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Suppose you want to estimate the parameters $\alpha$ and $\beta$ of a $\text{gamma}(\alpha, \beta)$ distribution where we know that $\alpha = \beta$. Would you treat this as a distribution with one or two unknown parameters? It seems that you can treat it as a distribution with 1 unknown parameter.

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    Are you saying that you know the mean of your gamma distribution is $1$? If so then yes, you can treat the shape and scale parameters as a single unknown, subject to that constraint.2012-04-07
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    @Henry: So if you have a $N(a, a^2)$ distribution with $a$ unknown, then if you find an estimator for $a$ you have found an estimator for $a^2$?2012-04-07
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    yes, and again in your $N(a,a^2)$ example it is subject to a constraint.2012-04-07
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    @Henry: I see. So you can find an estimator for the scale parameter (or shape parameter) and say that you have also found an estimator for the shape shape parameter (scale parameter)?2012-04-07
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    @Henry: In other words, you could say that the shape parameter is known and the scale parameter is unknown?2012-04-07
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    No - if the shape parameter is unknown (but known to be equal to the scale parameter) then it is unknown. Under this constraint, an estimator for one is an estimator for the other.2012-04-07
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    @Henry: I see. But you wouldn't have a jointly sufficient statistic given that constraint?2012-04-07
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    For your Gamma distribution you have a joint density $\left({1 \over \Gamma(\alpha) \alpha^{\alpha}}\right)^n \left(\prod_{i=1}^n x_i\right)^{\alpha-1} \exp \left({-{1 \over \alpha} \sum_{i=1}^n{x_i}}\right)$ so there is not an obviously simpler sufficient statistic than $\left( \prod_{i=1}^n{x_i} , \sum_{i=1}^n{x_i} \right)$2012-04-07
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    @Henry: So that would be the same jointly sufficient statistic if $\alpha$ and $\beta$ were not equal. So we don't get a simpler sufficient statistic even if $\alpha = \beta$?2012-04-07
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    @Henry: Couldn't you just say that a sufficient statistic for $\alpha$ is $\prod_{i=1}^{n} x_i$? Then you can just square this value to get an estimator of $\beta$?2012-04-07
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    You could do that produce an estimator, but it would not be a sufficient statistics and so cannot be optimal.2012-04-07

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