I am new to Bayesian statistics. I am trying to understand a certain passage in my course notes.
Excerpt:
Discussion:
I don't understand much about the above excerpt. I think that the goal here is to compute something along the lines of $\pi(\theta | x) = \frac{f(x | \theta) \pi(\theta)}{f(x)}$, where $f(x) = \int f(x | \theta) \pi(\theta)d\theta$. We are trying to turn our model for the data and the parameter into a posterior distribution for the parameter given the data.
I do not see what the trick is. What I see is that, for example, if $\theta | x \sim \text{Beta} (\alpha, \beta)$, then $\pi (\theta | x) = \frac{\theta^{\alpha - 1}(1 - \theta)^{\beta - 1}}{\text{B}(\alpha, \beta)}$, where the denominator is a constant (I think) so it follows that $\pi(\theta|x) \propto \theta^{\alpha - 1}(1 - \theta)^{\beta - 1}$.
I believe the same procedure works for the other two examples, in which you set up the equation for the known form of the density, strip out the constants, replace $=$ with $\propto$, and Bob's your uncle.
Or maybe the point is that if $\pi(\theta | x) = \frac{f(x | \theta) \pi(\theta)}{f(x)}$ where the denominator is a nasty integral, then it's simpler to write $\pi(\theta | x) \propto f(x | \theta) \pi(\theta)$. So maybe the upshot is that we perform the multiplication $f(x | \theta) \pi(\theta)$, see if it similar to a known distribution, and this gives us information about the type of distribution the posterior has.
I think the subsequent example in the notes seems consistent with my expectations. (Sorry, these notes seem sloppily written.)
I appreciate any help.