Let $\{X_i\}_{i=1}^{n}$ be iid according to the Poisson distribution P(λ). Use Method 1 to find the UMVU estimator of $\lambda^k$ for any positive integer k.

Method 1: If T is a complete sufficient statistic, the UMVU estimator of any U-estimable function $g(θ)$ is uniquely determined by the set of equations $E_{θ}[\delta(T)] = g(θ)$ for all $\theta$:

This is my attempt:

$\sum_{t=0}^{\infty}\delta(t)\frac{\lambda^t e^{-\lambda}}{t!}=\lambda^k$

through algebra equivalent to


After this I get stuck trying to manipulate the two series to compare coefficients to identify $\delta(t)$. Any help would be appreciated.

  • 1
    $\begingroup$ You can get displayed equations by using double instead of single dollar signs. This is especially relevant when you're mixing fractions, subscripts, superscripts and exponents. The equations are hard to read in the present form. $\endgroup$ – joriki Jan 7 at 2:15

$T=\sum\limits_{i=1}^{n} X_i$ is a complete sufficient statistic in this case. $T$ is a Poisson random variable with mean $n \lambda$. Let us find a nice function $g$ such that $$\mathbb{E}[g(T)]=\lambda^{k}$$. $\sum\limits_{t=0}^{\infty} g(t)e^{-n \lambda} \frac{n^t\lambda^{t}}{t!}=\lambda^{k}$. so $$\sum\limits_{t=0}^{\infty} g(t) \frac{n^{t}\lambda^t}{t!}=\lambda^{k} e^{n \lambda}$$, recall that $$\lambda^{k}e^{n \lambda}=\sum\limits_{t=0}^{\infty} \frac{n^{t} \lambda^{k+t}}{t!}=\sum\limits_{u=k}^{\infty} n^{u-k} \frac{\lambda^{u}}{(u-k)!}. $$Hence $g(t)=0$ for $0 \le t \le k-1$. What can you say about $g(t)$ for $t \ge k$?

  • $\begingroup$ $g(t)=\frac{t!}{n^k (t-k)!}$ Thank you!. $\endgroup$ – Noe Vidales Jan 7 at 2:42

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