This question deals with Bayesian updating with conjugate prior.Suppose we have a prior distribution of N~(5, 3) and then we observe 5 data points (8, 9, 10, 8, 7) (assumed to be taken randomly from a N~(9, 3) distribution). What would be the posterior after these observations in the form of N~(x, y)? I read the Wikipedia article on conjugate priors, but I want to have a more precise understanding of how to solve this specific problem. If there is no way to solve it without assuming some things, can you please explain what needs to be known and solve under very simple assumptions? Thank you in advance.

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    $\begingroup$ Where is the likelihood? and what are the parameters? $\endgroup$
    – niandra82
    Commented Sep 27, 2016 at 9:34
  • $\begingroup$ If the information is all that is given to work with, is there any way to approximate the posterior? $\endgroup$ Commented Sep 27, 2016 at 10:44
  • $\begingroup$ @Kelly if you know that the data comes from N(9, 3) distribution, then there is no need to estimate anything since you know the parameters in advance (9, 3). What niandra82 was saying is that you need to specify priors and likelihood, what you did is you only specified prior for mean. $\endgroup$
    – Tim
    Commented Sep 27, 2016 at 11:01
  • $\begingroup$ When you say N~(a,b) presumably that indicates a normal distribution with mean a ... but is b the variance or the standard deviation in this notation? $\endgroup$
    – Glen_b
    Commented Sep 27, 2016 at 11:15
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    $\begingroup$ @Kelly if you do not know the distribution of your data (i.e. you are not willing to assume it), then you cannot use classical Bayesian inference since for this you would need to assume a likelihood function (i.e. distribution) for your data. If you do not assume distribution for your data then what parameters you want to estimate? N(a,b) is prior for what? $\endgroup$
    – Tim
    Commented Sep 27, 2016 at 11:25

1 Answer 1


I am afraid that you are misunderstanding what Bayesian inference is about in general. Bayes theorem is

$$ \underbrace{p(\theta \mid X)}_\text{posterior} = \frac{\overbrace{p(X \mid \theta)}^\text{likelihood} \, \overbrace{p(\theta)}^\text{prior}}{\underbrace{p(X)}_\text{normalizing constant}} $$

To make it more concrete, you can estimate $\mu$ and $\sigma^2$ parameters from normal distribution (i.e. normal likelihood) using data $X$, assuming normal prior for $\mu$ with hyperparameters $\mu_0$ and $\sigma^2_0$, and uniform prior for $\sigma^2$ with hyperparameters $a$ and $b$, to obtain posterior distributions for $\mu$ and $\sigma^2$:

$$ X \sim \mathrm{Normal}(\mu, \sigma^2) \\ \mu \sim \mathrm{Normal}(\mu_0, \sigma^2_0) \\ \sigma^2 \sim \mathrm{Uniform}(a,b) $$

So priors are assigned to parameters of interest, not to data. You also have to specify assumed distribution of your data (likelihood). Finally, posterior is distribution over estimated parameters, so you need to specify what you are actually estimating.


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