Index: The Book of Statistical ProofsProbability Distributions ▷ Multivariate continuous distributions ▷ Normal-gamma distribution ▷ Mean

Theorem: Let $x \in \mathbb{R}^n$ and $y > 0$ follow a normal-gamma distribution:

\[\label{eq:ng} x,y \sim \mathrm{NG}(\mu, \Lambda, a, b) \; .\]

Then, the expected value of $x$ and $y$ is

\[\label{eq:ng-mean} \mathrm{E}[(x,y)] = \left[ \left( \mu, \frac{a}{b} \right) \right] \; .\]

Proof: Consider the random vector

\[\label{eq:rvec} \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} x_1 \\ \vdots \\ x_n \\ y \end{array} \right] \; .\]

According to the expected value of a random vector, its expected value is

\[\label{eq:mean-rvec} \mathrm{E}\left( \left[ \begin{array}{c} x \\ y \end{array} \right] \right) = \left[ \begin{array}{c} \mathrm{E}(x_1) \\ \vdots \\ \mathrm{E}(x_n) \\ \mathrm{E}(y) \end{array} \right] = \left[ \begin{array}{c} \mathrm{E}(x) \\ \mathrm{E}(y) \end{array} \right] \; .\]

When $x$ and $y$ are jointly normal-gamma distributed, then by definition $x$ follows a multivariate normal distribution conditional on $y$ and $y$ follows a univariate gamma distribution:

\[\label{eq:ng-def} x,y \sim \mathrm{NG}(\mu, \Lambda, a, b) \quad \Leftrightarrow \quad x \vert y \sim \mathcal{N}(\mu, (y \Lambda)^{-1}) \quad \wedge \quad y \sim \mathrm{Gam}(a,b) \; .\]

Thus, with the expected value of the multivariate normal distribution and the law of conditional probability, $\mathrm{E}(x)$ becomes

\[\label{eq:mean-x} \begin{split} \mathrm{E}(x) &= \iint x \cdot p(x,y) \, \mathrm{d}x \, \mathrm{d}y \\ &= \iint x \cdot p(x|y) \cdot p(y) \, \mathrm{d}x \, \mathrm{d}y \\ &= \int p(y) \int x \cdot p(x|y) \, \mathrm{d}x \, \mathrm{d}y \\ &= \int p(y) \left\langle x \right\rangle_{\mathcal{N}(\mu, (y \Lambda)^{-1})} \, \mathrm{d}y \\ &= \int p(y) \cdot \mu \, \mathrm{d}y \\ &= \mu \int p(y) \, \mathrm{d}y \\ &= \mu \; , \end{split}\]

and with the expected value of the gamma distribution, $\mathrm{E}(y)$ becomes

\[\label{eq:mean-y} \begin{split} \mathrm{E}(y) &= \int y \cdot p(y) \, \mathrm{d}y \\ &= \left\langle y \right\rangle_{\mathrm{Gam}(a,b)} \\ &= \frac{a}{b} \; . \end{split}\]

Thus, the expectation of the random vector in equations \eqref{eq:rvec} and \eqref{eq:mean-rvec} is

\[\label{eq:ng-mean-qed} \mathrm{E}\left( \left[ \begin{array}{c} x \\ y \end{array} \right] \right) = \left[ \begin{array}{c} \mu \\ a/b \end{array} \right] \; ,\]

as indicated by equation \eqref{eq:ng-mean}.


Metadata: ID: P237 | shortcut: ng-mean | author: JoramSoch | date: 2021-07-08, 09:40.