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

Theorem: Let $x$ and $y$ follow a normal-gamma distribution:

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

Then, the joint probability density function of $x$ and $y$ is

\[\label{eq:ng-pdf} p(x,y) = \sqrt{\frac{|\Lambda|}{(2 \pi)^n}} \frac{b^a}{\Gamma(a)} \cdot y^{a+\frac{n}{2}-1} \exp \left[ -\frac{y}{2} \left( (x-\mu)^\mathrm{T} \Lambda (x-\mu) + 2b \right) \right] \; .\]

Proof: The probability density of the normal-gamma distribution is defined as as the product of a multivariate normal distribution over $x$ conditional on $y$ and a univariate gamma distribution over $y$:

\[\label{eq:ng-pdf-w1} p(x,y) = \mathcal{N}(x; \mu, (y \Lambda)^{-1}) \cdot \mathrm{Gam}(y; a, b)\]

With the probability density function of the multivariate normal distribution and the probability density function of the gamma distribution, this becomes:

\[\label{eq:ng-pdf-s2} p(x,y) = \sqrt{\frac{|y \Lambda|}{(2 \pi)^n}} \exp \left[ -\frac{1}{2} (x-\mu)^\mathrm{T} (y \Lambda) (x-\mu) \right] \cdot \frac{b^a}{\Gamma(a)} y^{a-1} \exp\left[-by\right] \; .\]

Using the relation $\lvert y A \rvert = y^n \lvert A \rvert$ for an $n \times n$ matrix $A$ and rearranging the terms, we have:

\[\label{eq:ng-pdf-qed} p(x,y) = \sqrt{\frac{|\Lambda|}{(2 \pi)^n}} \frac{b^a}{\Gamma(a)} \cdot y^{a+\frac{n}{2}-1} \exp \left[ -\frac{y}{2} \left( (x-\mu)^\mathrm{T} \Lambda (x-\mu) + 2b \right) \right] \; .\]
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Metadata: ID: P44 | shortcut: ng-pdf | author: JoramSoch | date: 2020-02-07, 20:44.