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Math Help - Maximum likelihood estimator

  1. #1
    Senior Member chella182's Avatar
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    Maximum likelihood estimator

    Similar to another thread I've posted, only that one's asking how to sketch the functions, this time I'm stumped on the calculations.

    So the full Question (apart from part (i) because I've done that, and it doesn't relate) is...

    A manufacturer of watches claims that the weekly error, in seconds, of their watches follow a N(0,1) distribution. An inspection of the performance of 5 randomly chosen watches gave the following weekly errors -0.68, 0.25, 0.29, -1.41 and 1.59.

    (ii) Assume that the weekly errors follow a N(0,\sigma^{2}) distribution. Sketch the likelihood function and log-likelihood for \sigma and calculate the most likely value of \sigma.

    All I've got so far is
    f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\{{-\frac{1}{2}(\frac{x}{\sigma})^{2}}\}

    But then anything could be wrong after that I got this for the likelihood function:
    \sigma^{-n}(2\pi)^{-n/2}\exp\{-\frac{1}{2}\frac{\sum_{i=1}^{n}x_{i}^{2}}{\sigma^{  2}}\}

    Which gave me the log-likelihood function of:
    K-n\ln{\sigma}-\frac{\sum_{i=1}^{n}x_{i}^{2}}{2\sigma^{2}}
    And now I'm stumped. I don't think what I've done is right at all. Can anyone help?
    Last edited by chella182; April 29th 2009 at 05:13 PM. Reason: I want to subscribe to the thread so I know when someone's replied
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  2. #2
    MHF Contributor matheagle's Avatar
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    Rewrite as sigma squared and diff wrt sigma^2
    Quote Originally Posted by chella182
    (ii) Assume that the weekly errors follow a N(0,\sigma^{2}) distribution. Sketch the likelihood function and log-likelihood for \sigma and calculate the most likely value of \sigma.

    All I've got so far is
    f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\{{-\frac{1}{2}(\frac{x}{\sigma})^{2}}\}

    But then anything could be wrong after that I got this for the likelihood function:
    \sigma^{-n}(2\pi)^{-n/2}\exp\{-\frac{1}{2}\frac{\sum_{i=1}^{n}x_{i}^{2}}{\sigma^{  2}}\}

    Which gave me the log-likelihood function of:
    K-{n\over 2}\ln{\sigma^2}-\frac{\sum_{i=1}^{n}x_{i}^{2}}{2\sigma^{2}}


    And now I'm stumped. I don't think what I've done is right at all. Can anyone help?
    You should get {\sum_{i=1}^{n}x_{i}^{2}\over n} as the MLE of sigma^2.
    Hence the MLE of sigma is the square root of this.
    Last edited by mr fantastic; April 29th 2009 at 06:43 PM. Reason: Added open quote tag.
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  3. #3
    Senior Member chella182's Avatar
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    So is what I have right then?
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  4. #4
    MHF Contributor matheagle's Avatar
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    and what do you have now?
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  5. #5
    Senior Member chella182's Avatar
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    I have what's in my original post, which I don't think is right. I don't like that \sum^{n}_{i=1}x_{i}^{2} that I have in my log-likelihood function much... I'm gonna have another good look at it later.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I edited your post.
    The MLE is what you have divided by n.
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  7. #7
    Senior Member chella182's Avatar
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    I'm totally lost, sorry. I don't know what you mean.
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  8. #8
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    please see attached.
    EK
    Attached Files Attached Files
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  9. #9
    Senior Member chella182's Avatar
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    Thankyou. That's what I got. I think the \sum_{i=1}^{n}x_{i}^{2} just threw me a bit.
    Last edited by mr fantastic; May 1st 2009 at 01:32 AM.
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