My question:
Let $X_1, X_2\sim N(\theta,1)$. Let $\bar X = aX_1+(1-a)X_2$ for $0<a<1$. Find the distribution of $(X_1,X_2)$ given $\bar X$.
Update: My previous try is wrong. I delete it. Now based on the suggestion given below by @Did, I did the following:
Try to write done the distribution of $(X_1,\bar X)$. Again, both $X_1$ and $\bar X$ are normally distributed, where $\bar X\sim N(\theta,a')$ where constant $a'=\sqrt{a^2+(1-a)^2}$. Clearly, $X_1$ and $\bar X$ are not independent so we have to write done the correlation so that I can use bivariate normal density.
To compute correlation, we notice that $$ \operatorname{cov}(X_1,\bar X)=\operatorname{cov}(X_1,aX_1+(1-a)X_2) = a\operatorname{cov}(X_1,X_1)+(1-a) \operatorname{cov}(X_1,X_2) = a $$ Thus, we have the distribution of $(X_1,\bar X)$ is the bivariate normal distribution with mean vector $(\theta,\theta)$, variance $\sigma_1=1$ and $\bar \sigma = a'$, and $\rho = a$. Hence, it has pdf $$ f(x_1,\bar x) = \frac{1}{2\pi {a'}\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x_1-\theta)^2}{1} +\frac{(\bar x-\theta)^2}{{a'}^2} - \frac{2\rho(x_1-\theta)(\bar x-\theta)}{{a'}} \right] \right) $$ and hence the conditional distribution $f(x_1|\bar x)$ can be obtained by $$ \frac{f(x_1,\bar x)}{f(\bar x)}=\frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x_1-\theta)^2}{1} +\frac{(\bar x-\theta)^2}{{a'}^2} - \frac{2\rho(x_1-\theta)(\bar x-\theta)}{{a'}} \right]+\frac{(\bar x-\theta)^2}{2{a'}^2} \right) $$