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Math Help - Find the MLE

  1. #1
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    Find the MLE

    Hi, I'm trying to solve the following problem:

    Let Y1, Y2, ..., Yn denote a random sample from the probability density function
    <br />
f(y|\theta)={(\theta + 1)}{y}^{\theta} , for 0 < y < 1, \theta > -1,
    and 0 elsewhere.

    Find the MLE for \theta. Compare the answer to the method-of-moments answer: [2(Y bar) -1]/[1-(Y bar)

    Sorry for that last part. Y bar means Y with the line on top.

    I tried to do this by taking the lns of each term after multiplying the fy's, taking the derivative with respect to theta, and equating to zero but I end up with - (ln y + 1). Is this possible? I figure it's not since it asks to compare to the M.O.M. answer which seems totally different...

    Any ideas?
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  2. #2
    MHF Contributor matheagle's Avatar
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    I put that on my exam last week. you need the likelihood function, which is

    L=\prod_{i=1}^n = (\theta +1)^n \left(\prod_{i=1}^n y_i \right)^{\theta}

    you can differentiate this or take the log and differentiate that.

    \ln L=n\ln (\theta +1) +\theta \sum_{i=1}^n \ln y_i

    NOW differentiate and set equal to zero.
    Last edited by mr fantastic; April 14th 2009 at 01:17 AM. Reason: Made the brackets bigger around the product symbol
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  3. #3
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    Quote Originally Posted by Canadian0469 View Post
    I tried to do this by taking the lns of each term after multiplying the fy's, taking the derivative with respect to theta, and equating to zero

    That's what I was describing... It seems to have given me a fishy answer.

    Thanks for the quick response.
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  4. #4
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    What I did next was
    ({\theta + 1})^ny^{\theta n}
    (nln]({\theta + 1}) + {\theta n}ln y
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  5. #5
    MHF Contributor matheagle's Avatar
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    NO, you have n different y's, reread what I wrote.
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  6. #6
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    Ohhhh right... Forgot about that log rule... I know once I take the derivative I get:
    <br />
=n/{ (\theta +1)} + \sum_{i=1}^n \ln y_i<br />

    and that I'm supposed to get (Y bar) from the summation on the right side but how can I get rid of that ln? Because if I take the exp of both sides I'll end up with theta as a exponent... If I put n ln y_i and then {y_i}^n I won't get (Y bar) like I want... and it just occured to me that {y_i}^n doesn't even make sense..
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  7. #7
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    Anybody know how I can solve for theta after this step? I can't think of how to get rid of the ln y_i
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  8. #8
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    Quote Originally Posted by Canadian0469 View Post
    Anybody know how I can solve for theta after this step? I can't think of how to get rid of the ln y_i
    You don't 'get rid of' it.

    Do you know how to solve 0 = \frac{n}{\theta + 1} + B for \theta ?

    You can't get it in terms of \overline{y}.
    Last edited by mr fantastic; April 14th 2009 at 01:24 AM.
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  9. #9
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    Sure I'll end up with:
    \theta = -n/{\sum_{i=1}^n \ln y_i} - 1
    \theta = n/{\sum_{i=1}^n 1/y_i} - 1

    but I don't see any relation to the M.O.M. result. I was expecting it to be the same or similar...
    Last edited by Canadian0469; April 14th 2009 at 01:59 AM.
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  10. #10
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    Quote Originally Posted by Canadian0469 View Post
    Sure I'll end up with:
    \theta = -n/{\sum_{i=1}^n \ln y_i} - 1
    \theta = n/{\sum_{i=1}^n 1/y_i} - 1 Mr F says: Where has your log gone?
    \theta = n{\sum_{i=1}^n y_i} - 1 Mr F says: This is not a valid algebraic manipulation.
    \theta = {Y bar} - 1

    but I don't see any relation to the M.O.M. result. I was expecting it to be the same or similar...
    ..
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  11. #11
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    Yeah sorry about the algebra mistake...
    I think this is legit:
    n/{(\theta+1)} = -{\sum_{i=1}^n \ln y_i}
    {\theta+1} = -n/{\sum_{i=1}^n \ln y_i}
    {\theta} = -n/{\sum_{i=1}^n \ln y_i} -1

    But I still don't see any connection to the M.O.M. result...
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  12. #12
    MHF Contributor matheagle's Avatar
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    The MLE and the MOM are NOT the same.
    Here we have newly transformed y's, W=log y
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