# Homework Help: Sufficient Estimators

1. Jan 9, 2010

### cse63146

1. The problem statement, all variables and given/known data

Show that the product of the sample observations is a sufficient statistic for theta if the random sample is taken from a gamma distribution with parameters alpha = theta and beta = 6.

2. Relevant equations

3. The attempt at a solution

So I need to make sure that $$Y = \prod X_i$$ is a sufficient estimator for theta, which is true if:

$$\frac{f(x_1; \theta)...f(x_n; \theta)}{Y = \prod X_i}$$ does not depend on theta.

But I keep getting getting 1 for the ratio. Am I correct?

Last edited: Jan 9, 2010
2. Jan 9, 2010

You take samples from populations, not other samples.
You neglected to state the type of distribution (normal, gamma, uniform, ?) being sampled; until you do that not much can be done.

When you do use the actual distribution, think about writing out the joint distribution - as you started in the numerator of your first post, and work on the factorization criterion for identifying sufficiency.

3. Jan 9, 2010

### cse63146

Sorry about the type, I meant to say a sample is taken from a gamma distribution with parameters alpha = theta and beta = 6;not from a sample.

The theoroem I'm trying to use is:

$$\frac{f(x_1; \theta)...(f(x_n \theta)}{f_{Y_1}[u_1(x_1,...,x_n); \theta]} = h(x_1,...x_n)$$

and if h(x1,...,xn) doesn't depend on theta, then it's an sufficient estimator.

I get h(x1,...,xn)=1 and I'm wondering if this is correct.

4. Jan 10, 2010

Have you seen this result?

For a random sample from a distribution that has density $$f(x;\theta), \, \theta \in \Omega$$, a statistic
$$Y=u_1(X_1, X_2, \dots, X_n)$$

is sufficient for $$\theta$$ if, and only if there are two nonnegative functions $$k_1, k_2$$ such that

$$\prod_{i=1}^n f(x_i ; \theta) = k_1[u_1(x_1, x_2, \dots, x_n);\theta] k_2(x_1, x_2, \dots, x_n)$$

where $$k_2(x_1, x_2, \dots, x_n)$$ does not depend on $$\theta$$.

This is the factorization theorem I referred to in my earlier post.

To use this, write out the joint distribution and factor it so that the second factor does not depend on either unknown parameter. This is essentially a consequence of the approach you cited, but may be a bit easier to deal with.

5. Jan 10, 2010

### cse63146

That definition is in my textbook, but we haven't gotten to it yet and I'm unsure on how to use it; we've only done the definition I mentioned in my previous post, and I thought I could use it to solve the problem.

Guess I'll have to wait until tuesday to learn it. Thanks for your help.

6. Jan 10, 2010

May not need to wait. I'm away from my computer now, so typing is a pain: what do you get for the product of the densities? If you can, post it in its gory detail.

7. Jan 10, 2010

### cse63146

I was looking over my textbook and found an example that used it, and after a while I figured it out:

$$\frac{1}{\Gamma(\theta)^n 6^{\theta n}}\prod x_i^{\theta - 1}e^{-\Sigma x_i /6} = \frac{1}{\Gamma(\theta)^n 6^{\theta n}}\prod x_i^{\theta}e^{-\Sigma x_i /6} \frac{1}{\prod x_i}$$

And since $$\frac{1}{\prod x_i}$$ doesn't depend on theta, Y is a sufficient estimator of theta.

Thanks for all your help once again.

8. Jan 10, 2010