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Math Help - Gamma distribution, mgf and unbiased estimator

  1. #1
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    Gamma distribution, mgf and unbiased estimator

    Question.

    Let Y_1, ... Y_n be a random sample from a Gamma distribution \Gamma(k, \frac{1}{\theta}) with density function

    f_y(y)=\frac{{\theta}^k}{(k-1)!}y^{k-1}e^{{-\theta}y} for x>0,
     0 otherwise

    where k\geq1 is a known integer, and \theta>0 is an unknown parameter.

    i. Show that the moment generating function of Y_1+Y_2+ ... +Y_n is M_Y(t)=(1-\frac{t}{\theta})^{-nk}

    ii. Find the constant c such that c/{\bar}Y (that's "c over Y bar", if it does not show up properly on the screen) is an unbiased estimator for \theta, where {\bar}X (X bar) is the sample mean. Calculate the vairance of this estimator and compare it with the Cramer-Rao lower bound.

    (This is an exact wording from the book, including X bar which appears out of nowhere)

    Answer.

    (i) E(e^{t(Y_1+Y_2+...+Y_n)}=E(e^{ntY_i})=[M_Y(t)]^n since Ys are iid.

    =[(\frac{\theta}{\theta-t})^k]^n=(\frac{1}{1-\frac{t}{\theta}})^{nk}=(1-\frac{t}{\theta})^{-nk}

    (ii) I need to find E(c/{\bar}Y), and I don't think I can simply substitute the mean in the denominator E(1/{\bar}Y) (expected value of the 1 over Y bar) here, so what else can I do?

    I could also try E(\frac{c}{1/n\Sigma_{i=1}^nY_i}). Any pointers?
    Last edited by Volga; February 12th 2011 at 05:31 AM.
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  2. #2
    Moo
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    Hello,

    For the mgf, you got it wrong.
    \frac{\theta}{\theta-t}=\frac{1}{1-\frac{1}{\theta}}, so there's no factor \theta^n (you just got messed up in the algebra)
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  3. #3
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    Ah! cancels out from nominator and denominator, of course ))))
    I will adjust the original post
    thanks!
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  4. #4
    Moo
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    For the second question, all I can see is that you notice, from the mgf, that the sum Y=Y_1+\dots+Y_n follows a Gamma distribution \Gamma(nk,1/\theta), hence you have its density \tilde{f}_Y(y) ( y\in (0,\infty))

    So \displaystyle E[1/Y]=\int_0^\infty \frac{\tilde{f}_Y(y)}{y}~dy

    I didn't do the calculations, but this integral can be solved by retrieving the pdf of another gamma (which integral is 1)
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  5. #5
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    Thanks - let me try.
    Can I ask what you use the tilde over y for here? as in \tilde{f}_Y(y). To show that it is a distribution function?

    Now then,
    let U=Y_1+\dots+Y_n (to distinguish it from Y variable)
    then per (a), M_U(t)=(1-t{\theta})^{-nk} and we conclude that U\simGamma(nk, \frac{1}{\theta}) as Moo has pointed out and I should have seen myself ))) (probably it is a good idea to use part (a) in working out part (b))

    E(\frac{1}{\bar{Y}})=E(\frac{1}{1/n*U})=E(\frac{n}{U}) and I will use the distribution of U to find the expected value of U:

    E(1/U)=\int_{range of u}\frac{1}{u}f_u(u)du=\int_0^{\infty}\frac{1}{u}\f  rac{\theta^{nk}}{\Gamma(nk)}u^{nk-1}e^{-{\theta}u}du=\frac{{\theta}{\Gamma}(nk-1)}{\Gamma(nk)}\int_0^{\infty}\frac{{\theta}^{nk-1}}{\Gamma(nk-1)}u^{nk-2}e^{-{\theta}u}du=\theta\frac{1}{nk-1} since the integral is a Gamma distribution integral for Gamma(nk-1,\frac{1}{\theta}) and is equal to 1.

