1. ## Sufficient Estimators

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

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?

2. I'm not sure what you're doing with that ratio.
You need the conditional density.
Sufficient statistic - Wikipedia, the free encyclopedia

3. The theorem I'm using 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.

It looks roughly the same as the formula given on the wikipedia entry for sufficient estimators(By Hoggs and Craig which is the textbook I'm using right now):

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

YOU need to state the density
Seeing an alpha and a beta I'm guessing your underlying density is a gamma
so show me your work, especially the density

5. Can't believe I forgot the density; it's as you said a gamma.

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

I'm wondering would my $f_{Y_1}[u_1(x_1,...,x_n); \theta]$ be the same as the product of PDFs as I mentioned above in this post?

6. you're missing a negative sign in the exponent of the exponential
and IT's a 6 not a theta in that exponent, which solves your problem
The product of the x's and the theta cannot be separated, giving you the product as the suff stat.

7. So would $f_{Y_1}[u_1(x_1,...,x_n); \theta]= \frac{1}{\Gamma(\theta)^n 6^{\theta n}}\prod x_i^{\theta - 1}e^{-\Sigma x_i /6}$ cause that's where I'm having a problem with.

Since $Y = \prod x_i$

8. $f(x_1,...,x_n; \theta) = \frac{1}{\Gamma(\theta)^n 6^{\theta n}}\prod x_i^{\theta - 1}e^{-\Sigma x_i /6}$

FACTORS into $g(x_1,...,x_n; \theta)h(x_1,...,x_n)$

The function of the data (xi's) that canot be separated from theta is $\prod x_i^{\theta - 1}$

$= \left(\prod x_i\right)^{\theta - 1} =\left(\prod x_i\right)^{\theta}\left(\prod x_i\right)^{-1}$

Hence $\prod x_i$ is suff for theta

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

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

EDIT: Didn't see your previous post.

Saved my neck once again. Sorry for being such an idiot.

10. NO, its that part that is stuck with theta that is suff.

The product is suff for theta because of the theta in the expoenent of the product
that one over the product part that can go into h function (it's garbage)

The key term is $\left(\prod x_i\right)^{\theta}$

The $\left(\prod x_i\right)^{-1}$ is garbage that can go into h

11. Really? Cause that's what my book implies in a similar example:

$k_1[u_1(x_1,...,x_2); \theta] = \theta^n (\prod x_i)^n$ and $k_2(x_1,...,x_n) = \frac{1}{\prod x_i}$

Since $k_2(x1,...,x_n)$ does not depend on theta, $\prod x_i$ is a sufficient statistic for theta.

Unless I've misunderstood something again.

12. Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. If the probability density function is ƒθ(x), then T is sufficient for θ if and only if functions g and h can be found such that
i.e. the density ƒ can be factored into a product such that one factor, h, does not depend on θ and the other factor, which does depend on θ, depends on x only through T(x).

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THE SUFF STAT IS THE PART WITH X's that cannot be separated from the parameter (theta).
The h function is ONLY a function of just x's
WHILE g is a function of theta and THE SUFF STAT.

13. Got it.

Thanks for putting up with my stupidity.

14. $g= \frac{1}{\Gamma(\theta)^n 6^{\theta n}}\left(\prod x_i\right)^{\theta}$

$h=e^{-\Sigma x_i /6} \frac{1}{\prod x_i}I(x_1>0)\cdots I(x_n>0)$