Conditional probabilities and so on

Exercise 90 (Modular Equation)

Let $A, B$ be two events. Prove:

$${\mathbb P}[A \cup B] = {\mathbb P}[A] + {\mathbb P}[B] - {\mathbb P}[A\cap B]$$

Definition 6

Two events $A$ and $B$ are disjoint if $A \cap B = \emptyset$, and they are called almost disjoint if ${\mathbb P}[A\cap B] =0$.

Exercise 91

Define a probability space in which there exist two events $A,B$ that are not disjoint but almost disjoint.

Definition 7 (Conditional probability)

If $A$ and $B$ are events and ${\mathbb P}[B]>0$, we can write

$${\mathbb P}[A \mid B] = \frac{​{\mathbb P}[A\cap B]}{​{\mathbb P}[B]}$$

(this reads "probability of $A$ given $B$".)

Conditional probability gives us a way to focus on a smaller "world" (the world in which event $B$ happens). This is formalized in the following exercise.

Exercise 92

Let $(\Omega,{\mathbb P})$ be a finite probability space and ${\mathbb P}(B) >0$. Define ${\mathbb P}_B(\omega) = {\mathbb P}[\omega \mid B]$ for all $\omega \in \Omega$. Prove that $(B, {\mathbb P}_B)$ is a finite probability space.

If we rewrite ${\mathbb P}[A \cap B] = {\mathbb P}[A\mid B] \cdot {\mathbb P}[B]$, we can think of this intuitively as follows. The event that both $A$ and $B$ happen can be viewed "sequentially": First, $B$ happens with probability ${\mathbb P}[B]$, and afterwards, in the world where $B$ already happens, $A$ would happen with probability ${\mathbb P}[A \mid B]$.

Observation 2 (Bayes' equation)

${\mathbb P}[B \mid A] = {\mathbb P}[A \mid B] \cdot {\mathbb P}[B]/{\mathbb P}[A]$.

Exercise 93

Prove that ${\mathbb P}[A \cap B \cap C] = {\mathbb P}[A \mid B \cap C] \cdot {\mathbb P}[B\mid C] \cdot {\mathbb P}[C]$.

Example 35

We toss three unbiased coins. Given that the second coin is head, what is the probability that the total number of heads is $2$?

Proof:

Let us try to first derive the answers using only the definitions. Let $B$ be the event that the second coin is head, so we have $B= \{010,011,110,111\}$. Let $A$ be the event that there are two heads, so $A = \{110,011,101\}$. The set $A \cap B=\{110, 011\}$, so ${\mathbb P}[A \mid B] = 2/4 = 1/2$. Note that this is not the way a human being thinks intuitively.

Those who have done this kind of exercise before would say things like, once we fix the second coin being head, we can focus on the outcome of the first and third coins. The probability that one head occurs out of these two coins is just $(1/2)(1/2) + (1/2)(1/2) = 1/2$.

This intuitive process corresponds to shifting the probability space to $(B, {\mathbb P}_B)$ (as in Exercise 92) where each atomic event in $B$ occurs with probability $1/4$. In this space, we have that ${\mathbb P}_B[\{110,001\}] = 1/4+1/4 = 1/2$.

Working with probability is very prone to error, and therefore we should be aware what our intuitive calculation corresponds to in formal terms.

A partition of $\Omega$ is a collection of pairwise disjoint events $H_1,\ldots, H_k$ of positive probability so that $\Omega = \bigcup_j H_j$. Each set $H_j$ is referred to as a part of this partition.

Theorem 39

Let $(H_1,\ldots, H_k)$ be a partition of $\Omega$ into non-trivial events and $A \subseteq \Omega$ be an event. Then

$${\mathbb P}[A] = \sum_{i} {\mathbb P}[A \mid H_i] \cdot {\mathbb P}[H_i]$$

Exercise 94

Prove Theorem 39. The proof should be very simple.

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