ELEC 321
# Multivariate Normal

Updated 2017-12-07

Recall that a standard normal random variable has a expected value / mean of 0 and a variance of 1. Then a random vector of standard normal is denoted as \(\mathbf Z \sim N(\mathbf 0, I)\). Where \(Z\) is the random vector with a size of \(N\), \(\mathbf 0\) is a vector of zeroes, and \(I\) is a \(N\times N\) identity matrix.

Suppose there are *independent* standard normal random variables \(Z_1, Z_2, \dots, Z_n\) then their joint density is the product:

Recall that

\[\varphi (z)= \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\]Then the joint density can be simplified to

\[f(\mathbf z)=\frac{1}{(\sqrt{2\pi})^N}e^{-\frac{1}{2}\mathbf{z'z}}\]Where \(\mathbf z\) is a vector that contains \(z_1,\dots,z_N\), and \(\mathbf {z'z}\) is the dot product of itself.

The expected value of the random vector composed of standard multivariate normal is a vector of zeroes:

\[\boldsymbol \mu= \mathbb E\{\mathbf Z\}=\mathbf 0\]For \(N\) random variables, the covariance matrix is an \(N\times N\) identity matrix.

Given that \(\mathbf Z\) is the random vector of standard multivariate normal; any multivariate normal random vector can take the form:

\[\mathbf X=A\mathbf Z + \mathbf b\]Where \(\mathbf X\) is the random vector comprised of random variables with general normal distribution; vector \(\mathbf X\) has a length of \(N\) . \(A\) is an \(N\times N\) *invertible* matrix. And \(\mathbf b\) is a vector of constants.

Because \(A\) is invertible, then

\[\mathbf Z=A^{-1}(\mathbf X-\mathbf b)\]The mean of a generate multivariate normal is \(\mathbf b\) since the mean for the standard multivariate normal is \(\mathbf 0\):

\[\boldsymbol \mu = \mathbb E\{\mathbf X\}=\mathbb E\{A\mathbf Z+\mathbf b\}=\mathbf b\]The covariance is \(AA'\) where \(A'\) is the transpose of \(A\):

\[\Sigma=\text{Cov}(\mathbf X)=\text{Cov}(A\mathbf Z+\mathbf b)=\text{Cov}(A\mathbf Z)=A\text{Cov}(\mathbf Z)A'=AA'\]Notice that \(\mathbf X=A\mathbf Z + \mathbf b\) is in fact a transformation, therefore its corresponding Jacobian is simply

\[J=\vert \det(A^{-1})\vert =\frac{1}{\vert \det(A)\vert }\]Now that we know the Jacobian, it follows that the density is given by

\[f_\mathbf X(\mathbf x)=\frac{1}{\vert \det(A)\vert }f_{\mathbf Z}(A^{-1}(\mathbf x-\mathbf b))\]Plugging in \(f(\mathbf z)=\frac{1}{(\sqrt{2\pi})^N}e^{-\frac{1}{2}\mathbf{z'z}}\), and we obtain

\[\frac{1}{\vert \det(A)\vert } \frac{1}{(\sqrt{2\pi})^N}e^{-\frac{1}{2}\mathbf{(A^{-1}(\mathbf x-\mathbf b))'(A^{-1}(\mathbf x-\mathbf b))}}\]Note that the covariance matrix \(\Sigma=AA'\), which also implies \(\Sigma^{-1}=(A^{-1})'A^{-1}\).

Also note that the determinant of the covariance matrix is \(\det(\Sigma)=\det(AA')\), working it out we see that \(\sqrt{\det(\Sigma)}=\vert \det(A)\vert\).

Plugging these equations in, the above density simplify down to

\[f_\mathbf X(\mathbf x)=\frac{1}{\sqrt{(2\pi)^N\Sigma}}e^{-\frac{1}{2}(\mathbf x-\mathbf b)'\Sigma^{-1}(\mathbf x-\mathbf b)}\]**1. Linear transformation of normal vectors results in normal vectors**

Suppose we have a vector \(\mathbf X\sim N(\boldsymbol \mu, \Sigma)\), and a matrix \(C\) is full rank. then let \(\mathbf Y=C\mathbf X+\mathbf d\). The mean of \(\mathbf Y\) is \(C\boldsymbol \mu+\mathbf d\); the variance is \(C\Sigma C'\).

**2. Marginal distributions are normal**

Suppose we have a random vector that has size 2:

\[\mathbf X=\begin{bmatrix} X_1\\ X_2\\ \end{bmatrix} \sim N\left( \begin{bmatrix} \mu_1\\ \mu_2\\ \end{bmatrix} , \begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22}\\ \end{bmatrix} \right)\]Then we get

\[X_1\sim N(\mu_1,\Sigma_{11}),\qquad X_2\sim N(\mu_2,\Sigma_{22})\]**3. Conditional distributions are normal**

Using the previous case, suppose we have \(x_2\) as a realization for \(X_2\), then the conditional on \(X_1\) is

\[X_1\vert (X_2=x_2)\sim N(\mu_{1\vert 2}, \Sigma_{1\vert 2})\]Where \(\mu_{1\vert 2}=\mu_1+\frac{\Sigma_{12}}{\Sigma_{22}}(x_2-\mu_2)\) and \(\Sigma_{1\vert 2}=\Sigma_{11}-\frac{\Sigma_{12}\Sigma_{21}}{\Sigma_{22}}\).