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Math Help - integration and moment estimators

  1. #1
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    integration and moment estimators

    Well, the question and my working out is shown below. I think that I know how to do the moment estimator maybe.. but I'm just having trouble with the integration part. I have a STRONG feeling that I'm doing it wrong. Can someone please help me?

    Let X1 .. Xn be a random sample with pdf f(x) = e^-(x - theta), x> theta. Find the moment estimator and the maximum likelihood estimator of the parameter theta.

    Attached Thumbnails Attached Thumbnails integration and moment estimators-moments21.jpg   integration and moment estimators-moments22.jpg   integration and moment estimators-moments23.jpg  
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  2. #2
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    Quote Originally Posted by gconfused View Post
    Well, the question and my working out is shown below. I think that I know how to do the moment estimator maybe.. but I'm just having trouble with the integration part. I have a STRONG feeling that I'm doing it wrong. Can someone please help me?

    Let X1 .. Xn be a random sample with pdf f(x) = e^-(x - theta), x> theta. Find the moment estimator and the maximum likelihood estimator of the parameter theta.
    Where have the 0 and 1 come from? Are you saying that the support of the distribution is [0, 1]? In which case you get the parameter by putting the integral of the pdf over [0, 1] equal to 1.
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    Well, I'm not sure where the 0 and 1 came from. I'm really lost, and just put that there because all the examples in my book use 0 and 1 so i just assumed.. well i guess Im wrong then SIGH

    So um how do you know what the range is then? ><
    Last edited by mr fantastic; April 28th 2009 at 09:43 PM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by gconfused View Post
    Well, I'm not sure where the 0 and 1 came from. I'm really lost, and just put that there because all the examples in my book use 0 and 1 so i just assumed.. well i guess Im wrong then SIGH

    So um how do you know what the range is then? ><
    It looks like the support is [\theta, + \infty).
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  5. #5
    MHF Contributor matheagle's Avatar
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    yup, the support is theta to infinity.
    I noticed that immediately.
    It's a shift of an exponential with mean equal to 1.
    Y=X-theta.
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    Quote Originally Posted by matheagle View Post
    yup, the support is theta to infinity.
    I noticed that immediately.
    It's a shift of an exponential with mean equal to 1.
    Y=X-theta.

    Hi guys, also looking at this question.

    I got theta = X(bar) -1

    as my answer to the integral from theta to infinity.

    Also, how do I get started on finding the likelihood function?
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  7. #7
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    f_X(x)=\exp^{(x-\theta)}u(x-\theta) where u(.) is the unit step function.

    The likelihood function is
    L(\theta) = \prod_{i=1}^{n}f_X(x_i)

    Hint: Find the log-likelihood function. There is no need to differentiate to find the MLE(this is usually the case when the parameter is the upper/lower bound of the support)
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  8. #8
    MHF Contributor matheagle's Avatar
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    and in your integration you should use e^{a+b}=e^ae^b
    so you can pull the e^{\theta} out of the integral.
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  9. #9
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    Thanks guys, I've solved this problem
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