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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
    Commented May 16, 2018 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
    Commented 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
    Commented May 17, 2018 at 3:04

1 Answer 1

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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 $(-\theta_0, \theta_0)$.

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  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented May 17, 2018 at 4:27

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