Stochastic Modeling

Basic definitions¶

Random events¶

Special events¶

The following operations on events are fundamental:

$A \cap B$ : $A$ and $B$
$A \cup B$ : $A$ or $B$ (non-exclusive “or”)
$B \setminus A$ : $B$ without $A$
$A^c = \Omega \setminus A$ : the complementary event “not $A$ ”
$\bigcap_{i=1}^{\infty} A_i$ : all $A_i$
$\bigcup_{i=1}^{\infty} A_i$ : at least one $A_i$

Probability measure¶

Consider a random experiment with sample space $\Omega$ . Recall that we write $\mathcal{A}$ for the set of all possible events $E$ . A probability on $\Omega$ is a function which assigns to every event $E \in \mathcal{A}$ a real number $P(E)$ and satisfies the following properties known as the Axioms of Kolmogorov:

Definition 2 (Probability Measure)

A probability measure on $\Omega$ is a function $P : \mathcal{A} \to \mathbb{R}$ satisfying the following axioms:

A1) $0 \leq P(E) \leq 1$ for all events $E \in \mathcal{A}$

A2) $P(\Omega) = 1$

A3) For every sequence of mutually disjoint events $E_i$ , $i = 1, 2, \ldots$ (i.e., $E_i \cap E_j = \emptyset$ for $i \neq j$ ), we have

P\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} P(E_i)

(2)

We refer to $P(E)$ as the probability of the event $E \in \mathcal{A}$ .

Events $E \in \mathcal{A}$ with $P(E) = 0$ are called almost impossible, events $E \in \mathcal{A}$ with $P(E) = 1$ are called almost certain.

Remarks:

(Summation rule) If in Axiom A3) we choose a finite sequence of disjoint events, say $E_1, E_2$ and $E_3$ , we can simply set $E_i = \emptyset$ for all $i \geq 4$ to obtain
$P(E_1 \dot{\cup} E_2 \dot{\cup} E_3) = P(E_1) + P(E_2) + P(E_3)$
(3)
that is, probabilities of disjoint events sum up.
If the $n$ outcomes of a random experiment are equally likely to occur, then every single outcome, considered as an event, occurs with probability $1/n$ . If event $A$ is the union of $k$ elementary outcomes, we have by Axiom A3),
$P(A) = \frac{k}{n} = \frac{\text{number of good outcomes}}{\text{total number of possible outcomes}} = \frac{|A|}{|\Omega|}$
(4)
We refer to such a $P$ as the uniform probability measure on the sample space $\Omega$ .

We gather now some simple consequences which can be derived from the axioms of a probability measure.

Example 4 (Two Fair Coins)

Suppose that we toss two fair coins, so that each of the four outcomes in the sample space

\Omega = \{(\text{heads}, \text{tails}), (\text{heads}, \text{heads}), (\text{tails}, \text{heads}), (\text{tails}, \text{tails})\}

(7)

is equally likely to occur (with probability $1/4$ ). Let

E = \{(\text{heads}, \text{heads}), (\text{heads}, \text{tails})\} \quad \text{and} \quad F = \{(\text{heads}, \text{heads}), (\text{tails}, \text{heads})\}

(8)

That is, $E$ is the event that the first coin falls heads, and $F$ is the event that the second coin falls heads. By Proposition 1 d), we have for the probability that either the first or the second coin falls heads

P(E \cup F) = P(E) + P(F) - P(E \cap F) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4}

(9)

Alternatively, since every outcome is equally likely to occur, we see directly that

P(E \cup F) = P(\{(\text{heads}, \text{heads}), (\text{heads}, \text{tails}), (\text{tails}, \text{heads})\}) = \frac{3}{4}

(10)

Example 5 (Probability Measure Construction)

The foregoing exercise follows a typical modeling schema: First, select a suitable sample space $\Omega$ , second, define $P(A)$ for all events $A$ . For example if $\Omega$ is finite and all outcomes are equally likely we usually take $P(A) = \frac{|A|}{|\Omega|}$ . If it is not the case that all outcomes are equally likely, then $P(A)$ would be given by some other formula.

The general situation is the following: Suppose $\Omega = \{1, 2, \ldots, M\}$ be a finite sample space. Next let $p(1), p(2), \ldots, p(M)$ be non-negative real numbers such that $\sum_{i=1}^M p(i) = 1$ . For any subset $A \subseteq \Omega$ , put

P(A) := \sum_{i \in A} p(i)

(11)

This is a well defined probability measure on $\Omega$ . Analogously, one can construct a probability measure on an infinite but countable sample space.

Conditional probabilities¶

Example 6 (Two Dice Conditional Probability)

We toss two fair dice. Each of the 36 possible outcomes in $\Omega$ is equally likely to occur and hence has probability $\frac{1}{36}$ . Now suppose that we observe that the first die is a four. Then given this information, what is the probability that the sum of the two dice equals six?

To calculate this probability we reason as follows: Given that the initial die is a four, it follows that there can be at most six possible outcomes of our experiment, namely,

\Omega' = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}

(12)

Since each of these outcomes have the same probability of occurring, they have equal probabilities. That is, given that the first die is a four, then the (conditional) probability of each of the outcomes in $\Omega'$ is $\frac{1}{6}$ while the (conditional) probability of the other 30 elements from the sample space $\Omega$ is 0. Hence the wanted probability is $\frac{1}{6}$ .

In general, the conditional probability is defined in the following manner:

Example 7 (Family with Two Children)

A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy?

