# Finding the UMVUE of $e^{3\lambda}$ in Poi($\lambda$)

Let $$X = (X_1, ... , X_n)$$ iid variables coming from Poisson distribution with mean $$\lambda$$.
Find the UMVUE of $$e^{3\lambda}$$.

I tried understanding the solution below (in the possible duplicate link) but couldn't come up with appropriate weights. So I thought of using the moment-generating function and Lehman-Scheffe Theorem.

For starters, we know $$S_n = \sum_{i=1}^{n} X_i$$ is a complete and sufficient statistic (C.S.S) and has Poi(n$$\lambda$$) distribution.

Then the moment-generating function of $$M_{S_n}(t) = e^{n\lambda(e^t-1)} = E(e^{S_n t})$$.

So if we let $$n\lambda(e^t-1) = 3\lambda \iff t = \ln(3/n + 1)$$, we have $$E(e^{S_n(\ln(3/n+1)}) = e^{3\lambda}$$. Since it is a function of the C.S.S, we can say $$e^{S_n(\ln(3/n+1)}$$ is the UMVUE.

Am I missing or overlooking anything here? Would appreciate some advice or insights.

Possible duplicate

• You can't have $E[e^{S_n\log(3/n+1)}]$ as the UMVUE because it's not an estimator; it's an expectation using the unknown $\lambda$. If you drop the expectation then it's an estimator and seems to be unbiased Commented Feb 16, 2021 at 3:46