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

• May 5th 2009, 06:19 PM
survivor1980
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

• May 5th 2009, 07:34 PM
mr fantastic
Quote:

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

• May 5th 2009, 08:45 PM
matheagle
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
• May 5th 2009, 09:31 PM
survivor1980
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
• May 6th 2009, 04:20 PM
matheagle
I said that yesterday, nor do I know what '1-α = alpha, Β=beta' means.
• May 6th 2009, 04:32 PM
survivor1980
• May 6th 2009, 04:34 PM
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.
• Feb 28th 2010, 09:22 AM
ricer
Quote:

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~(Thinking)
• Apr 12th 2010, 07:06 PM
palabine
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.
• Apr 12th 2010, 07:32 PM
ricer
Thanks(Clapping)
• Apr 12th 2010, 08:42 PM
matheagle
Quote:

Originally Posted by survivor1980

THE mean here is $\displaystyle {\alpha\over \beta}$ since you're writing the beta in the numerator instead of in the denominator.
• May 3rd 2011, 02:45 AM
mnazam
Quote:

Originally Posted by survivor1980

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)?
• May 3rd 2011, 03:19 AM
Moo
Quote:

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.
• May 3rd 2011, 04:15 AM
mnazam
Quote:

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?