# Sufficient Statistic for a Geometric R.V.

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

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the ratio you end up with is an incomplete statement of Bayes' theorem. Actually, it must be: $$\frac{P(X=x)P(T=t \mid X=x)}{P(T=t)}.$$
But now you see that $P(T=t \mid X=x)$ works like an indicator function taking as value $1$ in the case x is such that $T(x)=t$, and $0$ otherwise.
In both cases you see that the distribution of $X$ does not depend on $\theta$ and now you're done.