Multivariate normal distribution, calculating transform distribution and conditional expectations

I have a problem following the solution to the following problem.

Let $$X\in N(0,\Lambda)$$, where

$$\Lambda=\begin{pmatrix} 1&2 &-1 \\ 2&6 &0 \\ -1&0 &4 \end{pmatrix}.$$

Set $$Y_1=X_1+X_3,\quad Y_2=2X_1-X_2,\quad Y_3=2X_3-X_2$$. Compute the conditional expectations $$E[Y_3|Y_1=3]$$ and $$E[Y_3|Y_2=-1]$$.

The solution is given by the following.

We have $$\pmb{Y}\in N(\pmb{0},\pmb{\Sigma})$$ where

$$\pmb{\Sigma}=\begin{pmatrix} 3&-2 &4 \\ -2&2 &-2 \\ 4&-2 &22 \end{pmatrix}.$$

Thus,

$$E[Y_3|Y_1=3]=0+\frac{4}{3}(3-0)=4,\quad E[Y_3|Y_2=-1]=0+\frac{-2}{2}(-1-0)=1.$$

When cacluating the $$\pmb{Y}$$ distribution, what rule/method do they use? Is the characteristic function an option? And finally, how do they calculate the conditional expectations from this?

• The vector $Y=(Y_1,Y_2,Y_3)$, being a linear transformation of the multivariate normal vector $X$, is itself multivariate normal. This implies $(Y_1,Y_3)$ and $(Y_2,Y_3)$ are also bivariate normal, from which it follows that the conditional distributions are univariate normal. More details here. – StubbornAtom Oct 3 '19 at 8:29
• Yes, that is true. I have made some progress, the covariance matrix Sigma is given by ALambdaA' (where ' denotes transponate). And I believe what you told me was the missing puzzle piece I was looking for. I tried to partition it, but that way I had do put conditions on all of the other r.v.s, which didn't work. Yes, I also read on the Wikipedia page, now I understand that I can find the bivariate distributions for these combinations by applying the "general form" of the covariance matrix. Thank you. – SimpleProgrammer Oct 3 '19 at 8:58

So, I finally got it right. Here is the full solution.

We are given that the transform of $$(X_1,X_2,X_3)$$ to $$(Y_1,Y_2,Y_3)$$ is

$$\begin{pmatrix} Y_1\\ Y_2\\ Y_3 \end{pmatrix}=\begin{pmatrix} 1&0 &1 \\ 2&-1 &0 \\ 0&-1 &2 \end{pmatrix}\begin{pmatrix} X_1\\ X_2\\ X_3 \end{pmatrix}.$$

Since $$\pmb{Y}$$ is a linear transformation of multivariate normal vector $$\pmb{X}$$, as StubbornAtom commentated, then $$\pmb{Y}$$ is also a multivariate normal vector. We may then combine the components anyway we want and that (vector) combination will aslo be normally distributed. We choose the normal bivariate vectors as $$(Y_1,Y_3)$$ and $$(Y_2,Y_3)$$. From here we can easily calculate the conditional distribution of $$Y_3$$ given $$Y_1=y_1$$ and $$Y_2=y_2$$, respectively. The conditional distributions will also be (univariate) normally distributed with mean (or expected value, which we are looking for!) given as

$$\mu_3 + \rho \frac{\sigma_3}{\sigma_{i}}(y_{i}-\mu_{i})=\mu_3 + \frac{\text{Cov}(Y_i,Y_3)}{\sigma_{i}^2}(y_{i}-\mu_{i}),\quad i=1,2 .$$

Here $$\text{Cov}(Y_i,Y_3)$$ is the covariance between $$Y_i$$ and $$Y_3$$, and can be found as the $$i,3$$ (or $$3,i$$ since symmetry) element of the covariance matrix $$\Sigma$$.

From here we can calculate the desired results.

• You might also add the part about $\Sigma=A\Lambda A^T$ where $A$ is the transformation matrix. – StubbornAtom Oct 3 '19 at 9:48