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Math Help - Confidence Interval

  1. #1
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    Confidence Interval

    Suppose that Y is normally distributed with mean 0 and unknown variance \sigma ^2. Then \frac{Y^2}{\sigma^2} has a \chi^2 with 1 df. Use the pivotal quantity \frac{Y^2}{\sigma^2} to find:

    a) 95% confidence interval for \sigma^2
    b) 95% upper confidence limit of \sigma^2

    Solution:
    I based my work on http://www.mathhelpforum.com/math-he...-interval.html which gave me

    f(y) = \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}}

    then P(Y<\alpha) = \int^{\alpha}_0 \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025

    P(Y<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 y^{(1/2)-1}e^{-y/2} \ dy = 0.025

    if I make the substitution y=2x and \frac{dy}{dx} =2 \longrightarrow dy = 2dx then I get:

    P(2X<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025

    P(2X<\alpha) = \frac{1}{\sqrt{\pi}} \int^{\alpha}_0 x^{(1/2)-1}e^{-x} \ dx = 0.025

    now this looks a lot like a Gamma function, but my upper bound is \alpha and not \infty, which is throwing me off.

    P(Y>\beta) =  \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025

    P(Y>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.025

    same substitution as before

    P(2X>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025

    P(2X>\beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.025

    again this is looking a lot like a Gamma function, but this this my lower bound is not 0 it's \beta

    the answer in the back of the book is \left(\frac{Y^2}{5.02389},\ \frac{Y^2}{0.71072} \right)<br />

    b) P(Y \geq \beta) =  \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.95

    P(Y \geq \beta) =  \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.95

    same substitution and same argument as before. I get:

    P(2X \geq \beta) =  \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.95

    with the solution in the back of the book being \frac{Y^2}{0.0039321}
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  2. #2
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    Quote Originally Posted by lllll View Post
    Suppose that Y is normally distributed with mean 0 and unknown variance \sigma ^2. Then \frac{Y^2}{\sigma^2} has a \chi^2 with 1 df. Use the pivotal quantity \frac{Y^2}{\sigma^2} to find:

    a) 95% confidence interval for \sigma^2
    b) 95% upper confidence limit of \sigma^2

    Solution:
    I based my work on http://www.mathhelpforum.com/math-he...-interval.html which gave me

    f(y) = \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}}

    Mr F says: All corrections are given in red:

    then P({\color{red}U}<\alpha) = \int^{\alpha}_0 \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025 Mr F says: No. The random variable here is not Y. It's {\color{red}\frac{Y^2}{\sigma^2} = U}, say.

    P( {\color{red}U} <\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 y^{(1/2)-1}e^{-y/2} \ dy = 0.025

    if I make the substitution y=2x and \frac{dy}{dx} =2 \longrightarrow dy = 2dx then I get:

    P({\color{red}U}<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha{\color{red}/2}}_0 (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025 Mr F says: The substitution is a change of variable only in the integral, NOT a change of random variable! That 2X you had should still be a U. Also, if you introduce a change of variable, you have to change the integral terminals too!

    Actually, my next comment (below) shows why there's actually no point in making a substitution.

    P({\color{red}U}<\alpha) = \frac{1}{\sqrt{\pi}} \int^{\alpha{\color{red}/2}}_0 x^{(1/2)-1}e^{-x} \ dx = 0.025

    now this looks a lot like a Gamma function, but my upper bound is \alpha and not \infty, which is throwing me off.

    Mr F says: This equation can only be readily solved for {\color{red}\alpha} by using technology. In fact, I'm not sure it can be solved at all without using technology.

    Everything I've said above applies below too.

    P(Y>\beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025

    P(Y>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.025

    same substitution as before

    P(2X>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025

    P(2X>\beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.025

    again this is looking a lot like a Gamma function, but this this my lower bound is not 0 it's \beta

    the answer in the back of the book is \left(\frac{Y^2}{5.02389},\ \frac{Y^2}{0.71072} \right)<br />

    b) P(Y \geq \beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.95

    P(Y \geq \beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.95

    same substitution and same argument as before. I get:

    P(2X \geq \beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.95

    with the solution in the back of the book being \frac{Y^2}{0.0039321}
    ..
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