So $X_1,...,X_n$ forms a random sample from a normal dist with unknown mean $µ$ and variance $σ^2$. So we have the prior distribution with mean $0$ and variance $σ^2$. We then need to show that if $n$ is large then the posterior distribution of $µ$ given that $X_i=x_i(i=1,…,n)$ will be approximately a normal distribution with mean $\bar{x_n}$ and variance $\sigma^2/n$.

I've found solutions from various places, but I don't understand some of them. I mean, from bayes' theorem, the posterior is proportional to product of prior and likelihood function, because the denominator in the formula just goes to constant with respect to mu, so we just look at the proportionality. Also, once you get the product, the constant at the front isn't important and also drops off.

I found similar questions and solutions, but I'm not sure how to apply them here: http://lausanne.isb-sib.ch/~darlene/gda/add/BayesConjugateNormal.pdf Bayes Estimator of normal distribution and normal prior

I wasn't sure why the $\tau$ notation for variance was introduced, and how to express the argument that what we get at the end is our $\bar{x_n}$ and variance $\sigma^2/n$.

  • $\begingroup$ Your prior should give a (possibly improper) distribution for $\mu$ and a distribution for $\sigma^2$ (or a joint distribution for the two of them). So you should have two 1-D prior distributions or one 2-D prior distribution, but you do not. $\endgroup$ – Henry Jun 2 '13 at 15:20

I assume your $\sigma^2$ is a given constant. You have the model,

$$ \begin{aligned} X_1, \cdots, X_n | \mu &\sim Normal(\mu, \sigma^2) \\ \mu &\sim Normal(0, \sigma^2) \\ \therefore \ \ \mu | X_1, \cdots, X_n &\sim \prod_{i=1}^n \frac{1}{\sqrt{2\pi}} \exp \left( - \frac{1}{2\sigma^2} (x_i - \mu)^2 \right) \times \frac{1}{\sqrt{2\pi}} \exp (-\frac{1}{2\sigma^2} \mu^2) \\ &\propto \exp \left(-\frac{n+1}{2\sigma^2}(\mu-\frac{\bar{x}}{n+1})^2 \right) \\ &\sim Normal(\frac{\bar{x}}{n+1}, \sigma^2/(n+1)) \end{aligned} $$


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