Finding UMVUE of function of poisson parameter

Question

I am to estimate $exp(-\lambda)\lambda^2/2$ from the distribution $Exp(\lambda) \sim \frac{e^{-\lambda}\lambda^x}{x!}$

I used the Indicator function $I_{X_1=2}(X)$ as an initial unbias estimator. statistic $T=\sum{X_i}$. Using Rao-Blackwell, $\hat{W} = E(W|T) = \frac{t(t-1)(n-1)^{t-2}}{2n^t}$

However, when I check $E(\hat{W}) = \sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}e^{-n\lambda}(n\lambda)^t}{2n^tt!}=\frac{1}{2}\exp(-n\lambda)\lambda^2$.

Where is the n coming from and how do i get rid of it? Where did i made a mistake?

Answer 1

Sorry but following your procedure (that is correct)

$$W=\mathbb{1}_{\{2\}}(X_1)$$

thus using the law of total expectation

$$\mathbb{E}[\hat{W}]=\mathbb{E}[\mathbb{E}[W|T]]=\mathbb{E}[W]=1\cdot \frac{e^{-\lambda}\lambda^2}{2}+0\cdot\left[1-\frac{e^{-\lambda}\lambda^2}{2}\right]=\frac{e^{-\lambda}\lambda^2}{2}$$

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Direct calculation

$$\mathbb{E}[\hat{W}]=\sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}}{2\cdot n^t}\cdot\frac{e^{-n\lambda}(n\lambda)^t}{t!}=$$

$$=\frac{e^{-n\lambda}\lambda^2}{2}\sum_{t-2=0}^{n}\frac{[(n-1)\lambda]^{t-2}}{(t-2)!}=\frac{e^{-n\lambda}\lambda^2}{2}\cdot e^{(n-1)\lambda}=\frac{e^{-\lambda}\lambda^2}{2}$$

Answer 2

your initial unbiased estimator looks fine to me...using standard procedure I get exactly the resutl you will expect.

$$\mathbb{E}[W|T]=\mathbb{P}[W=1|T=t]=\frac{\mathbb{P}[X_1=2]\cdot\mathbb{P}[T=t|X_1=2]}{\mathbb{P}[T=t]}=$$

$$=\frac{\frac{e^{-\lambda}\lambda^2}{2!}\cdot\mathbb{P}[\Sigma_{i=2}^n=t-2]}{\frac{e^{-n\lambda}(n\lambda)^t}{t!}}=$$

$$=\frac{\frac{e^{-\lambda}\lambda^2}{2!}\cdot\frac{e^{-(n-1)\lambda}[(n-1)\lambda]^{t-2}}{(t-2)!}}{\frac{e^{-n\lambda}(n\lambda)^t}{t!}}=\frac{t(t-1)(n-1)^{t-2}}{2\cdot n^t}$$