Hi guys. I have a question and I hope someone can help me out
Let X1,.....Xn be a random sample from gamma(a,b) with a known.
Find the best unbiased estimator of 1/b
Waiting for your response as soon as you can
Thanks in advance
Hi guys. I have a question and I hope someone can help me out
Let X1,.....Xn be a random sample from gamma(a,b) with a known.
Find the best unbiased estimator of 1/b
Waiting for your response as soon as you can
Thanks in advance
Read this: Best Unbiased Estimators
OK, I'll work with that, BUT that is not how most people write the gamma density. I place beta in the denomiator.
The likelihood function is $\displaystyle L(\beta )={\beta^{n\alpha}\over \bigl(\Gamma (\alpha )\bigr)^n}\bigl(\Pi_{i=1}^nx_i\bigr)^{\alpha-1}e^{-\beta\sum_{i=1}^nx_i}$.
Thus with $\displaystyle \alpha$ known we have $\displaystyle \sum_{i=1}^nx_i$ suficient for $\displaystyle \beta$.
And since $\displaystyle E\biggl(\sum_{i=1}^nX_i\biggr)={n\alpha\over \beta}$ we have $\displaystyle {\sum_{i=1}^nX_i\over n\alpha}$ UMVUE for $\displaystyle {1\over \beta}$.
I could have done this yesterday, but the mean of my gamma's is $\displaystyle \alpha\beta$
and getting an unbiased estimator of $\displaystyle 1/\beta$ is a lot harder in that case.
And I wasn't going to do this until I knew how you were writing your density.
It can be shown that Sum(Xi) is not only sufficient but complete for beta (as w(beta) of exponential function contains an open set), so try 1/sum(xi) to estimate 1/beta. Lehmann-Scheffe tells us that an unbiased estimator that is a function of a complete statistic is the best unbiased estimator.
It can be shown that, assuming iid xi, Sum(xi) is distributed as Gamma(n*alpha,beta).
Given the above, let Y=(1/sum(xi)). Y is distributed as an inverted gamma(n*alpha, 1/beta) with mean=(1/beta)/(n*alpha-1). Thus, E(1/sum(xi)) = E(Y) = (1/beta)/(n*alpha-1), which is obviously a biased estimator of 1/beta.
Now let T = (n*alpha - 1)/sum(xi) be the unbiased estimator of 1/beta which is a function of a complete statistic. Thus, T is the best unbiased estimator of 1/beta, but it can be shown that it does not attain the lower Cramer-Rao bound.
I see. Thanks for the info. If the pdf can be written like this, it'd be so easy. Like we can show that it is an exponential family thus it is a complete sufficient estimator for beta. Then we can use MLE to get 1/beta(^). Then to show that 1/beta(^) is an unbiased estimator we calculate the mean of it and we get 1/beta. So 1/beta(^) in unbiased estimator.
So we can conclude that 1/beta(^) is the best estimator. Is it right?