta +  \sum_i \ln(x!)
    Log L(\theta) = \sum_i x\ln(\theta) -\sum_i \theta +  \sum_i \ln(x!)
    Log L(\theta) = \ln(\theta)\sum_i x -n \theta +  \sum_i \ln(x!)


    try and finish yourself from there. if stuck, the solution is in the spoiler.
    Spoiler:

    Differentiate with respect to theta

    \frac{d(Log L(\theta))}{d \theta} = \frac{\sum_i x}{\theta}  -n  +  0

    Set the derivative equal to 0 and solve to geth the MLE ( \hat{\theta}})

    0 = \frac{\sum_i x}{\hat{\theta}} - n
    \hat{\theta} = \frac{\sum_i x}{n}
    \hat{\theta} =  \bar{x}
    Your professor may expect you to prove that the turning point is a maximum, which you can do using whatever methods you prefer.
    Last edited by SpringFan25; March 27th 2013 at 04:58 PM.
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