    Therefore

    E(\frac{1}{\bar{Y}})=nE(\frac{1}{U}=\frac{n\theta}  {nk-1}

    and E(\frac{c}{\bar{Y}})=\frac{nc{\theta}}{nk-1}=\theta

    c=\frac{nk-1}{n}, and the unbiased estimator is \frac{c}{\bar{Y}}=\frac{nk-1}{n\bar{Y}}
    Last edited by Volga; February 13th 2011 at 03:04 AM.
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  6. #6
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    I now need to find variance of that unbiased estimator

    Var(\frac{c}{\bar{Y}})=\frac{(nk-1)^2}{n^2}Var(\frac{1}{\bar{Y}})=\frac{(nk-1)^2}{n^2}n^2Var(\frac{1}{U})

    Var(\frac{1}{U})=E[(\frac{1}{U})^2]-E[\frac{1}{U}]^2=\frac{\theta^2}{(nk-2)(nk-1)}-\frac{\theta^2}{(nk-1)^2}=\frac{\theta^{2}(nk-1)-{\theta}^2(nk-2)}{(nk-1)^{2}(nk-2)}=\frac{\theta^2}{(nk-1)^{2}(nk-2)}

    Here is how I came up with

    E[(\frac{1}{U})^2]=\int_0^{\infty}\frac{1}{u^2}\frac{\theta^{nk}}{\G  amma(nk)}u^{nk-1}e^{-{\theta}u}du=\frac{\Gamma(nk-2)\theta^2}{\Gamma(nk)}\int_0^{\infty}\frac{\theta  ^{nk-2}}{\Gamma(nk-2)}u^{nk-3}e^{-{\theta}u}du=\frac{\Gamma(nk-2)\theta^2}{\Gamma(nk)}=\frac{\theta^2}{(nk-2)(nk-1)}

    Then finally the variance of the unbiased estimator

    Var(\frac{c}{\bar{Y}})=\frac{(nk-1)^2}{n^2}Var(\frac{1}{\bar{Y}})=\frac{(nk-1)^2}{n^2}n^2Var(\frac{1}{U})==\frac{(nk-1)^2}{n^2}n^2\frac{\theta^2}{(nk-1)^2(nk-2)}=\frac{\theta^2}{nk-2}

    Not sure if it is OK to have theta in the formula?

    Next, I must figure out Cramer-Rao lower bound for variance...
    Last edited by Volga; February 12th 2011 at 10:27 PM.
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  7. #7
    Moo
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    The tilde was here because I name the rv Y. So when I saw that there could be a confusion with the very first pdf you defined, I chose to call the pdf differently instead of calling the rv differently

    Your calculations for finding c are correct

    It's ok for the variance too, and it's not a problem if there is \theta. You can simplify from the beginning the n²... It's too bad you want to keep it 'til the end

    For Cramér-Rao's bound, you have to compute Fisher's information, right ? So just try it ^^ It's not that difficult.
    FYI : I(\theta)=-E\left[\frac{\partial^2}{\partial \theta^2} \ln(f_Y(Y;\theta))\right]
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  8. #8
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    I am sorry, I was just building up to it - I am terribly insecure when it comes to Cramer-Rao

    Log-likelihood

    l(\theta, y)=kln(\theta)-ln(k-1)!+(k-1)lny-{\theta}y

    Information

    I_Y(\theta)=-E(\frac{\partial^2}{\partial\theta^2}l_y(\theta;y)  )=E(\frac{k}{\theta^2})=\frac{k}{\theta^2}, for a single observation

    Then for a random sample of iid nI_Y(\theta)=\frac{kn}{\theta^2}

    Var({\theta}hat){\geq}\frac{1}{nI_Y(\theta)}

    \frac{\theta^2}{nk-2}{\geq}\frac{\theta^2}{nk}, as expected.

    I think I am done here. By the way, how do you place a 'hat' on the top of the variable in Latex code?
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  9. #9
    Moo
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    It's correct

    By the way, it's l_y(Y;\theta) : you consider the rv Y, and the second member is for the parameter, which is \theta here.

    \hat{something}
    \tilde{something}
    \bar{something} or \overline{something}
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