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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 $$ \frac{f(x_1,\bar x)}{f(\bar x)}=\frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x_1-\theta)^2}{1} +\frac{(\bar x-\theta)^2}{{a'}^2} - \frac{2\rho(x_1-\theta)(\bar x-\theta)}{{a'}} \right]+\frac{(\bar x-\theta)^2}{2{a'}^2} \right) $$

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  • $\begingroup$ "I'm wondering is that my argument make sense?" Not much. If $p_{X_1,X_2,\bar X}$ refers to the PDF of $W=(X_1,X_2,\bar X)$, then the trouble is that $W$ has no PDF since $W\in H$ almost surely, where $H$ denotes the plane in the $(x,y,z)$-space of equation $z=ax+(1-a)y$. On the other hand, assuming $(X_1,X_2)$ are i.i.d. then $(X_1,\bar X)$ has a PDF. Can you compute the conditional distribution of $X_1$ conditioned on $\bar X$? It might help to introduce $Z=(1-a)X_1-aX_2$. $\endgroup$ – Did Oct 7 '15 at 20:19
  • $\begingroup$ @Did I upload my post. Please have a look! $\endgroup$ – spatially Oct 8 '15 at 2:57
  • $\begingroup$ If the distributions of $(X_1,\bar X$ is $\mu$ then the distribution of $W=(X_1,X_2,\bar X)$ is $\nu$ defined by $$\nu(dxdydz)=\mu(dxdz)\delta_{(z-ax)/(1-a)}(dy).$$ The Dirac part, which reflects the fact that $W\in H$ almost surely, might be some reason why you find this nondirect. $\endgroup$ – Did Oct 8 '15 at 5:56
  • $\begingroup$ By the way, did you actually compute the distribution of $(X_1,\bar X)$ then the conditional distribution of $X_1$ conditioned on $\bar X$? $\endgroup$ – Did Oct 8 '15 at 6:03
  • $\begingroup$ Yea I did, it is another normal distribution right? But, could you write more information about the equation you are writing there? I kind of lost though... $\endgroup$ – spatially Oct 8 '15 at 13:19

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