I have a problem that I know I am very close to the solution for, but I think I just need some more formatting to make it a really clean proof.
The problem goes like this:
Suppose X is a discrete r.v. with the pmf p(x) = $(1-\theta)^{(x-1)}*\theta, x=1,2,3,...,$ and $0\le\theta\le1.$ Consider the statistic $T=\Sigma X_i$
i) Use the definition of a sufficient statistic to show that T is a sufficient statistic for theta.
Looking at this pmf, (and a hint from my professor saying," Does the pmf look familiar? It should."), I saw that this is a geometric distribution.
So going by the definition of sufficiency: $\frac{(P(X_1=x_1)P(X_2=x_2)***P(X_1=x_1)}{P(T=t)}$=H
Where H does not contain $\theta$.
For the top half of my equation, I get:
$\theta^n(\theta-1)^{(\Sigma X_i-1)}$
And for the denominator, I get:
$\theta(\theta-1)^{(T-1)}$
So, I end up with:
$\frac{\theta^n(\theta-1)^{(\Sigma X_i-1)}}{\theta(\theta-1)^{(T-1)}}$
I know this does not cancel out to where my statistic is not dependent upon $\theta$.
I am not sure where I am going wrong with this one, but if someone could let me know, that would be greatly appreciated.
b. The other half is to prove this by factorization:
I get my factorization down to:
$(1-\theta)^{T-1}\theta^n$
I am having the same problem as I am with a problem I posted earlier about a Beta distribution. I have g(T,$\theta$), but what would be my h? I know that the sum of geometric r.v.'s add up to one, so is my answer one again? :S