Let $X_1,...X_n$ be a random sample with distribution $\text{Normal}(\theta,1)$. Find the UMVUE for $\theta^2$
What I´ve done so far:
I have already shown that $T=\sum_{i=1}^nX_i$ is a complete sufficient statistic for $\theta$. Let $\widehat{\theta^2}=\bar {X}^2-\frac{1}{n}$ be an estimator for $\theta^2$
$$E[\widehat{\theta^2}]=E[\bar {X}^2-\frac{1}{n}]=E[\bar{X}^2]-E^2[\bar X]+E^2[\bar X]+\frac{1}{n}=Var(\bar X)+E^2[\bar X]+\frac{1}{n}$$
I know that $\bar X$ has distribution $\text{Normal}(\theta, \frac{1}{n})$ It follows that :$$E[\widehat{\theta^2}]=\theta^2$$ Hence $\widehat{\theta^2}=\bar {X}^2-\frac{1}{n}$ is unbiased estimator
Know I have to compute $g(T)=E[\widehat{\theta^2}|T]$ and by Lehmann-Scheffé this will be the UMVUE for $\theta^2$ but the problem is how can I compute $$E[\widehat{\theta^2}|T]=E[\bar {X}^2-\frac{1}{n}|\sum_{i=1}^nX_i]$$ I know I need to find the joint density of $\bar {X}^2-\frac{1}{n}$ and $\sum_{i=1}^nX_i$ but is there an easy way to do it?
Or is there another way to find the UMVUE? I would really appreciate if you can help me with this problem