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Math Help - Maximum Likelihood.

  1. #1
    Junior Member
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    Nov 2008
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    Maximum Likelihood.

    Hello, I have a couple of exercise questions which I attempted and failed to complete, so the help I could get from you guys would be greatly appreciated!

    1. It is known that the probability p of a head on a biased coin is either 1/4 or 3/4. The coin is tossed twice and a value for the number of heads Y is observed.

    (a) For each possible value of Y, which of the two possible values for p maximizes the probability that Y = y.

    (b) Depending on the value y actually observed, what is the maximum likelihood estimate of p (ie. 1/4 or 3/4)?

    ---------------------

    2. Suppose that \hat{\theta} is the maximum likelihood estimator of \theta. Let t(\theta) be a function of \theta that possesses a unique inverse. (That is, if \beta = t(\theta) then \theta = t^{-1}(\beta)) . Show that t(\hat{\theta}) is the maximum likelihood estimator of t(\theta)

    Thank you!
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  2. #2
    Newbie
    Joined
    Aug 2009
    Posts
    9

    Just maximize likelihood over two possible values of p

    As far as there are two trials and y is observed number of heads likelihood function looks as follows:
    L(p)=C^y_2*p^y*(1-p)^(2-y)
    To obtain estimate based on likelihood maximize L() over two possible values for p=1/4 or 3/4. For
    y=0 you have L(p)=(1-p)^2 which is maximized by p=1/4. For y=1 we have L(p)=2*p*(1-p) is to be maximized
    by both p=1/4 and p=3/4. Analogously for y=2 the p=3/4 would be maximum likelihood estimate.

    Second is invariance which is one of basic properties of maximum likelihood estimates. If L(teta) is likelihood function
    then L'=L( t^(-1) (beta) ) is likelihood function for beta. As far as teta' is maximum of L the beta'=t(teta') is maximum of L'.

    As a consequence of invariance you have that ML estimate is generally biased.
    Last edited by kobylkinks; August 15th 2009 at 09:15 AM.
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