How is it possible to calculate the variance $\sigma^2$ for the Normal distribution if only $\alpha$ and $\beta$ (based on data) from the Inverse-gamma distribution are available? I followed the definitions on the Wiki and found out that for a Normal distribution in case of unknown variance and known mean the cojugate distribution is the Inverse-gamma with the updates $\alpha = \alpha + \frac{1}{2}$ and $\beta = \beta + \frac{(x - \mu)^2}{2}$.

I tried $\sigma^2 = \frac{\beta^2}{(\alpha - 1)^2 (\alpha - 2)} \text{for } \alpha > 2$ but $\alpha$ becomes less than $2$ after an update, and the variance returns NaN.

Thanks for any suggestions.

  • $\begingroup$ What do you mean by "only"? You do not have data? $\endgroup$
    – Tim
    Nov 8, 2016 at 11:40
  • $\begingroup$ Edited the question and added the fact that data is used. $\endgroup$
    – NumbThumb
    Nov 8, 2016 at 12:00
  • $\begingroup$ But do you want to estimate variance of the distribution of $\sigma^2$, or do you want to estimate $\sigma^2$ itself? $\endgroup$
    – Tim
    Nov 8, 2016 at 12:14
  • $\begingroup$ I want to estimate $\sigma^2$ in order to calculate the Normal distribution using Conjugate families. In case of Gaussian with known mean and unknown variance the conjugate resembles the Inverse-gamma distribution with the already mentioned update rules (see original post). $\endgroup$
    – NumbThumb
    Nov 8, 2016 at 12:21

1 Answer 1


When estimating the variance of normal distribution with known mean $\mu$ using conjugate inverse gamma prior, the model is

$$ X \sim \mathrm{Normal}(\mu, \sigma^2) \\ \sigma^2 \sim \mathrm{IG}(\alpha, \beta) $$

So we assume an inverse gamma prior parametrized by $\alpha$ and $\beta$ for $\sigma^2$. When we observe some data $x_1,\dots,x_n$ we can update our prior, to get posterior distribution of $\sigma^2$ and since inverse gamma is a conjugate prior, the update has a simple closed-form solution. The updated parameters become

$$ \alpha' = \alpha + \frac{n}{2}, \qquad \beta' = \frac{\sum_{i=1}^n (x_i - \mu)^2}{2} $$

and the posterior distribution of $\sigma^2$ is inverse gamma parametrized by $\alpha'$ and $\beta'$. So by updating the $\alpha$ and $\beta$ parameters you already estimated the distribution of $\sigma^2$. If you want to obtain a point estimate of $\sigma^2$, then you can calculate mean $\tfrac{\beta'}{\alpha'-1}$ or mode $\tfrac{\beta'}{\alpha'+1}$ of it's posterior distribution. What you were trying to do, is you were trying to calculate the variance of $\sigma^2$, that is certainly not it's point estimate.

Check also Bayesian updating with conjugate priors using the closed form expressions and Bayesian updating with new data for similar questions.

  • $\begingroup$ Thanks, this sounds plausible. Will have a more detailed look in a couple of minutes and accept your answer afterwards. Maybe I will come up with some questions if I work through the similar questions. $\endgroup$
    – NumbThumb
    Nov 8, 2016 at 12:59
  • $\begingroup$ The mean or mode calculation of the point estimate of $\sigma^2$ both take the form $\frac{\beta'}{\alpha'-1}$, is this correct or just a typo? $\endgroup$
    – NumbThumb
    Nov 8, 2016 at 15:00
  • $\begingroup$ @NumbThumb they differ by using + or - sign see en.wikipedia.org/wiki/Inverse-gamma_distribution $\endgroup$
    – Tim
    Dec 20, 2016 at 12:38

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