Index: The Book of Statistical ProofsProbability Distributions ▷ Multivariate continuous distributions ▷ Multivariate normal distribution ▷ Mean

Theorem: Let $x$ follow a multivariate normal distribution:

$\label{eq:mvn} x \sim \mathcal{N}(\mu, \Sigma) \; .$

Then, the mean or expected value of $x$ is

$\label{eq:mvn-mean} \mathrm{E}(x) = \mu \; .$

Proof:

1) First, consider a set of independent and standard normally distributed random variables:

$\label{eq:zi} z_i \overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0,1), \quad i = 1,\ldots,n \; .$

Then, these variables together form a multivariate normally distributed random vector:

$\label{eq:z} z \sim \mathcal{N}(0_n, I_n) \; .$ $\label{eq:mean-rvec} \mathrm{E}(z) = \mathrm{E}\left( \left[ \begin{array}{c} z_1 \\ \vdots \\ z_n \end{array} \right] \right) = \left[ \begin{array}{c} \mathrm{E}(z_1) \\ \vdots \\ \mathrm{E}(z_n) \end{array} \right] \; .$

Because the expected value of all its entries is zero, the expected value of the random vector is

$\label{eq:z-mean} \mathrm{E}(z) = \left[ \begin{array}{c} \mathrm{E}(z_1) \\ \vdots \\ \mathrm{E}(z_n) \end{array} \right] = \left[ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} \right] = 0_n \; .$

2) Next, consider an $n \times n$ matrix $A$ solving the equation $A A^\mathrm{T} = \Sigma$. Such a matrix exists, because $\Sigma$ is defined to be positive definite. Then, $x$ can be represented as a linear transformation of $z$:

$\label{eq:x-z} x = Az + \mu \sim \mathcal{N}(A 0_n + \mu, A I_n A^\mathrm{T}) = \mathcal{N}(\mu, \Sigma) \; .$

Thus, the expected value of $x$ can be written as:

$\label{eq:x-mean} \mathrm{E}(x) = \mathrm{E}( Az + \mu ) \; .$

With the linearity of the expected value, this becomes:

$\label{eq:mvn-mean-qed} \begin{split} \mathrm{E}(x) &= \mathrm{E}( Az + \mu ) \\ &= \mathrm{E}(Az) + \mathrm{E}(\mu) \\ &= A \, \mathrm{E}(z) + \mu \\ &\overset{\eqref{eq:z-mean}}{=} A \, 0_n + \mu \\ &= \mu \; . \end{split}$
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Metadata: ID: P339 | shortcut: mvn-mean | author: JoramSoch | date: 2022-09-15, 02:22.