Normal-inverse-gamma distribution

${\displaystyle f(x,\sigma ^{2}\mid \mu ,\lambda ,\alpha ,\beta )={\frac {\sqrt {\lambda }}{\sigma {\sqrt {2\pi }}}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1}\exp \left(-{\frac {2\beta +\lambda (x-\mu )^{2}}{2\sigma ^{2}}}\right)}$

For the multivariate form where ${\displaystyle \mathbf {x} }$ is a ${\displaystyle k\times 1}$ random vector,

${\displaystyle f(\mathbf {x} ,\sigma ^{2}\mid \mu ,\mathbf {V} ^{-1},\alpha ,\beta )=|\mathbf {V} |^{-1/2}{(2\pi )^{-k/2}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1+k/2}\exp \left(-{\frac {2\beta +(\mathbf {x} -{\boldsymbol {\mu }})^{T}\mathbf {V} ^{-1}(\mathbf {x} -{\boldsymbol {\mu }})}{2\sigma ^{2}}}\right).}$

where ${\displaystyle |\mathbf {V} |}$ is the determinant of the ${\displaystyle k\times k}$ matrix ${\displaystyle \mathbf {V} }$. Note how this last equation reduces to the first form if ${\displaystyle k=1}$ so that ${\displaystyle \mathbf {x} ,\mathbf {V} ,{\boldsymbol {\mu }}}$ are scalars.

It is also possible to let ${\displaystyle \gamma =1/\lambda }$ in which case the pdf becomes

${\displaystyle f(x,\sigma ^{2}\mid \mu ,\gamma ,\alpha ,\beta )={\frac {1}{\sigma {\sqrt {2\pi \gamma }}}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1}\exp \left(-{\frac {2\gamma \beta +(x-\mu )^{2}}{2\gamma \sigma ^{2}}}\right)}$

In the multivariate form, the corresponding change would be to regard the covariance matrix ${\displaystyle \mathbf {V} }$ instead of its inverse ${\displaystyle \mathbf {V} ^{-1}}$ as a parameter.

${\displaystyle F(x,\sigma ^{2}\mid \mu ,\lambda ,\alpha ,\beta )={\frac {e^{-{\frac {\beta }{\sigma ^{2}}}}\left({\frac {\beta }{\sigma ^{2}}}\right)^{\alpha }\left(\operatorname {erf} \left({\frac {{\sqrt {\lambda }}(x-\mu )}{{\sqrt {2}}\sigma }}\right)+1\right)}{2\sigma ^{2}\Gamma (\alpha )}}}$

Given ${\displaystyle (x,\sigma ^{2})\sim {\text{N-}}\Gamma ^{-1}(\mu ,\lambda ,\alpha ,\beta )\!.}$ as above, ${\displaystyle \sigma ^{2}}$ by itself follows an inverse gamma distribution:

${\displaystyle \sigma ^{2}\sim \Gamma ^{-1}(\alpha ,\beta )\!}$

while ${\displaystyle {\sqrt {\frac {\alpha \lambda }{\beta }}}(x-\mu )}$ follows a t distribution with ${\displaystyle 2\alpha }$ degrees of freedom. [1]

Proof for ${\displaystyle \lambda =1}$

For ${\displaystyle \lambda =1}$ probability density function is

${\displaystyle f(x,\sigma ^{2}\mid \mu ,\alpha ,\beta )={\frac {1}{\sigma {\sqrt {2\pi }}}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1}\exp \left(-{\frac {2\beta +(x-\mu )^{2}}{2\sigma ^{2}}}\right)}$

Marginal distribution over ${\displaystyle x}$ is

${\displaystyle {\begin{aligned}f(x\mid \mu ,\alpha ,\beta )&=\int _{0}^{\infty }d\sigma ^{2}f(x,\sigma ^{2}\mid \mu ,\alpha ,\beta )\\&={\frac {1}{\sqrt {2\pi }}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\int _{0}^{\infty }d\sigma ^{2}\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1/2+1}\exp \left(-{\frac {2\beta +(x-\mu )^{2}}{2\sigma ^{2}}}\right)\end{aligned}}}$

Except for normalization factor, expression under the integral coincides with Inverse-gamma distribution

${\displaystyle \Gamma ^{-1}(x;a,b)={\frac {b^{a}}{\Gamma (a)}}{\frac {e^{-b/x}}{{x}^{a+1}}},}$

with ${\displaystyle x=\sigma ^{2}}$, ${\displaystyle a=\alpha +1/2}$, ${\displaystyle b={\frac {2\beta +(x-\mu )^{2}}{2}}}$.

