Index: The Book of Statistical Proofs ▷ Probability 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

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

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}.

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