We have $(X_1,...,X_n)$ random sample of distribution $N(m,1)$. I need to find UMVUE of $g(m)=e^m$.
The natural guess was that the unbiased estimator which I need to find may be something like $e^T$, where $T=\frac{1}{n} \sum\limits_{i=1}^n X_i$. So I calculated $$\mathbb{E}[e^{T}]=exp \left[m+\frac{1}{2n}\right]$$ therefore the unbiased estimator that I was looking for is $g(T)=e^{T-\frac{1}{2n}}.$
My question is: is there a way to get this result in any other way than just guessing? I considered solving: $$\mathbb{E}g(T)=e^m$$ so I would get
$$\int\limits_0^\infty g(t) \frac{1}{\sqrt{\frac{2 \pi}{n}}}e^{-\frac{(t-m)^2}{\frac{2}{n}}} dt=e^m$$ but that doesn't seem to get me anywhere.
What do you think?