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Finding the MLE of a Poisson Distribution

  1. Jul 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that X has a poisson distribution with parameter [tex] \lambda [/tex]. Given a random sample of n observations, find the MLE of [tex] \lambda [/tex], [tex] \hat{\lambda} [/tex].


    2. Relevant equations
    The MLE can be found by
    [tex] \Sigma^{n}_{i=1} \frac{e^{- \lambda} \lambda^{x_{i}}}{x_{i}!}[/tex]
    = [tex] e^{- \lambda} \times ( \frac{\lambda^{x_1}}{x_{1}!} + \frac{\lambda^{x_2}}{x_{2}!} + . . . )[/tex]
    =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 text book which has no examples!

    Thanks heaps





    3. The attempt at a solution
     
    Last edited: Jul 28, 2008
  2. jcsd
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