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Math Help - Confidence interval

  1. #1
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    Confidence interval

    Suppose that a random variable Y has a gamma distribution w. parameters \alpha = 2 and an unknown \beta. We have earlier proved that 2Y/\beta has a chi-square distribution w. 4 df. Using 2Y/\beta as a pivotal quantity. derive a 90% confidence interval for \beta.

    I don't know if I'm on the right way. I create a new random variable U = 2Y/\beta and look for P(a\leq U\leq b) = 0.90

    =

     P(a\leq 2Y/\beta\leq b) = 0.90

    I then have a similar example in the book that tells me to set it up like this:

    P(U<a) = \int_{0}^{a}{} f(u) du = 0.05

    P(U>b) = \int_b^\infty f(u) du = 0.05

    Must I derive the pdf of U manually or can I use that 2Y/\beta is chi-square distributed and put in the pdf of a chi-square distribution?

    That would be:
    (y)^{(\upsilon /2)-1}e^{-y/2}/2^{\upsilon/2}\Gamma (\upsilon /2) but with y=u and \upsilon = 4
    I've tried to do that but I don't get the right answer, which is 2Y/9.49, 2Y/.711.

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by approx View Post
    Suppose that a random variable Y has a gamma distribution w. parameters \alpha = 2 and an unknown \beta. We have earlier proved that 2Y/\beta has a chi-square distribution w. 4 df. Using 2Y/\beta as a pivotal quantity. derive a 90% confidence interval for \beta.

    I don't know if I'm on the right way. I create a new random variable U = 2Y/\beta and look for P(a\leq U\leq b) = 0.90

    =

     P(a\leq 2Y/\beta\leq b) = 0.90

    I then have a similar example in the book that tells me to set it up like this:

    P(U<a) = \int_{0}^{a}{} f(u) du = 0.05

    P(U>b) = \int_b^\infty f(u) du = 0.05

    Must I derive the pdf of U manually or can I use that 2Y/\beta is chi-square distributed and put in the pdf of a chi-square distribution?

    That would be:
    (y)^{(\upsilon /2)-1}e^{-y/2}/2^{\upsilon/2}\Gamma (\upsilon /2) but with y=u and \upsilon = 4
    I've tried to do that but I don't get the right answer, which is 2Y/9.49, 2Y/.711.
    Thanks for your help.
    The pdf of U is f(u) = \frac{u e^{-u/2}}{4}.

    Your trouble is solving:


    \Pr(U < a) = \int_0^a \frac{u e^{-u/2}}{4} \, du for a. I get a = 0.710723.


    \Pr(U >  b) = \int_b^{+\infty} \frac{u e^{-u/2}}{4} \, du for b. I get b = 9.48772.


    So your trouble is calculus, not statistics. Now that you know that what you're trying to do is correct, I suggest you go back and re-check your calculations. Then:

    0.9 = \Pr \left( 0.710723 \leq U \leq 9.48792 \right) = \Pr \left( 0.710723 \leq \frac{2Y}{\beta} \leq 9.48792 \right)

     = \Pr \left( \frac{0.710723}{2Y} \leq \frac{1}{\beta} \leq \frac{9.48792}{2Y} \right)

    where the direction of the inequalities is maintained (that is, they are NOT reversed) because Y > 0. Y > 0 because Pr(Y > 0) = 1 (since Y has a gamma distribution)

    = \Pr \left( \frac{2Y}{9.48792} \leq \beta \leq \frac{2Y}{0.710723} \right) .
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  3. #3
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    Seems like I can't get the right answer after all. Can someone please check my calculations to see where I go wrong?

