posterior Gaussian distribution I have quite a newbie doubt about Bayesian inference. Let's say that my prior data is composed by a Gaussian distribution (mean1, standard deviation1).
My likelihood is another Gaussian with mean2, std deviation2.
Now, my question is how can I get the posterior, please? The most clear tutorial I found is this, in eq(28), but I do not have clear whether the author is applying Monte Carlo there or not.
 A: Finding the posterior distribution does not involve Monte Carlo, but involves the Bayes rule. If the posterior turns out to be complicated (usually meaning that it is not a known distribution), then Monte Carlo methods are used to sample from the posterior. Thus, the first step is to always try and write down the posterior.
Notationally, your likelihood is $Y_i|\mu_1 \sim N(\mu_2, \sigma_2^2)$ assuming $\sigma_2^2 > 0$ is known. Let the prior on $\mu_2$ be $N(\mu_1, \sigma_1^2)$ where $\mu_1 \in \mathbb{R}, \sigma_1^2 > 0$ are fixed.
The posterior distribution is found by the Bayes Rule.
$$f(\mu_2|y) = \dfrac{f(\mu_2) \prod_{i=1}^{n}f(y_i|\mu_2)}{\prod_{i=1}^{n}f(y_i|\mu_2))}. $$
Since we condition on $y$, $f(y_i)$ is a constant.
\begin{align*}
f(\mu_2|y) & \propto f(y|\mu_2)f(\mu_2)\\
&= \prod_{i=1}^{n} \left(\frac{1}{\sqrt{2\pi \sigma_2^2}} \exp \left\{-\dfrac{(y_i - \mu_2)^2}{2\sigma_2^2} \right\} \right) \frac{1}{\sqrt{2\pi \sigma_1^2}} \exp \left\{-\dfrac{(\mu_2 - \mu_1)^2}{2\sigma_1^2} \right\}\\
& \vdots
\end{align*}
If you keep solving this like the link you shared, you will see that this takes the form of a Normal distribution in $\mu_2$ with the parameters as indicated in the link. You can find more details in the following links
https://www.youtube.com/watch?v=c-d05z0_5mw
http://www.people.fas.harvard.edu/~plam/teaching/methods/conjugacy/conjugacy_print.pdf
