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

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


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

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?

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 3 '19 at 8:29
  • 1
    $\begingroup$ 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. $\endgroup$ 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.