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Thread: Point Estimates

  1. #1
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    Point Estimates

    Point Estimate Question

    Hi,
    May I confirm if my answers to the questions are correct?

    For Q1) I used the following limits for my integral
    $\displaystyle \int_{x_{0}}^{0} \alpha x_{0}^{\alpha}\ x^{-\alpha-1}\ dx = 1$

    Q2) My final answer is:
    $\displaystyle \alpha = \frac{\mu_{1}}{\mu_{1} - x_{0}}$

    Q3) My answer after solving the derivative of the log likelihood
    $\displaystyle \alpha = \frac{1}{\log \left( \frac{x}{x_{0}} \right)}$

    Thank you in advance
    Linda
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  2. #2
    MHF Contributor matheagle's Avatar
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    Re: Point Estimates

    the upper bound should be infinity not zero
    I get a slightly differenet answer for part 2
    and for the MLE, do we have a sample or just one observation?
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  3. #3
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    Re: Point Estimates

    Hi Matt,

    Thank you for pointing out that the limit bound should be infinity!

    May I ask what you got for part 2?
    They way I arrived at my answer was I found the expectation of the distribution first:
    $\displaystyle E[X] = \alpha x_{0}^{\alpha} \int_{x_{0}}^{\infty} x \times x^{-\alpha-1}\ dx$
    $\displaystyle =\frac{\alpha x_{0}}{\alpha -1}$

    I then substituted the $\displaystyle E[X]$ with the first moment $\displaystyle \mu_{1}$ and then rearranged the above term to obtain $\displaystyle \alpha$

    w.r.t MLE, that was how the question was given, we had no further data to work with.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Re: Point Estimates

    it's math eagle
    the mean is correct, next you solve for alpha

    $\displaystyle \frac{\alpha x_{0}}{\alpha -1}=\bar X$

    $\displaystyle \frac{\alpha -1}{\alpha}= {x_0\over \bar X} $

    $\displaystyle 1-\frac{1}{\alpha}= {x_0\over \bar X} $

    ok, now you finish
    Last edited by matheagle; Jun 21st 2011 at 07:59 AM.
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  5. #5
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    Re: Point Estimates

    Quote Originally Posted by matheagle View Post
    it's math eagle
    Oh no! I am so sorry about getting you're alias incorrect. Forgive me for my dumbness.
    I know someone with the surname Heagle, and so originally read it as Mat Heagle. My apologies!!!


    $\displaystyle 1-\frac{1}{\alpha}= {x_0\over \bar X} $

    ok, now you finish
    $\displaystyle \\ -\frac{1}{\alpha} = \frac{x_{0}}{\bar{X}} - 1 \\ \frac{1}{\alpha} = 1 - \frac{x_{0}}{\bar{X}} \\ \frac{1}{1 - \frac{x_{0}}{\bar{X}}} = \alpha \\ \alpha = \frac{\bar{X}}{\bar{X}-x_{0}}$

    I see that I should have substituted $\displaystyle \bar{X}$ for $\displaystyle \mu_{1}$. Thank you!!

    As a further I have updated my answer for the MLE to:
    $\displaystyle \alpha = \frac{n}{\sum\log \left( \frac{x_{i}}{x_{0}} \right)}$
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  6. #6
    MHF Contributor matheagle's Avatar
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    Re: Point Estimates

    its ok, I've been called much worse, its just not matt
    again with the MLE, usually you have a sample of i.i.d. rvs.
    But sometimes it's just one observation.
    It isn't clear, but most likely its the sample of n observations.
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