Let $X$ be a single observation such that given $\Lambda=\lambda$, $X|\Lambda=\lambda$ follows the Poisson distribution with mean $\lambda$. Put a prior distribution on $\Lambda$, say $\pi(\lambda)$, with pdf:
$$\pi(\lambda)=\frac{\lambda^{\tau-1}e^{-\frac{\lambda}{\mu}}}{\Gamma(\tau)\mu^\tau}$$
Find the Bayes estimator of $\lambda$ under squared error loss.
I find that the distribution of $\Lambda|X$ ends up being $GAMMA(\tau+\sum x_i, (\frac{1}{\mu}+n)^{-1})$, which implies that the Bayes estimator under squared error loss is simply the product of the parameters:
$$\lambda_{Bayes}=\frac{\tau+\sum x_i}{\frac{1}{\mu}+n}$$
Where $n$, in this context, is the number of collected samples. Is this correct?