    P(U<a) = \int_0^a \frac{ue^{-u/2}}{4} du = 0.05 =

    1/4\int_0^a {ue^{-u/2}} du] = 0.05 =

    1/4[u*(-2e^{-u/2}]_0^a - \int_0^a 1*(-2e^{-u/2})\, du = 0.05 =

    1/4[[-u2e^{-u/2}]_0^a - \int_0^a -2e^{-u/2} du] = 0.05

     1/4[[-a2e^{-a/2}] - [4e^{-u/2}]_0^a] = 0.05 =

    1/4[-a2e^{-a/2} - [4(e^{-a/2}-e^{-0})] = 0.05

    1/4[-a2e^{-a/2} - [4(e^{-a/2}-1)] = 0.05 =

    1/4[-a2e^{-a/2} - 4e^{-a/2} + 4)] = 0.05
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  4. #4
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    Quote Originally Posted by approx View Post
    Seems like I can't get the right answer after all. Can someone please check my calculations to see where I go wrong?

    P(U<a) = \int_0^a \frac{ue^{-u/2}}{4} du = 0.05 =

    1/4\int_0^a {ue^{-u/2}} du] = 0.05 =

    1/4[u*(-2e^{-u/2}]_0^a - \int_0^a 1*(-2e^{-u/2})\, du = 0.05 =

    1/4[[-u2e^{-u/2}]_0^a - \int_0^a -2e^{-u/2} du] = 0.05

     1/4[[-a2e^{-a/2}] - [4e^{-u/2}]_0^a] = 0.05 =

    1/4[-a2e^{-a/2} - [4(e^{-a/2}-e^{-0})] = 0.05

    1/4[-a2e^{-a/2} - [4(e^{-a/2}-1)] = 0.05 =

    1/4[-a2e^{-a/2} - 4e^{-a/2} + 4)] = 0.05
    Your calculations are correct. Now solve for a (numerically, I'd imagine) and get an answer to the required accuracy. (An exact answer in terms of the Lambert W-function is possible but neither useful nor required).
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  5. #5
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    Sorry for going about this forever. I also posted this in the math part, but no one was able to help me out. I go on like this:

    1/4[-a2e^{-a/2} - 4e^{-a/2} + 4)] = 0.05<br />

    1/4(-a2e^{-a/2}) -e^{a/2} +1 = 0.05<br />

    -0.5a(e^{-a/2}) -e^{a/2} +1 = 0.05<br />

    -0.5a(e^{-a/2}) -e^{a/2} = -0.95<br />

    -e^{-a/2} (1+0.5a)= -0.95<br />

    -e^{-a/2} = \frac{-0.95}{(1+0.5a)}<br />

    e^{-a/2} = \frac{0.95}{-(1+0.5a)}<br />

    But how do I solve for a from here? Thanks for your help.

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  6. #6
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    Quote Originally Posted by approx View Post
    Sorry for going about this forever. I also posted this in the math part, but no one was able to help me out. I go on like this:

    1/4[-a2e^{-a/2} - 4e^{-a/2} + 4)] = 0.05 <br />

    1/4(-a2e^{-a/2}) -e^{a/2} +1 = 0.05" alt="
    1/4(-a2e^{-a/2}) -e^{a/2} +1 = 0.05" />

    -0.5a(e^{-a/2}) -e^{a/2} +1 = 0.05
    " alt="
    -0.5a(e^{-a/2}) -e^{a/2} +1 = 0.05" />

    -0.5a(e^{-a/2}) -e^{a/2} = -0.95
    " alt="
    -0.5a(e^{-a/2}) -e^{a/2} = -0.95" />

    -e^{-a/2} (1+0.5a)= -0.95
    " alt="
    -e^{-a/2} (1+0.5a)= -0.95" />

    -e^{-a/2} = \frac{-0.95}{(1+0.5a)}
    " alt="
    -e^{-a/2} = \frac{-0.95}{(1+0.5a)}" />

    e^{-a/2} = \frac{0.95}{-(1+0.5a)}
    " alt="
    e^{-a/2} = \frac{0.95}{-(1+0.5a)}" />

    But how do I solve for a from here? Thanks for your help.

    As I said earlier, you will have to solve it numerically. That means you have to use technology to get an approximate solution. eg. I used a TI-89 calculator to get my answers.
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