I have one two data sets of scalar values: one large data set (about 700 data points) and one small data set (80 data points). I would like to update the large data set with the small one using the Bayes’ theorem, and so create another large data set (posterior).

The large data set serves as prior, and it is assumed to be normally distributed and so the posterior. This was motivated by the existence of the closed-form expression of the posterior distribution parameters https://en.wikipedia.org/wiki/Conjugate_prior (the first row in the table for Continuous distributions) for the conjugate prior.

However, if I substitute into the closed-form expressions for posterior mean and variance, using the mean and variance values of prior (inferred from the large data set) and local data (inferred from the small data), the resulting posterior distribution does not make sense.

Do I misunderstand that I can simply substitute into these closed-form expressions the known values in order to get the posterior distribution?

  • $\begingroup$ Could you give an example of what you are actually doing..? What do you substitute? $\endgroup$
    – Tim
    Commented Sep 1, 2016 at 10:11
  • $\begingroup$ From the large set I observed mean=20.1, std=7.9 and from the small set: mean=20.6, std=9.5. I've simply taken these observed values, considering the large set as my prior, and substitute them into the formulas for the posterior's mean and std. $\endgroup$
    – Vasek
    Commented Sep 1, 2016 at 10:20
  • $\begingroup$ What exactly and how do you substitute? And why "doesn't it make sens"? $\endgroup$
    – Tim
    Commented Sep 1, 2016 at 10:29
  • $\begingroup$ Using the same notations as in en.wikipedia.org/wiki/Conjugate_prior, I substitute mu0=20.1, sigma0=7.9 and mu=20.6, sigma=9.5, n=70 (n is the size of the small sample), sum(xi)= 1448.7 into the formulations for the posterior parameters (the first row in the table Continuous distributions) and receive the posterior parameters mu=20.7 and sigma=1.2. Now, mu seems to follow some expected value, however, sigma is far off comparing to the values from the data sets. $\endgroup$
    – Vasek
    Commented Sep 1, 2016 at 11:19

1 Answer 1


First of all, the formulas are defined in terms of variance, not standard deviations.

Second, the variance of the posterior is not a variance of your data but variance of estimated parameter $\mu$. As you can see from the description ("Normal with known variance $\sigma^2$"), this is formula for estimating $\mu$ when $\sigma^2$ is known. The prior parameters $\mu_0$ and $\sigma_0^2$ are parameters of distribution of $\mu$, hence the assumed model is

$$ \begin{align} X_i &\sim \mathrm{Normal}(\mu, \sigma^2) \\ \mu &\sim \mathrm{Normal}(\mu_0, \sigma_0^2) \end{align} $$

When both $\mu$ and $\sigma^2$ are unknown and are to be estimated, then you need slightly more complicated model (in Wikipedia table under "$\mu$ and $\sigma^2$ Assuming exchangeability"):

$$ \begin{align} X_i &\sim \mathrm{Normal}(\mu, \sigma^2) \\ \mu &\sim \mathrm{Normal}(\mu_0, \tfrac{\sigma^2}{n+\nu}) \\ \sigma^2 &\sim \mathrm{IG}(\alpha, \beta) \end{align} $$

where first we need to update parameters of inverse gamma distribution to obtain $\sigma^2$:

$$ \begin{align} \alpha' &= \alpha + \frac{n}{2} \\ \beta' &= \beta + \frac{1}{2}\sum_{i=1}^n (x_i -\bar x)^2 + \frac{n\nu(\bar x -\mu_0)^2}{2(n+\nu)} \end{align} $$

and then we can proceed to calculate $\mu$ and MAP point estimate for $\sigma^2$:

$$ \begin{align} \mu &= \frac{ \mu_0\nu + \bar x n }{\nu + n} \\ \operatorname{Mode}(\sigma^2) &= \frac{ \beta' }{ \alpha' + 1 } \end{align} $$

For learning more, refer to "Conjugate Bayesian analysis of the Gaussian distribution" paper by Kevin Murphy, or "The Conjugate Prior for the Normal Distribution" notes by Michael Jordan (notice that there are slight differences between those two sources and that some formulas are given for precision $\tau$ rather then variance) and M. DeGroot Optimal Statistical Decisions, McGraw-Hill, 1970 (pp. 169-171).

  • $\begingroup$ Thank you. This is useful!! My follow-up question is, how should one decide on σ2 value? In my application where I'm "updating" the large data set with the properties of the small data set, the posterior (resulting) variance is not known. $\endgroup$
    – Vasek
    Commented Sep 1, 2016 at 14:59
  • 1
    $\begingroup$ @Vasek If it is not known then you should not use this approach... Basically, this is just a simple "textbook" example of usage of conjugate priors and most real-life application would require more complicated models (e.g. estimated using MAP or MCMC). $\endgroup$
    – Tim
    Commented Sep 1, 2016 at 15:02
  • $\begingroup$ Thanks for explaining. I'll then have to find some "textbook" (meaning something practical) explanation on how to proceed such more complicated approach for the application in interest. It's sort surprising conclusion; I'm basically replicating the methodology of some authors who (in multiple papers) used this conjugate priors closed-form expressions, explicitly mentioning that this way they might overcome the need of employing MCMC samplers. $\endgroup$
    – Vasek
    Commented Sep 1, 2016 at 15:17
  • $\begingroup$ Well, here we want to "simulate" an outcome (posterior). Apriori, we do not know any of its parameters such that the closed form expressions could be operationalized (as you explained previously). Unless one assumes some asymptotical cases when one of the posterior parameters is exactly equal to the related parameter in the observed data sets (and the remaining parameter could be then computed). So it's not really obvious how they come up with the results without using some samplers. But this is perhaps a question to the authors of methodology that I mentioned earlier. $\endgroup$
    – Vasek
    Commented Sep 1, 2016 at 16:18
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Vasek
    Commented Sep 2, 2016 at 6:59

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