I am trying to derive the conjugate prior of the univariate Gaussian distribution over both the mean and the precision. I know that the prior I'm looking for is the normal-gamma distribution, but the idea is to derive this result.

It seems that you can write down the conjugate prior of an exponential family distribution immediately if you represent it in the canonical exponential family form. This link suggests a way to do just that. So far I have written the Gaussian probability function in this form and equated the following, but I don't see how the form of the conjugate prior is immediately obvious.

$$p(\mu, \lambda | D) \propto p(\mu,\lambda)p(D | \mu, \lambda)$$

  • $\begingroup$ @Xi'an, the OP seems to be asking how to derive the prior, but I don't see any derivations at the linked thread. Am I missing something? $\endgroup$ Commented Jan 29, 2015 at 20:49
  • $\begingroup$ @gung, given that the definition on Wikipedia is both generic and straightforward, it seemed to me that this was a duplicate of many earlier questions on how to apply the general formula in the canonical form to this and to that distribution. I acknowledge I did not check this specific link in details. $\endgroup$
    – Xi'an
    Commented Jan 29, 2015 at 20:58
  • $\begingroup$ Given @Xi'an's objections, can you clarify what you are stuck on in the Wikipedia presentation, Mohamed? $\endgroup$ Commented Jan 29, 2015 at 21:03
  • $\begingroup$ @Xi'an, the Wikipedia definition is too generic. Also, they seem to only state the form of the conjugate prior rather than deriving it. I don't see how it is so obvious as to not require justification. $\endgroup$
    – Mohamed
    Commented Feb 7, 2015 at 17:29
  • $\begingroup$ A (full) derivation of the conjugate prior for this specific case is what i would ideally like to figure out given some clarity on the definition. I've also found quite a few contradictions as to whether the conjugate prior should be of the same form as the posterior or the likelihood. $\endgroup$
    – Mohamed
    Commented Feb 7, 2015 at 17:32

1 Answer 1


The intuitive approach to conjugate priors is to try to deduce a family of distributions from the likelihood function. In the normal case, the likelihood is \begin{align*}\ell(\mu,\Sigma|x_1,\ldots,x_n)&\propto|\Sigma|^{-n/2}\,\exp\left\{ -\frac{1}{2}\sum_{i=1}^n (x_i-\mu)^\text{T}\Sigma^{-1}(x_i-\mu)\right\}\\ &\propto|\Sigma|^{-n/2}\,\exp\left\{-\frac{1}{2}\sum_{i=1}^n (x_i-\bar{x})^\text{T}\Sigma^{-1}(x_i-\bar{x})\right\}\\&\qquad \times \exp\left\{-\frac{n}{2}(\bar{x}-\mu)^\text{T}\Sigma^{-1}(\bar{x}-\mu)\right\}\\ &\propto |\Sigma|^{-n/2}\,\exp\left\{-\frac{1}{2}\text{tr}\left(\Sigma^{-1}S_n \right)-\frac{n}{2}(\bar{x}-\mu)^\text{T}\Sigma^{-1}(\bar{x}-\mu)\right\}\\ \end{align*}where $$S_n=\sum_{i=1}^n (x_i-\bar{x})(x_i-\bar{x})^\text{T}$$So we have three items in this likelihood:

  1. a power of $|\Sigma|$;
  2. an exponential of a trace of $\Sigma^{-1}$ times another matrix;
  3. an exponential of a quadratic in $\mu$ with matrix $\Sigma^{-1}$.

And all three terms are stable by multiplication, i.e.

  1. $|\Sigma|^a\times |\Sigma|^b = |\Sigma|^{a+b}$;
  2. $\exp\left\{-\text{tr}\left(\Sigma^{-1}A\right)\right\}\times\exp\left\{-\text{tr}\left(\Sigma^{-1}B\right)\right\}=\exp\left\{-\text{tr}\left(\Sigma^{-1}[A+B]\right)\right\}$;
  3. $\exp\left\{-(a-\mu)^\text{T}\alpha\Sigma^{-1}(a-\mu)\right\}\times\exp\left\{-(b-\mu)^\text{T}\beta\Sigma^{-1}(b-\mu)\right\}$ remains an exponential of a quadratic in $\mu$ with matrix $\Sigma^{-1}$ (with an extra term of the form $\exp\left\{-\text{tr}\left(\Sigma^{-1}A\right)\right\}$ as this is not a perfect quadratic term).

This means that the likelihood induces a shape of prior that remains stable by multiplication with another term with this shape. Which is a way of defining conjugacy. So, if I take my prior to be $$\pi(\mu,\Sigma)\propto|\Sigma|^{-\gamma/2}\,\exp\left\{-\frac{1}{2}\text{tr}\left(\Sigma^{-1}\Xi \right)-\frac{\nu}{2}(\xi-\mu)^\text{T}\Sigma^{-1}(\xi-\mu)\right\}$$the posterior will look the same, except that $\gamma,\Xi,\nu,\xi$ will change.


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