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Consider a random sample of size $n$ froma gamma distribution, $X_i\sim GAM(\theta, \kappa)$, with $\kappa$ being the shape parameter and $\theta$ being the scale parameter and let $\bar X=\dfrac{1}{n}\sum X_i$ and $\tilde X=(\prod X_i)^{1/n}$ be the sample mean and geometric mean, respectively.

Show that the conditional distribution of $\bar X |_{\tilde X = \tilde x}$ does not depend on $\kappa$.

I have absolutely no idea how to find this conditional distribution. Even if there are some strategies to go about solving this problem without solving for the conditional distribution explicitly, I don't know what they are. How would you show this? Is there a general strategy for finding conditional distributions of statistics that I don't know about? I'm so lost.

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  • $\begingroup$ See here $\endgroup$
    – zhoraster
    Mar 5 '18 at 18:47
  • $\begingroup$ @zhoraster could you elaborate on how to use the observation that $\bar{X}/\tilde{X}$ does not depend on the scale parameter $\theta$ to conclude that the conditional distribution of $\bar{X}$ given $\tilde{X}=\tilde{x}$ does not depend on the shape parameter $\kappa$? $\endgroup$
    – John
    Mar 5 '18 at 19:36
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    $\begingroup$ @zhoraster. Yeah, the fact that it's the shape parameter instead of the scale really has me stumped. $\endgroup$
    – jippyjoe4
    Mar 7 '18 at 7:15
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    $\begingroup$ Anyway. For fixed (known) $\theta$, prove that $X_1\cdots X_n$ is a sufficient statistic, which immediately gives the result. $\endgroup$
    – zhoraster
    Mar 7 '18 at 7:40
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    $\begingroup$ Shape parameter, you mean? And why ratio? We're talking about sample mean, not ratio. Anyway, this does not matter at all. Sufficiency of $T$, by definition, means that the distribution of the sample given $T$ is independent of the parameter. Then, obviously, the same is true for any function of the sample: sample mean, sample variance, $X_1$ etc. $\endgroup$
    – zhoraster
    Mar 7 '18 at 11:11

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