Assume that the sample space $\Omega$ is given by $\Omega = \{(b, b), (b, g), (g, b), (g, g)\}$ , all the outcomes are equally likely. ( $(b, g)$ means for example that the older child is a boy and the younger a girl).

Solution. Letting $B$ denote the event that both children are boys, and $A$ the event that at least one of them is a boy, then the desired probability is given by

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(\{(b, b)\})}{P(\{(b, b), (b, g), (g, b)\})} = \frac{1}{3}

(14)

In the preceding display, we have used that $P(E \cap F) = P(F) P(E|F)$ . This rule is called multiplication rule and generalizes to the intersection of three (and in fact more) events:

The concept of conditional probabilities can be used to calculate probabilities in the following manner:

Let $\{A_k\}_{1 \leq k \leq n}$ be a finite partition of the sample space $\Omega$ (i.e., $\biguplus_{k=1}^n A_k = \Omega$ , $A_j \cap A_i = \emptyset$ for $i \neq j$ and $P(A_j) > 0$ for every $1 \leq j \leq n$ ).

If $E$ is an arbitrary event, we may write

E = E \cap \Omega = E \cap \left(A_1 \dot{\cup} A_2 \dot{\cup} \cdots \dot{\cup} A_n\right) = (E \cap A_1) \dot{\cup} (E \cap A_2) \dot{\cup} \cdots \dot{\cup} (E \cap A_n)

(17)

If we apply the multiplication rule $n$ times, we obtain the law of total probabilities:

Another useful formula is named after Thomas Bayes. This formula is applied when we have to quantify the likelihood of a hypothesis $A$ given an event $E$ (i.e., instead of $P(E|A)$ we are interested in $P(A|E)$ ).

A simple application of the definition of conditional probabilities followed by a formal multiplication by one yields

P(A|E) = \frac{P(A \cap E)}{P(E)} \cdot \frac{P(A)}{P(A)} = \frac{P(E|A)P(A)}{P(E)}

(21)

Now, if we partition $\Omega$ into the events $A$ and $A^c$ and calculate $P(E)$ using the law of total probabilities, we get Bayes’ rule:

Example 11 (The Monty Hall Problem)

In a game show, a contestant is told the rules as follows: There are three doors, labelled 1, 2, 3. A single prize is hidden behind one of them. You get to select one door. Initially your chosen door will not be opened. Instead, the gameshow host will open one of the two other doors, and he will do so in such a way as not to reveal the prize.

At this point, you will be given a fresh choice of door: you can either stick with your first choice, or you can switch to the other closed door. All the doors will then be opened and you will receive whatever is behind your final choice of door.

Imagine that the contestant chooses door 1 first; then the gameshow host opens door 3, revealing nothing behind the door, as promised. Should the contestant (a) stick with door 1, (b) switch to door 2, or (c) does it make no difference?

Solution. Let $H_i$ denote the hypothesis that the prize is behind door $i$ , $i = 1, 2, 3$ . We make the following assumptions: The three hypotheses $H_i$ are equiprobable a priori, i.e.:

P(H_i) = \frac{1}{3}, \quad i = 1, 2, 3

(25)

The datum we receive, after choosing door 1, is one of $D = 2$ and $D = 3$ (meaning door 2 or 3 is opened, respectively). If the prize is behind door 1 then the host has a free choice; in this case we assume that the host selects at random between $D = 2$ and $D = 3$ . Otherwise the choice of the host is forced and the probabilities are 0 and 1: $$

\begin{align} P(D = 2|H_1) &= \frac{1}{2}, & P(D = 2|H_2) &= 0, & P(D = 2|H_3) &= 1 \\ P(D = 3|H_1) &= \frac{1}{2}, & P(D = 3|H_2) &= 1, & P(D = 3|H_3) &= 0 \end{align}

(26)

Now, using Bayes’ rule, we evaluate the posterior probabilities of the hypotheses:

P(H_i|D = 3) = \frac{P(D = 3|H_i)P(H_i)}{P(D = 3)}

(27)

{#eq-monty-hall}

Hence $$

\begin{align} P(H_1|D = 3) &= \frac{(1/2)(1/3)}{1/2} = \frac{1}{3} \\ P(H_2|D = 3) &= \frac{1 \cdot (1/3)}{1/2} = \frac{2}{3} \\ P(H_3|D = 3) &= \frac{0 \cdot (1/3)}{1/2} = 0 \end{align}

(28)

So the contestant should switch to door 2 in order to have the biggest chance of getting the prize.

Bayesian Interpretation

Bayes’ formula allows for a fruitful interpretation. Suppose that $D = \{d_1, \ldots, d_n\}$ is observed data (e.g., measurements) and $A$ a specific set of parameters (e.g., measurement conditions). Then Bayes’ rule

P(A|D) = \frac{P(D|A)}{P(D)} P(A)

(29)

tells us how probable the parameter set $A$ is in view of the measurements $D$ : One has to multiply the likelihood $P(D|A)$ of the observed data $D$ given the parameter set $A$ by the a priori probability $P(A)$ and normalize the expression by $P(D)$ . The likelihood $P(D|A)$ can be viewed as a function of $A$ . It expresses how probable the observed data is for different parameter sets $A$ .

In view of this interpretation, Bayes’ rule may be stated in words as follows:

\text{posterior probability} \propto \text{likelihood} \times \text{a priori probability}

(30)

where all of these terms are viewed as functions of $A$ .