Expectation si the main tool to translate a posterior distribution into the various outputs of Bayesian inference (point estimate, credible intervals, prediction, action).
Expectation of a single random variable
Recall: \[\mathbb{E}[X] = \sum_x x p_X(x),\] where the sum is over the point masses of \(X\), i.e. \(\{x : p_X(x) > 0\}\).
Test yourself: compute \(\mathbb{E}[X]\) if \(X \sim {\mathrm{Bern}}(p)\), with \(p = 0.8\).
Law of the Unconscious Statistician
Proposition: if \(g\) is some function, \[\mathbb{E}[g(X)] = \sum_x g(x) p_X(x).\]
Test yourself: compute \(\mathbb{E}[X^2]\) if \(X \sim {\mathrm{Bern}}(p)\), and hence \(\operatorname{Var}[X] = \mathbb{E}[X^2] - (\mathbb{E}[X])^2\).
Expectation of a function of several random variables
Let us go back to our running example:
Imagine a bag with 3 coins each with a different probability parameter \(p\)
Coin \(i\in \{0, 1, 2\}\) has bias \(i/2\)—in other words:
First coin: bias is \(0/2 = 0\) (i.e. both sides are “heads”, \(p = 0\))
Second coin: bias is \(1/2 = 0.5\) (i.e. standard coin, \(p = 1/2\))
Third coin: bias is \(2/2 = 1\) (i.e. both sides are “tails”, \(p = 1\))
Consider the following two steps sampling process
Step 1: pick one of the three coins, but do not look at it!
Step 2: flip the coin 4 times
Mathematically, this probability model can be written as follows: \[
\begin{align*}
X &\sim {\mathrm{Unif}}\{0, 1, 2\} \\
Y_i | X &\sim {\mathrm{Bern}}(X/2)
\end{align*}
\tag{1}\]
Example: computing \(\mathbb{E}[X (Y_1+1)]\) (similar to what you will be doing in the exercise in question 1.1)
Note: this is of the form \(\mathbb{E}[g(\dots)]\), so we can use the Law of the Unconscious Statistician.
How to do it:
first, identify \(g\), here it is \(g(x, y_1, \dots, y_4) = x(y_1+1)\) (in the exercise it is slightly different)
denote by \(p\) the joint PMF of all the random variables in the model