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Math Help - LRT, exponential

  1. #1
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    LRT, exponential

    Could you please take a look if my answer makes sense- I don't have a solution to this practice question. Also, I am not quite sure how to read the data in part (c).

    Question

    Let Y_1,...Y_n be a random sample from an exponential distribution with density function

    f_Y(y)={\theta}e^{-{\theta}y}, y>0

    where \theta>0 is an unknown parameter.

    (a) Obtain the likelihood ratio test statistic for testing H_0:\theta=\theta_0 against H_1:\theta=\ne{\theta_0}.

    (b) State the assymptotic distribution of twice the logarithm of the likelihood ratio test statistic under Ho.

    (c) If n=100,Y=31, \theta_0=3, will you reject the null hypothesis Ho at a significance level of 5%?


    Answer.

    (a)

    the MLE \hat{\theta}}for \theta is \frac{1}{\bar{y}} (I checked against a few sources).

    Therefore my likelihood ratio

    r=\frac{L_y(\hat{\theta};y)}{L_Y(\theta_0;y)}=\fra  c{\hat{\theta}e^{-\hat{\theta}\Sigma_{i=1}^nY_i}}{{\theta_0}e^{{-\theta_0}\Sigma_{i=1}^nY_i}}=\frac{{(\frac{1}{\bar  {Y}})^n}e^{-\frac{1}{\bar{Y}}n{\bar{Y}}}}{{\theta_0}^ne^{{-\theta_0}n\bar{Y}}}=(\frac{1}{\bar{Y}\theta_0})^ne  ^{n{\bar{Y}(-\frac{1}{\bar{Y}}+\theta_0)}



    (b) the assymptotic distribution

    2ln(r(Y))->\chi^2_1 (coverges in distribution)



    (c) calculate r from the given test results.

    Now I am not sure what is "Y" - if that is \Sigma_{i=1}^nY_i then why the MLE realisation ( \frac{1}{31/100}) is so far away from \theta_0=3?

    Ideally I would use (b) to test the hypothesis.

    Does the part (c) data make sense to you?
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  2. #2
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    Part (c) is a little unclear. They could mean \bar Y or \sum Y_i which will lead to different answers obviously. There's nothing wrong with the MLE being far away from the null value; that's what we want, since we are testing hypotheses.

    It might not be worth the trouble, but you can do better in part (a). You don't need to use asymptotics to solve this problem.
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  3. #3
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    Well, if you don't mind can you show me how you'd approach (a)? I am always open to improvements!

    Let me finish (c) assuming 31 is the sample mean.

    ln(r)=ln[(\frac{1}{\bar{Y}\theta_0})^ne^{n{\bar{Y}}(-\frac{1}{\bar{Y}}+\theta_0)}]=n*(-ln(\bar{Y}\theta_0))+n\bar{Y}(-\frac{1}{\bar{Y}}+\theta_0)}=100(-ln(3.1*3)-100+100*3.1*3=607 or thereabouts, ie some ridiculously high number.

    I assume the double of this is wa-ay outside the tails of chi square distribution, so reject the null.
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  4. #4
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    Quote Originally Posted by theodds View Post
    There's nothing wrong with the MLE being far away from the null value; that's what we want, since we are testing hypotheses.
    Well I guess the real world tests may have all kinds of results, but I wonder why bother run a test with x% confidence level if it is obvious to the naked eye that the null does not make sense?

    Could be a typo in the book... this question is from the same chapter I posted a while ago (page 18)
    http://www.mathhelpforum.com/math-he...tml#post621601
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  5. #5
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    There's nothing unusual about largely significant tests. It happens in practice all the time. Just because it is "obvious" that some parameters are (say) not equal to 0 that doesn't excuse the statistician from testing it typically.

    To do better on (a), it turns out that you can write the test as reject when \bar Y > c_1 or \bar Y < c_2 for appropriately chosen c_1, c_2. You need P_{H_0}(\mbox{Reject}) = \alpha and you also need c_1, c_2 to chosen so that if \bar Y = c_1 or if \bar Y = c_2 then you get the same value of the LR test statistic (the second restriction is correct up-to a stupid mistake on my part). You should think along the lines of the inverse image of the LR and think in terms of what values you \bar Y you are rejecting. You have to solve these equations numerically I think.
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