\[\begin{align} \renewcommand{\Pr}{\mathbb{P}} \newcommand{\Bern}{\text{Bern}} \newcommand{\Unif}{\text{Unif}} \newcommand{\Bet}{\text{Beta}} \newcommand{\Bin}{\text{Bin}} \newcommand{\Categorical}{\text{Categorical}} \newcommand{\Gamm}{\text{Gamma}} \newcommand\prior{\text{prior}} \newcommand\like{\text{likelihood}} \newcommand{\E}{\mathbf{E}} \newcommand{\var}{\text{Var}} \newcommand{\sd}{\text{SD}} \newcommand{\R}{\mathbf{R}} \newcommand{\1}{{\bf 1}} \newcommand{\argmin}{\textrm{argmin}} \newcommand{\argmax}{\textrm{argmax}} \newcommand{\Ascr}{\mathcal{A}} \newcommand{\ud}{\text{d}} \newcommand\deq{\stackrel{\scriptscriptstyle d}{=}} \newcommand{\Xscr}{\mathscr{X}} \newcommand{\Zscr}{\mathscr{Z}} \newcommand{\randindex}{\textrm{rand}} \newcommand{\rhot}{\rho} \newcommand{\rhop}{{\rho'}} \end{align}\]

Tutorial on the Bayes estimator

In this tutorial, we will go over a simple example of how to apply a Bayes estimator in practice. We will use a dataset of rocket launches to illustrate the concepts.

Application context

You are consulting for a satellite operator company. The company is about to about to send a $50M satellite on a Delta 7925H rocket. Data: as of November 2020 has been launched 3 times, with 0 failed launches. Should you recommend buying a $1M insurance policy?

Maximum likelihood would therefore give an estimate of 0% for the chance of failure, and hence that no insurance should be purchased. Obviously we will not use this crazy estimate today and instead use Bayesian methods.

Let us use the simplest Bayesian model for this task (in a later tutorial, we will improve this model): we assume the four launches are independent, biased coin flips, all with a shared probability of failure (bias) given by an unknown parameter failureProbability. In other words, we assume a binomial likelihood. We also assume a beta prior on the failureProbability, the parameter $p$ of the binomial (the parameter $n$ is known and set to the number of observed launches, numberOfLaunches).

Let us walk over the step involved into solving this problem using the Bayes estimator and the Blang Probabilistic Programming Language.

Creating a joint distribution (step 2)

• The laws { ... } block specifies a joint distribution ("law") over a collection of random variables, namely those marked with the keyword random when they are declared above. Each line starting in the form variable ~ ... declares a distribution or conditional distribution.

• We haven't introduced observation yet (we will do this when conditioning, step 2), so for now we treat all variable as unobserved (marked using ?: latentReal and ?: latentInt).

• Click on "See output" above to see the main plots created by running the code. You can find in particular a visualization of the marginal distribution on the random variable failureProbability by following the tab "Plots", then "posteriorPlot" located in the top right of the page linked by "See output". Right now, we are not conditioning on any of the random variables so the "posterior" reduces to the joint density.

Conditioning (step 2)

We will now introduce the actual dataset (3 successes, 0 failures).

Note that we haven't changed the description of the joint distribution (in "laws"). Rather, we changed ?: latentInt into ?: 0. The effect of this is that running the code will now approximate the posterior distribution of the variable that are unknown (latent), given those that now have a known value. (launch1, launch2, launch3)

Look at the marginal distribution of nextLaunch. It is now a posterior marginal. Notice that it has been updated

Bayes estimator (step 3)

In the following, we let $x = (x_1, x_2, x_3)$ denote the observed data, and $z = (p, x_4)$, the unobserved data.

To make a decision, we need to encode the following into a loss function (all numbers in \$M):

• With insurance $(a=1)$: loss of 0 if success, loss of 50 if failure.

• Without $(a=0)$: loss of 1 if success, loss of 1 if failure.

We can encode this as $L(a, z) = \1[a = 1] (\$1M) + \1[a = 0] \1[x_4 = 1] (\$50M)$

The Bayes estimator tells us we should use

\[\begin{align} \newcommand\testddd{OK} \argmin \{ \E[L(a, Z) | X] : a \in \mathcal{A} \} \end{align}\]

This can be simplified into:

\[\begin{align} \argmin \{ \E[L(a, Z) | X] : a \in \Ascr \} &= \argmin \{ \E[\1[a = 1] + 50 \1[a = 0] \1[X_4 = 1] | X] : a \in \Ascr \} \\ &= \argmin \{ \E[\1[a = 1] | X] + 50 \E[ \1[a = 0] \1[X_4 = 1] | X] : a \in \Ascr \} \\ &= \argmin \{ \1[a = 1] + 50 \1[a = 0] \Pr[X_4 = 1 | X] : a \in \Ascr \} \\ &= \1[p_\text{fail}(X) > 1/50]], \end{align}\]