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!

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    $\begingroup$ Is this homework? If so, then add the [self-study] tag. Also add the source of the exercise. Thank you. – Reviewer $\endgroup$ – Jim May 16 '18 at 10:40
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    $\begingroup$ 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 $\theta > 0$. $\endgroup$ – Zhanxiong May 16 '18 at 14:20
  • $\begingroup$ Hi Jim, this is a question from a practice test that my school provided, so I'm not really sure how to cite that. $\endgroup$ – Nguyen May 17 '18 at 3:04

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 $$f(\theta) \equiv \sum_{k = 0}^\infty \frac{g(k)n^k}{k!}\theta^k = 0, \quad \theta \in (0, \infty). \tag{1}$$

Extend the domain of $f(\theta)$ to $\mathbb{R}^1$ so that it can be viewed as a power series. $(1)$ says that $f(\theta)$ converges at every point in $(0, \infty)$, which further implies $f(\theta)$ converges everywhere in $\mathbb{R}^1$, for the convergence interval of a power series is always an interval that is symmetric about $0$. It then follows that $0 \equiv f^{(k)}(0) = g(k)n^k, k = 0, 1, 2, \ldots$ (see, for example, Corollary 8.10 of Rudin's Principles of Mathematical Analysis). From here to completeness of $T$ is immediate.

  • $\begingroup$ 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. $\endgroup$ – Nguyen May 17 '18 at 3:03
  • $\begingroup$ I made a mistake, will edit. After that, you just divide $e^{-n\theta}$ on both sides. $\endgroup$ – Zhanxiong May 17 '18 at 3:27
  • $\begingroup$ 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. $\endgroup$ – Nguyen May 17 '18 at 4:27

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