# Thread: Finding the Sufficient Statistic of Weibull Distribution.

1. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

The joint pdf is correct now and you are ready to claim it as an member of the exponential family.

http://en.wikipedia.org/wiki/Exponen...alar_parameter

In the exponential, there is only 1 term which is the product of the sufficient statistics and the (transformation of) parameter. Hopefully it is easy to figure out.

2. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

No it's not for me. Could you explain it a bit please?

3. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

Can you see it is an exponential family? So I quote the same link again here:

http://en.wikipedia.org/wiki/Exponen...alar_parameter

there is only 1 term which is the product of the sufficient statistics and the (transformation of) parameter
You have listed 3 terms inside the exponential. Which one is it? You are just like doing a matching - which term is

(refer to the notation in the wiki link I posted above)

The sufficient statistic of course is a function of the sample

4. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

So, it is sum of x_i^b since this is being multiplied by the thing that has alpha in it?

5. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

?

Don't want to be offensive or anything here but I think the help can be a lot better. I know you guys want people to try themselves and I have been trying, and when you're not getting something, small hints don't really add much to your understanding.

6. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

And what would the ideal help look like in this situation? I thought BGM was doing a good job.

7. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

Don't get me wrong. I do appreciate the help.

I am aware it is in the exponential family yes but given I know that, I am struggling to find the sufficient statistic for alpha given beta is known, which is the question I have to answer, but I would also like to know how the sufficient statistics would be different if alpha was known and i had to find the sufficient statistic of beta, and if both were known and both unknown. All of four scenarios..

....

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9. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

well thanks guys!

10. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

Sorry I miss your previous response. I have not subscribe any post so perhaps sometimes the new posts are sunk too quickly.

So, it is sum of x_i^b since this is being multiplied by the thing that has alpha in it?
And I think you got the right answer.

11. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

Thanks for the reply. much appreciated, but could you try to explain a bit more about the different scenarios and how I can find other sufficient statistics based on this exponential, if i knew alpha and not beta for example?

12. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

In general, if you have the one-parameter exponential family, it is pretty much the same as you did in this question - express the joint in the required form and figure out which one is the sufficient statistic. Usually only exponential family can provide this kind of dimension reduction in the sufficient statistic.

Refer to your example, when beta is not known, then it no longer belongs to the exponential family. The same argument no longer apply, and you may try to find it via the Neyman-Fisher factorization theorem. Not much reduction you can achieve.

13. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

Ah ok I get it now! Thanks dude.. but I was informed by somebody that you can't express Weibull as expontential family, is this true? i.e. even when beta is known..?

14. ## Re: Finding the Sufficient Statistic of Weibull Distribution.

When is known, it belongs to the exponential family as you have already shown it by yourself.

From the structure of the pdf, you can see it is a power transformation from the exponential distribution, and "fixing" makes it look quite similar to the exponential distribution.