Show $ \sum_{i=0}^n X_{i} $ is a complete sufficient statistic.

Question

We're given a random sample of $X_{1}, X_{2}, .... , X_{n}$ from a distribution with a pdf of

$f(x;\theta) = \theta^x(1-\theta), x = 0,1,2,.... $ and $ 0 < \theta < 1 $

We're asked to show that that $Tn = \sum_{i=0}^n X_{i} $ is a complete sufficient statistic for $\theta$.

I am able to show that $Tn$ is a sufficient statistic for $\theta$, I am just stuck on how to show that it is complete.

To show that $Tn$ is complete, we must show $E(g(Tn)) = 0 $. Where I am stuck is I do not know the distribution of Tn so I am not sure what to use for $f_{Tn}(t)$ for this expectation.

Answer 1

$X_i$ is a geometric random variable. Thus $T_n$ is the number of tails that appear before the $n$th heads, when repeatedly flipping a coin that has a $\theta$ chance of being tails. You can show that the PMF is $$P(T_n = k) = \binom{n+k-1}{k}\theta^k (1-\theta)^n$$

Answer 2

The definiton of Completeness is not the one you posted. It is a very difficult property to prove.

A statistic $T$ is complete if, $\forall \theta$ and $\forall g$ misurable function

$$\mathbb{E}_{\theta}[g(T)]=0$$

implies

$$\mathbb{P}_{\theta}[g(T)=0]=1$$

Good for you that in your case it is not necessary to show completeness with the definition.

It is enough to get a sufficient estimator by showing that the density belongs to the exponential family

$$f(x;\theta)=(1-\theta)Exp(x\cdot log\theta)$$

$T(x)=x$ and thus $\Sigma_i X_i$ is Sufficient and Complete (Exponential Family Canonical Statistic)