I'm stuck with the following problem: Suppose you have a n-variate random vector like $$Y = S + \epsilon, $$ $$S = LZ $$ where $\epsilon \sim N_n(0,I\sigma^2)$, $L$ is a known singular matrix, and $Z \sim N_n(0,I)$ ($N_n$ stands for an n-variate Normal distribution). In this case, the values of $S$ belongs to a space with dimension less than $n$, but since there's an error term $\epsilon$, $Y$ goes back to $\mathbb R^n$. How can I find the conditional distribution of $S|Y$? The fact that $LL'$ has no inverse makes the usual method fail because $S$ has no pdf.
As an example, suppose $n=2$, and $L =\bigl( \begin{smallmatrix}1 & 0\\1 & 0 \end{smallmatrix}\bigr)$, then $S$ must be in the $x=y$ line. My intuition says $S|Y$ that it has to be Normal Distributed, with mean $\frac{Y_1 + Y_2}{2}$ and applying the transformation $L$ to something. But I can't figure out the math behind this.
The "usual formula" that I know requires only that the covariance matrix of $Y$, namely $K:=LL'+\sigma^2I$, is invertible. Writing $M:=LL'$, you have that the conditional distribution of $S$, given that $Y=y$, is $n$-variate normal with mean vector $MK^{-1}y$ and covariance matrix $M-MK^{-1}M$. In your example $M=\left( \begin{smallmatrix}1 & 1\\1 & 1 \end{smallmatrix}\right)$ and $K=\left( \begin{smallmatrix}1+\sigma^2 & 1\\1 & 1+\sigma^2 \end{smallmatrix}\right)$, so the conditional mean vector is $\left( \begin{smallmatrix}(y_1+y_2)/(2+\sigma^2)\\(y_1+y_2)/(2+\sigma^2) \end{smallmatrix}\right)$ and the conditional covariance is $\left( \begin{smallmatrix}\sigma^2/(2+\sigma^2) & \sigma^2/(2+\sigma^2)\\\sigma^2/(2+\sigma^2) & \sigma^2/(2+\sigma^2) \end{smallmatrix}\right)$. (I am assuming that $\epsilon$ and $Z$ are independent.)