# Complete and Sufficient Statistics

Question: Let $X_1, X_2, ... , X_n$ be a random sample from a Poisson distribution with parameter $\theta$. Show that $Y = $$\sum_{i=1}^n X_i is a complete and sufficient statistic for \theta . I used factorization theorem to prove that Y =$$\sum_{i=1}^n X_i$ is a a sufficient statistic. By definition,$Y = $$\sum_{i=1}^n X_i is complete if E_\theta$$\sum_{i=1}^n X_i=0$ for all $\theta$. However, I just don't know where to start to show that it is complete.

Any suggestion or step-by-step answer is really appreciated. Thanks all!

• Is this homework? If so, then add the [self-study] tag. Also add the source of the exercise. Thank you. – Reviewer
– Jim
Commented May 16, 2018 at 10:40
• Check the definition of completeness. Instead of beginning with $E_\theta(\sum_{i = 1}^n X_i) = 0$, you should begin with assuming $E_\theta(g(\sum_{i = 1}^n X_i)) = 0$ for all $\theta > 0$ and try to deduce $P_\theta(g(\sum_{i = 1}^n X_i) = 0) = 1$ for all $\theta > 0$. Commented May 16, 2018 at 14:20
• Hi Jim, this is a question from a practice test that my school provided, so I'm not really sure how to cite that. Commented May 17, 2018 at 3:04

## 1 Answer

For simplicity, denote $$\sum_{i = 1}^n X_i$$ by $$T$$. By assumption, $$T \sim \text{Poisson}(n\theta)$$. Thus for any measurable function $$g: \mathbb{R}^1 \to \mathbb{R}^1$$ such that $$E_\theta[g(T)] = 0$$, we have $$\sum_{k = 0}^\infty g(k) e^{-n\theta}\frac{(n\theta)^k}{k!} = 0, \quad \forall \theta > 0,$$ which implies $$\sum_{k = 0}^\infty c_k\theta^k = 0, \quad \theta \in (0, \infty), \tag{1}$$ where $$c_k = g(k)n^k/k!, k = 0, 1, 2, \ldots$$.

Define a power series $$f$$ on $$\mathbb{R}$$ by $$f(x) = \sum_{k = 0}^\infty c_k x^k$$, then $$(1)$$ implies that the convergence interval of $$f$$ is the entire real line$$^\dagger$$. Since $$\sum c_k x^k = \sum 0 x^k$$ on $$E = (0, \infty)$$ by $$(1)$$, and $$E$$ contains a limit point, it follows by Theorem 8.5 in Principles of Mathematical Analysis by Walter Rudin that $$c_k = 0, k = 0, 1, 2, \ldots$$, whence $$g(0) = g(1) = \cdots = 0$$. This implies that \begin{align} P_\theta[g(T) = 0] = \sum_{k: g(k) = 0}e^{-n\theta}\frac{(n\theta)^k}{k!} = \sum_{k = 0}^\infty e^{-n\theta}\frac{(n\theta)^k}{k!} = 1, \end{align} i.e., $$T$$ is complete.

$$^\dagger$$: $$(1)$$ implies that for each $$\theta_0 > 0$$, the series $$\sum_k c_k\theta_0^k$$ converges. Hence there exists $$M > 0$$ such that $$|c_k\theta_0^k| \leq M$$ for all $$k$$. Therefore, for all $$x$$ such that $$|x| < \theta_0$$, \begin{align} \sum_{k = 0}^\infty \left|c_k x^k\right| = \sum_{k = 0}^\infty \left|c_k \theta_0^k\right|\left|\frac{x}{\theta_0}\right|^k \leq M\sum_{k = 0}^\infty \left|\frac{x}{\theta_0}\right|^k < \infty, \end{align} which shows that $$\sum_k c_kx^k$$ converges (absolutely) on $$(-\theta_0, \theta_0)$$.

• How exactly did you go from the first function to (1)? It would be very helpful if you can elaborate a little bit. Thank you. Commented May 17, 2018 at 3:03
• I made a mistake, will edit. After that, you just divide $e^{-n\theta}$ on both sides. Commented May 17, 2018 at 3:27
• I'm sorry that my question was not clear. I didn't understand how you can say that (1) is identical to f(θ). Let say if I have a different distribution (either continuous or discrete), how exactly can I find f(θ)? Thank you. Commented May 17, 2018 at 4:27