Talk:Normal-inverse-gamma distribution

The CDF does not look right, if sigma^2 to Infty, the CDF goes to zero if I am not mistaken, which does not make sense, no? DoubleMatchPoint (talk) 19:57, 8 November 2017 (UTC)

We probably ought to make the parameters match with those listed in the Conjugate prior article or vice-versa. --Rhaertel80 (talk) 16:56, 29 May 2008 (UTC)

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Notation is inconsistent with regard to the distribution of $\mu$. In the definition, mean is $\lambda$ and variance is $\sigma ^{2}/v$, but in the section about generating values from the distribution, mean is $v$ and variance is $\sigma ^{2}/\lambda$. I don't have a reference handy so I'm not sure which one is correct. Ksimek (talk) 23:40, 12 June 2009 (UTC)

User:rewtnode: I'm trying to verify the formula for the Expectation of \sigma^2 E[sigma^2] which is given now as \beta/(\alpha-1/2), but was just recently changed. Just a few days ago it was \beta/(2 (\alpha-1)). But I tried to carry out the integral myself and come to the old result. The issue is whether we assume this is a density function of (x, \sigma^2) or of (x, \sigma). If I assume it's a function of (x,\sigma) I get the old result E[\sigma^2] = \beta/(2(\alpha-1)). However if I assume that it is a function of (x,\sigma^2) I need to do a different variable substitution in the integral and get the result E[\sigma^2] = \beta^{1/2} \Gamma(\alpha-3/2)/\Gamma(\alpha). Very confusing. So I wonder if this distribution should be indeed seen as function of (x, \sigma), that is, a joint density over the domain (x, \sigma) — Preceding unsigned comment added by Rewtnode (talk • contribs) 00:40, 8 June 2018 (UTC)

Derivation of expected value of $\sigma ^{2}$

There seems to be some disagreement about whether $E[\sigma ^{2}]$ is $\beta /(\alpha -1)$ (correct) or $\beta /(\alpha -0.5)$ (incorrect). To get this right, you have to remember to integrate with respect to $\sigma ^{2}$ and not $\sigma$ since the support of the distribution is defined in terms of $x$ and $\sigma ^{2}$.

To make the above clear, all instances of $\sigma ^{2}$ below are written as $[\sigma ^{2}]$ to indicate that $\sigma ^{2}$ is our variable and not $\sigma$.

From the definition of $E[\sigma ^{2}]$:

$E[[\sigma ^{2}]]=\int _{0}^{\infty }\int _{-\infty }^{\infty }[\sigma ^{2}]\ {\text{Normal-inverse-gamma}}\left(x,[\sigma ^{2}]\ |\ \mu ,\lambda ,\alpha ,\beta \right)\ {\text{d}}x\,{\text{d}}[\sigma ^{2}]$

From the definition of the normal-inverse-gamma distribution:

$=\int _{0}^{\infty }\int _{-\infty }^{\infty }[\sigma ^{2}]{\frac {\sqrt {\lambda }}{\sqrt {2\pi [\sigma ^{2}]}}}{\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}[\sigma ^{2}]^{-(\alpha +1)}\exp \left(-{\frac {2\beta +\lambda (x-\mu )^{2}}{2[\sigma ^{2}]}}\right)\ {\text{d}}x\,{\text{d}}[\sigma ^{2}]$

Rearrange:

$={\frac {{\sqrt {\lambda }}\beta ^{\alpha }}{{\sqrt {2\pi }}\Gamma (\alpha )}}\int _{0}^{\infty }[\sigma ^{2}]{\frac {1}{\sqrt {[\sigma ^{2}]}}}[\sigma ^{2}]^{-(\alpha +1)}\exp \left(-{\frac {\beta }{[\sigma ^{2}]}}\right)\int _{-\infty }^{\infty }\exp \left(-{\frac {(x-\mu )^{2}}{2\lambda ^{-1}[\sigma ^{2}]}}\right)\ {\text{d}}x\,{\text{d}}[\sigma ^{2}]$

Integrate out $x$, which appears as a squared exponential function (proportional to the pdf of the normal distribution):

$={\frac {{\sqrt {\lambda }}\beta ^{\alpha }}{{\sqrt {2\pi }}\Gamma (\alpha )}}\int _{0}^{\infty }[\sigma ^{2}]{\frac {1}{\sqrt {[\sigma ^{2}]}}}[\sigma ^{2}]^{-(\alpha +1)}\exp \left(-{\frac {\beta }{[\sigma ^{2}]}}\right){\sqrt {2\pi \lambda ^{-1}[\sigma ^{2}]}}\ {\text{d}}[\sigma ^{2}]$

Simplify:

$={\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\int _{0}^{\infty }[\sigma ^{2}]^{-\alpha }\exp \left(-{\frac {\beta }{[\sigma ^{2}]}}\right)\ {\text{d}}[\sigma ^{2}]$

Integrate out $[\sigma ^{2}]$, which appears in the same form as the pdf of an inverse-gamma distribution with argument $[\sigma ^{2}]$:

$={\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}{\frac {\Gamma (\alpha -1)}{\beta ^{\alpha -1}}}$

$={\frac {\beta }{\alpha -1}}$

We can further confirm this result by generating samples from the normal-inverse-gamma distribution (the sampling procedure is described in the article) and estimating the expected value of the $\sigma ^{2}$ argument empirically. It converges to ${\frac {\beta }{\alpha -1}}$.

--CarlS (talk) 14:05, 27 September 2018 (UTC)

On the marginal distribution of x in the univariate case

The formula for the marginal distribution of $x$ is wrong (which can be readily observed by comparison with the result of the proof reported for the case $\lambda =1$ below) and is taken from the posterior predictive formula that one can find in the table of Conjugate prior.

I think the correct formula is $t_{2\alpha }\left(\mu ,{\frac {\beta }{\alpha \lambda }}\right)$. A derivation with a different notation for the parameters can be found in [1]https://bookdown.org/aramir21/IntroductionBayesianEconometricsGuidedTour/sec42.html#sec42:~:text=%2C%20respectively.-,The%20marginal%20posterior%20of,is,-%CF%80 Br1 Ursino (talk) 19:05, 21 September 2023 (UTC)