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!

  • 4
    $\begingroup$ 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, 2018 at 10:40
  • 2
    $\begingroup$ 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, 2018 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, 2018 at 3:04

1 Answer 1


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, 2018 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, 2018 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, 2018 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.