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Math Help - Exponential distribution

  1. #1
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    Exponential distribution

    Suppose that  \theta \sim \text{Exp}(\lambda) . An estimator for  \theta is  \bar{X} . Suppose that we know that  \frac{2n \bar{X}}{\theta} \sim \chi^{2}_{2n} . Construct a  100(1-\alpha) \% confidence interval for  \theta .

    Would it be  \frac{2n \bar{X}}{\chi^{2}_{\alpha/2, 2n}} < \theta < \frac{2n \bar{X}}{\chi^{2}_{1-\alpha/2, 2n}} ?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Suppose that  \theta \sim \text{Exp}(\lambda) . An estimator for  \theta is  \bar{X} . Suppose that we know that  \frac{2n \bar{X}}{\theta} \sim \chi^{2}_{2n} . Construct a  100(1-\alpha) \% confidence interval for  \theta .

    Would it be  \frac{2n \bar{X}}{\chi^{2}_{\alpha/2, 2n}} < \theta < \frac{2n \bar{X}}{\chi^{2}_{1-\alpha/2, 2n}} ?
    Now I think that those inequalities are the wrong way around.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Now I think that those inequalities are the wrong way around.

    RonL
    Actually aren't they correct?

    Wouldn't it be  P \left(\chi^{2}_{1- \alpha/2, 2n} < \frac{2n \bar{X}}{\theta} < \chi^{2}_{\alpha/2, 2n} \right) = 1- \alpha ?

    And this is equivalent to what I have written? Or maybe I made a mistake.
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  4. #4
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    Quote Originally Posted by heathrowjohnny View Post
    Actually aren't they correct?

    Wouldn't it be  P \left(\chi^{2}_{1- \alpha/2, 2n} < \frac{2n \bar{X}}{\theta} < \chi^{2}_{\alpha/2, 2n} \right) = 1- \alpha ?

    And this is equivalent to what I have written? Or maybe I made a mistake.
    If you want a 95\% interval \alpha=5\%, so


     \chi^{2}_{1- \alpha/2, 2n}=\chi^2_{97.5\%,2n} > \chi^{2}_{2.5\%, 2n} =\chi^{2}_{\alpha/2, 2n} ?

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    If you want a 95\% interval \alpha=5\%, so


     \chi^{2}_{1- \alpha/2, 2n}=\chi^2_{97.5\%,2n} > \chi^{2}_{2.5\%, 2n} =\chi^{2}_{\alpha/2, 2n} ?

    RonL
    But now that we are dividing those terms, the sign changes? Isn't it similar to the following:  P \left(\chi^{2}_{1- \alpha/2, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{\alpha/2, n-1} \right) = 1- \alpha , and so  \frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2, n-1}} < \sigma^{2} < \frac{(n-1)S^{2}}{\chi^{2}_{1- \alpha/2, n-1}} ?
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  6. #6
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    Quote Originally Posted by heathrowjohnny View Post
    But now that we are dividing those terms, the sign changes? Isn't it similar to the following:  P \left(\chi^{2}_{1- \alpha/2, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{\alpha/2, n-1} \right) = 1- \alpha , and so  \frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2, n-1}} < \sigma^{2} < \frac{(n-1)S^{2}}{\chi^{2}_{1- \alpha/2, n-1}} ?
     P \left(\chi^{2}_{97.5\%, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{2.5\%, n-1} \right) = 0

     P \left(\chi^{2}_{2.5\%, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{97.5\%, n-1} \right) \ne 0

    because:

     \chi^{2}_{2.5\%, n-1} < \chi^{2}_{97.5\%, n-1}

    RonL
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  7. #7
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    So the book probably made a blatant error. They said the first one was  1- \alpha .
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