# Limit of conditional distribution

I have bivariate normal prior distribution over $[x_1 x_2]$:

$$\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \Sigma\right),$$ and conditional distribution $y|x_1, x_2$: $$y | x_1, x_2 \sim N(x_1+x_2,\sigma^2).$$

I want to derive the reverse conditional distribution $x_1, x_2 | y$ as $\sigma \rightarrow 0$, ie when we have deterministic $y=x_1+x_2$. In other words, I want to write the conditional distribution as a function of $\sigma$, and then show that it converges to the standard result at $\sigma=0$. Is it possible to show that this function converges?

• When $\sigma=0$, $y=x_1+x_2$ hence, conditionally on $y$, the random variables $x_1$ and $x_2$ are both conditionally normaly distributed with means $\nu_1y$ and $\nu_2y$ and variances $\tau_1^2$ and $\tau_2^2$. The computation of the $\nu$ and $\tau$ values is standard, is this your question? – Did Oct 28 '16 at 14:40
• I'll rephrase the question. I'm really trying to write the conditional distribution as a function of $\sigma$, and then show that it converges to the standard result. – Betterthan Kwora Oct 28 '16 at 15:18
• The joint distribution of $(x_1,x_2,y)$ is known hence there is no problem in applying the standard procedure. The result will depend continuously on the parameter $\sigma^2$. – Did Oct 28 '16 at 18:13