Since ${\displaystyle \int _{0}^{\infty }dx\Gamma ^{-1}(x;a,b)=1,\quad \int _{0}^{\infty }dxx^{-(a+1)}e^{-b/x}=\Gamma (a)b^{-a}}$, and

${\displaystyle \int _{0}^{\infty }d\sigma ^{2}\left({\frac {1}{\sigma ^{2}}}\right)^{\alpha +1/2+1}\exp \left(-{\frac {2\beta +(x-\mu )^{2}}{2\sigma ^{2}}}\right)=\Gamma (\alpha +1/2)\left({\frac {2\beta +(x-\mu )^{2}}{2}}\right)^{-(\alpha +1/2)}}$

Substituting this expression and factoring dependence on ${\displaystyle x}$,

${\displaystyle f(x\mid \mu ,\alpha ,\beta )\propto _{x}\left(1+{\frac {(x-\mu )^{2}}{2\beta }}\right)^{-(\alpha +1/2)}.}$

Shape of generalized Student's t-distribution is

${\displaystyle t(x|\nu ,{\hat {\mu }},{\hat {\sigma }}^{2})\propto _{x}\left(1+{\frac {1}{\nu }}{\frac {(x-{\hat {\mu }})^{2}}{{\hat {\sigma }}^{2}}}\right)^{-(\nu +1)/2}}$.

Marginal distribution ${\displaystyle f(x\mid \mu ,\alpha ,\beta )}$ follows t-distribution with ${\displaystyle 2\alpha }$ degrees of freedom

${\displaystyle f(x\mid \mu ,\alpha ,\beta )=t(x|\nu =2\alpha ,{\hat {\mu }}=\mu ,{\hat {\sigma }}^{2}=\beta /\alpha )}$.

In the multivariate case, the marginal distribution of ${\displaystyle \mathbf {x} }$ is a multivariate t distribution:

${\displaystyle \mathbf {x} \sim t_{2\alpha }({\boldsymbol {\mu }},{\frac {\beta }{\alpha }}\mathbf {V} )\!}$

Suppose

${\displaystyle (x,\sigma ^{2})\sim {\text{N-}}\Gamma ^{-1}(\mu ,\lambda ,\alpha ,\beta )\!.}$

Then for ${\displaystyle c>0}$,

${\displaystyle (cx,c\sigma ^{2})\sim {\text{N-}}\Gamma ^{-1}(c\mu ,\lambda /c,\alpha ,c\beta )\!.}$

Proof: To prove this let ${\displaystyle (x,\sigma ^{2})\sim {\text{N-}}\Gamma ^{-1}(\mu ,\lambda ,\alpha ,\beta )}$ and fix ${\displaystyle c>0}$. Defining ${\displaystyle Y=(Y_{1},Y_{2})=(cx,c\sigma ^{2})}$, observe that the PDF of the random variable ${\displaystyle Y}$ evaluated at ${\displaystyle (y_{1},y_{2})}$ is given by ${\displaystyle 1/c^{2}}$ times the PDF of a ${\displaystyle {\text{N-}}\Gamma ^{-1}(\mu ,\lambda ,\alpha ,\beta )}$ random variable evaluated at ${\displaystyle (y_{1}/c,y_{2}/c)}$. Hence the PDF of ${\displaystyle Y}$ evaluated at ${\displaystyle (y_{1},y_{2})}$ is given by :${\displaystyle f_{Y}(y_{1},y_{2})={\frac {1}{c^{2}}}{\frac {\sqrt {\lambda }}{\sqrt {2\pi y_{2}/c}}}\,{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{y_{2}/c}}\right)^{\alpha +1}\exp \left(-{\frac {2\beta +\lambda (y_{1}/c-\mu )^{2}}{2y_{2}/c}}\right)={\frac {\sqrt {\lambda /c}}{\sqrt {2\pi y_{2}}}}\,{\frac {(c\beta )^{\alpha }}{\Gamma (\alpha )}}\,\left({\frac {1}{y_{2}}}\right)^{\alpha +1}\exp \left(-{\frac {2c\beta +(\lambda /c)\,(y_{1}-c\mu )^{2}}{2y_{2}}}\right).\!}$

The right hand expression is the PDF for a ${\displaystyle {\text{N-}}\Gamma ^{-1}(c\mu ,\lambda /c,\alpha ,c\beta )}$ random variable evaluated at ${\displaystyle (y_{1},y_{2})}$, which completes the proof.

Normal-inverse-gamma distributions form an exponential family with natural parameters ${\displaystyle \textstyle \theta _{1}={\frac {-\lambda }{2}}}$, ${\displaystyle \textstyle \theta _{2}=\lambda \mu }$, ${\displaystyle \textstyle \theta _{3}=\alpha }$, and ${\displaystyle \textstyle \theta _{4}=-\beta +{\frac {-\lambda \mu ^{2}}{2}}}$ and sufficient statistics ${\displaystyle \textstyle T_{1}={\frac {x^{2}}{\sigma ^{2}}}}$, ${\displaystyle \textstyle T_{2}={\frac {x}{\sigma ^{2}}}}$, ${\displaystyle \textstyle T_{3}=\log {\big (}{\frac {1}{\sigma ^{2}}}{\big )}}$, and ${\displaystyle \textstyle T_{4}={\frac {1}{\sigma ^{2}}}}$.

Measures difference between two distributions.