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Math Help - construct an interval estimator (exponential)

  1. #1
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    construct an interval estimator (exponential)

    I think I solved it but I don't have the answer in my book to check against.

    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. Define X=Y_1+...+Y_n. Find the moment generating function of X. Show that 2X\theta is a pivotal function, and hence construct an interval estimator for \theta with confidence coefficient 0.95. (Note. You may use the fact that the \chi^2 distribution with k degrees of freedom has moment generating function (1-2t)^{-k/2}.)

    Answer.

    M_X(t)=E(e^{Xt})=E(e^{Y_1t}...e^{Y_nt})=(M_Y(t))^n  =(\frac{\theta}{\theta-t})^n=(\frac{1}{(1-t/n)})^n which, I think, accidentally means X{\sim}Gamma(n,\theta)

    Now we show that the distribution of 2X\theta does not depend on \theta, so that it can be a pivotal function

    Let U=2X\theta

    M_U(t)=E(e^{Ut})=E(e^{X2{\theta}t})=M_X(2{\theta}t  )=(\frac{1}{(1-\frac{2{\theta}t}{\theta}})^n=(\frac{1}{1-2t})^n which is mgf for a chi-square distribution with 2n degrees of freedom.

    So U=2X\theta\sim\chi^2(2n) and the distribution of U does not depend on \theta.

    Now we construct an interval estimator for \theta with confidence coefficient 0.95. Using the distribution of U,

    P(t_{0.025}(2n){\leq}2X{\theta}{\leq}t_{0.975}(2n)  )=0.95

    I finally got the interval estimator [\frac{t_{0.025}(2n)}{2X}; \frac{t_{0.975}(2n)}{2X}].

    Appreciate your feedback.
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  2. #2
    Moo
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    Hello,

    It looks OK
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  3. #3
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    Thank you!
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