How to simulate from a gaussian conditional density? Here's the problem: 
Suppose that $x$ and $y$ are random vectors which are jointly normally distributed with density $p(x,y)$. I wish to draw samples from the density $p(x|y)$. Denote a draw from $p(x,y)$ as $x^{+}$ and $y^{+}$ and let $\hat{x}=E(x|y)$ and $\hat{x}^+=E(x^{+}|y^{+})$. I want to prove that the vector $\tilde{x}$, defined  as 
 $$ \tilde{x} = x^{+} - \hat{x}^+ + \hat{x},$$ 
is a draw from $p(x|y)$, by proving that $E(\tilde{x}|y)= E(x|y)$ and $V(\tilde{x}|y) = V(x|y).$
This algorithm appears in Durbin and Koopman(2002), Biometrika, under the title 'A simple and efficient simulation smoother for state space time series analysis'. 
Thanks,
Aqua.
 A: Since
\begin{align*}p(x|y) &= \frac{1}{(2\pi)^{n/2}|A|^{1/2}}\\
& \times\exp\left\{-\frac{1}{2}\left(x-  \mu_x - \Sigma_{xy} \Sigma_{yy}^{-1} (y - \mu_y)\right)A^{-1}\left(x-  \mu_x - \Sigma_{xy} \Sigma_{yy}^{-1} (y - \mu_y)\right)'\right\},\end{align*}
the random variable $X_y$ corresponding to this density can be written as the linear transform of a standard Normal vector $\epsilon_X$ with the same dimension $d$as $X$:
$$X_y=\mu_x + \Sigma_{xy} \Sigma_{yy}^{-1} (y - \mu_y)+A^{1/2}\epsilon_X
\qquad\epsilon_X\sim\mathcal{N}_d(0,\mathbf{I}_d)$$
which can also be written as
$$X_y=\mathbb{E}[X|y]+A^{1/2}\epsilon_X
\qquad\epsilon_X\sim\mathcal{N}_d(0,\mathbf{I}_d)$$
Therefore, switching from $Y=y_0$ to $Y=y_1$ simply means that the random variable or a realisation of that random variable is translated by $\Sigma_{xy} \Sigma_{yy}^{-1}(y_1-y_0)$. Hence, if $x⁺$ is a realisation from $p(x|y⁺)$, of if $(x⁺,y⁺)$ a realisation from $p(x|,y)$,then 
$$x_y=x⁺+\mathbb{E}[X|y]-\mathbb{E}[X|y^+]=x⁺+\Sigma_{xy} \Sigma_{yy}^{-1}(y-y⁺)$$is a realisation from $p(x|y)$.
