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Math Help - Maximum Likehood Estimator

  1. #1
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    Maximum Likehood Estimator

    If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

    a) Find the maximum likelihood estimator for theta if information has been collected on a random sample of 25 people.

    b) Find the method of moments estimator for theta if information has been collected on a random sample of 25 people.

    Assume k is unknown.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

    a) Find the maximum likelihood estimator for theta if information has been collected on a random sample of 25 people.

    [snip]

    Assume k is unknown.

    Any help would be appreciated.
    For an example of what to do, read this: http://www.mathhelpforum.com/math-he...estimator.html
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  3. #3
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    Quote Originally Posted by eigenvector11 View Post
    If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

    [snip]

    b) Find the method of moments estimator for theta if information has been collected on a random sample of 25 people.

    Assume k is unknown.

    Any help would be appreciated.
    The only thought I have at the moment is that E(Y) = \frac{k \theta}{\theta - 1} and so let \frac{k \theta}{\theta - 1} = \overline{y} and solve for \theta.
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  4. #4
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    For question a), I attached what I am getting so far. Am I on the right track here? I'm not sure where to go now.

    Edit merge: Actually I already spotted an error, it should be 25lnk in the derivative..oops.
    Attached Files Attached Files
    Last edited by mr fantastic; February 7th 2009 at 11:57 AM. Reason: Merged posts
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  5. #5
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    How did you come up with that expected value in question b)? And if you solve for theta in that equation, won't you get theta=(ybar-ybar)/k, which means that theta equals zero?
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  6. #6
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    Quote Originally Posted by eigenvector11 View Post
    How did you come up with that expected value in question b)? And if you solve for theta in that equation, won't you get theta=(ybar-ybar)/k, which means that theta equals zero?
    \overline{y} is the sample mean. E(Y) is NOT the same as \overline{y}. I had thought this would be clear. So you get \theta in terms of the sample eman.

    I found E(Y) using the usual formula: E(Y) = \int_k^{+\infty} y \, f_Y (y, \theta) \, dy. Again, this is something that I thought would be clear.
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