Source of the question: the exercise 6.17 of Statistical Inference Book by Casella and Berger.
Let $X_1,...,X_n$ be iid with geometric distribution
$$P_\theta (X=x)=\theta (1-\theta)^{x-1}, x=1,2,..., 0<\theta<1.$$
Show that $\Sigma X_i$ is sufficient for $\theta$, and find the family of distributions of $\Sigma X_i$.
My attempt:
I use factorization theorem. The joint pmf is
$$P (x_1,...,x_n | \theta)=\theta^n (1-\theta)^{\Sigma x_i-n}=\theta^n (1-\theta)^{-n}(1-\theta)^{\Sigma x_i}$$
The factor that contains $\theta$ depends on the sample $(x_1,...,x_n)$ only through the function $T(X)=\Sigma x_i$. Thus, $\Sigma x_i$ is sufficient for $\theta$.
My question is the the family of distributions of $\Sigma X_i$. I know the answer is negative binomial. But here, I think the idea of this question is to get pmf of $\Sigma X_i$ from the joint pmf above. By factorization theorem $f(X|\theta)=g(T(X)|\theta)h(X)$, we know the kernel of the distribution of the sufficient statistics $T(X)$ is $g(T(X)|\theta)$.
Thus, the kernel of the distribution of $T(X)=\Sigma x_i$ here must be $\theta^n (1-\theta)^{t-n}$.
The answer is $P(\Sigma X_i=t)={t-1\choose n-1} (1-\theta)^{t-n}$. But my gap is I don't know how to get ${t-1\choose n-1}$.
