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Math Help - Pivotal Method

  1. #1
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    Pivotal Method

    Suppose that we are to obtain a single observation Y from an exponential distribution with mean theta.

    a) Find the distribution of U = Y/theta? Is U a pivotal quantity?

    b) Use U = Y/theta and the pivotal method, to form a confidence interval formula for theta with a confidence level of 90%.


    I know that the exponential distribution is:

    f(x) = (1/theta) * e^(-x/theta)

    What I'm confused about is how to find a distribution of U = Y/theta and if it is a pivotal quantity. Any help would be appreciated.
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    Re: Pivotal Method

    .
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  3. #3
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    Re: Pivotal Method

    How are you being given these questions without being taught the basics behind them?
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    Re: Pivotal Method

    Quote Originally Posted by romsek View Post
    How are you being given these questions without being taught the basics behind them?
    Y is exponential with parameter $\theta$ i.e. $\large p_Y(y)=\frac 1 \theta e^{-y/\theta}$

    $Pr[Y<y] = F_Y(y)=1-e^{-y/\theta}$

    $U=\dfrac Y \theta$

    $Pr[U < u] = Pr[\dfrac Y \theta < u] = Pr[Y< u \theta]$

    $Pr[Y< u \theta]=$ $\large 1-e^{-\frac{u \theta}{\theta}}=1 - e^{-u}$

    $p_U(u) = \dfrac{d}{du} 1 - e^{-u}= e^{-u}$

    Thus $p_U(u)$ is independent of the parameter $\theta$ and is thus a pivotal quantity.
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    Re: Pivotal Method

    For part b. would this be the correct answer?

    P[ -z(0.05) ≤ (1/θ)*e^(-y/θ) ≤ z(0.05) ] = 0.90

    => P[ -1.64 ≤ (1/θ)*e^(-y/θ) ≤ 1.64 ] = 0.90

    => P[ -1.64 ≤ (1/θ)*e^(-u) ≤ 1.64 ] = 0.90

    => P[ -e^(-u) / 1.64 ≥ θ ≥ e^(u)/1.64 ] = 0.90
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    Re: Pivotal Method

    ??
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  7. #7
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    Re: Pivotal Method

    Actually would this be correct?

    For random variable X ~ U(0,θ) we need to construct a 90% confidence interval for θ.

    Yn would be the largest order statistic from a sample of size n from this distribution and has a pdf:

    f(y) = [n*y(n-1)]/θ for 0 ≤ y ≤ θ

    Let U = Yn/θ

    f(u) = n*u^(n-1) for 0 ≤ u ≤ 1

    F(u) = u^n for 0 ≤ u ≤ 1

    Then,

    P(a < Yn/θ < b) = 0.90

    F(a) = 0.05 and F(b) = 0.95

    a^n = 0.05 and b^n = 0.95

    giving:

    a = the nth root of(0.05)

    b = the nth root of (0.95)

    so,

    P(the nth root of(0.05) < Yn/θ < the nth root of (0.95))

    giving the confidence interval:

    (Yn/the nth root of(0.05), Yn/the nth root of (0.95))
    Last edited by AwesomeHedgehog; April 13th 2014 at 06:30 AM.
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    Re: Pivotal Method

    No. That's not how you do it.

    For a 90% confidence interval we can assign 5% to the lower tail and 5% to the upper tail.

    The interval will be $[a,b]$ such that

    $F_U(a)=0.05$

    $F_U(b)=0.95$

    $F_U(u) = 1-e^{-u}$ as determined earlier. So

    $1-e^{-a}=0.05$

    $0.95 = e^{-a}$

    $a=-\ln(0.95) \approx 0.0513$

    $1-e^{-b}=0.95$

    $0.05 = e^{-b}$

    $b=-\ln(0.05)\approx 2.9957$

    So a 90% confidence interval for U is $[0.0513, 2.9957]$

    Now $\theta =\dfrac Y U$ so our confidence interval for $\theta$ is given by $\left[ \dfrac Y {2.9957},\dfrac Y {0.0513}\right]$

    Note this is not the only $90\%$ confidence interval. Since the distribution isn't symmetric picking one interval over the other is a bit problematic. It's common to make the tails the same probability.
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