# Thread: Find the best unbiased estimator of 1/b of gamma dist.

1. ## Find the best unbiased estimator of 1/b of gamma dist.

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

2. Originally Posted by survivor1980
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

3. 1 I need to know how you're writing your gamma density. Sometimes the b is in the numerator, sometimes it in the denomiator.

2 is best=min variance? Hence UMVUE

4. Thank you so much guys for speedy response.

Let X₁. X₂, X₃,…, Xn be a random sample from gamma(α, Β) with α known. Find the best unbiased estimator of 1/ Β

Note 1-α = alpha, Β=beta

2- Yes the best unbiased estimatore is the same as UMVUE

thanks

I said that yesterday, nor do I know what '1-α = alpha, Β=beta' means.

6. Hi all

This is the function

7. 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.

8. Originally Posted by matheagle
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.
I met the same problem with this question.
if the the mean of gamma's is $\displaystyle \alpha\beta$. How to get an unbiased estimator of $\displaystyle 1/\beta$? Anyone can help? THANKS~

9. 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.

10. Thanks

11. Originally Posted by survivor1980
Hi all

This is the function

THE mean here is $\displaystyle {\alpha\over \beta}$ since you're writing the beta in the numerator instead of in the denominator.

12. Originally Posted by survivor1980
Hi all

This is the function

But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?

13. Originally Posted by mnazam
But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?
There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.

14. Originally Posted by Moo
There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.
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?

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