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Math Help - Maximum Log-Likelihoods

  1. #1
    Super Member Deadstar's Avatar
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    Maximum Log-Likelihoods

    Suppose r successes were obtained in a sequence of m Bernoulli trials with success probability {\Theta}_1, and that in an independent second sequence of n Bernoulli trials with success probability {\Theta}_2, there were s successes.

    Show that the maximum of the log-likelihood of all obervations is

    In{\binom{m}{r}} + In{\binom{n}{s}} + <br />
rIn(r) + sIn(s) + (m-r)In(m-r) +  (n-s)In(n-s) - mIn(m) - nIn(n).

    Is there a formula for this and its then just a case of plugging in the variables?
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  2. #2
    Super Member Deadstar's Avatar
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    Ive sort of advanced a little on this...

    Since the two trials are independent, the total max log-likelihood is the sum of the two individual trials i.e.

    Trial 1; max log-likelihood is In \binom{m}{r} + rIn(r) + (m-r)In(m-r) - mIn(m)

    Trial 2; max log-likelihood is In \binom{n}{s} + sIn(s) + (n-s)In(n-s) - nIn(n)

    Adding these together gives total max log-likelihood.

    Is the max log-likelihood for each individual trial found by doing something like finding the derivative of the log-likelihood and setting it to 0? I'm mostly unsure as to where the binomial expression comes from...
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  3. #3
    Super Member Deadstar's Avatar
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    Gotten further now...

    In(L) = In(L({\Theta}_1 | m,r))

    = In \binom{m}{r} + rIn({\Theta}_1) + (1 - {\Theta}_1)In(m-r)...
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