STT 861 Theory of Prob and STT I Lecture Note - 2

2017-09-13

Some of the basic probability and statistics concepts. joint probabilities, combinatories; conditional probabilities and independence and their examples; Bayes' rule.

Portal to all the other notes

Lecture 02 - Sept 13 2017

Recall: If A & B are 2 disjoint events, then $P [A \cup B] = P [A] + P [B]$ , $P [A \cap B] = P [A] \cdot P [B]$ .

If $A \cup B \neq \emptyset$ , then $P [A \cap B] = P [A] \cdot P [B] - P [A \cap B]$ .

General formula (“inclusion-exclusion formula”)

P [A_{1} \cup A_{2} \cup . . . \cup A_{n}] = \sum_{\emptyset \neq J \subseteq {1, 2, . . ., n}} (- 1)^{| J | - 1} ⋂_{j \in J} A_{j}

Example 1

Weather data for 2 consecutive days in some month in some location. We aggregate this data and find: ‘R1 is rain on 1st day’, ‘R2 is rain on 2nd day’. $P [R 1] = 0.6, P [R 2] = 0.5, P [R 1 \cap R 2^{C}] = 0.2$ . Find $P [R 1 \cap R 2]$ and $P [R 1 \cup R 2]$ .

Answer:

$P [(R 1 \cap R 2^{C}) \cup (R 1 \cap R 2)] = P [R 1]$ .

$P [R 1 \cap R 2^{C}] + P [R 1 \cap R 2] = P [R 1] \to P [R 1 \cap R 2] = 0.4$

$P [R 1 \cup R 2] = P [R 1] + P [R 2] - P [R 1 \cap R 2] = 0.6 + 0.5 - 0.4 = 0.7$

Example to introduce the use of some combinatories

Form a jury with 6 people. We choose from a group with 8 men & 7 women. Find the prob that there are exactly 2 women in the selected jury.

Need an assumption about the relative prob. of people being picked: 6/15. Every one has equal chance of being picked.

P=({# of ways to pick exactly 2 women})/({# of ways to pick 6 out of 15})

$(C_{7}^{2} \times C_{8}^{4}) / C_{15}^{6}$

(tree idea)

(\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!}

$C_{15}^{6} = 5005$

numerator $= 21 \times 70 = 1470$

$P = 1470 / 5005 = 0.2937$

After the break.

Chapter 1.3 Conditional prob. & independence

Definition: A & B are independence events, if $P [A \cap B] = P [A] \times P [B]$ .

The definition comes from the definition of conditional prob.

Definition: Let A & B be two events. The conditional prob of B given A denoted by $P [B | A]$ , is $P [A \cap B] / P [A]$ .

Note: if $P [B | A] = P [B]$ (given information of A does not affect/influence the chance of B), then the definition is proven.

Example 2

$P [d i s e a s e] = 0.006$ . Test: positive vs negative. $P [p o s i t i v e | d i s s e a s e] = 0.98$ (sensitivity). $P [p o s | n o d i s e a s e] = 0.01$ .

Q1: Find prob P[pos].

$P [p o s] = P [p o s \cap d i s e a s e] + P [n e g \cap n o d i s e a s e]$

$P [p o s | d i s e a s e] = P [p o s \cap d i s e a s e] / P [d i s e a s e]$

$P [p o s \cap d i s e a s e] = 0.98 \times 0.006$

similarly, $P [n e g \cap n o d i s e a s e] = (1 - 0.99) \times (1 - 0.006)$

$P [p o s] = 0.98 \times 0.006 + (1 - 0.99) \times (1 - 0.006) = 1.582 %$

Q2: Find $P [d i s | p o s]$ .

$P [d i s | p o s] = P [d i s \cap p o s] / P [p o s] = 0.00588 / 0.01582 = 0.3717$

The law of total probabilities:

Let $A_{1}, A_{2}, \dots, A_{n}$ be a partition of $Ω$ . ( $A_{i}$ are all mutually incompatible, and $\sum A_{i} = Ω$ )

then for any $B \in Ω$ , $P [B] = \sum_{i} P [B | A_{i}] P [A_{i}]$ .

Try to prove it at home.

Bayes’ rule

P [A | B] = \frac{P [A \cap B]}{P [B | A] P [A] + P [B | A^{C}] P [A^{C}]} = P [B | A] \times P [A] / P [B]

denominator: from law of total prob when n=2, $A_{1} = A, A_{2} = A^{C}$

General Bayes’ Rule:

For every fixed $i$ :

P [A_{i} | B] = \frac{P [B | A_{i}] P [A_{i}]}{\sum_{j = 1}^{n} P [B | A_{j}] P [A_{j}]}

Role of different pieces of information in the formula:

$P [A | B]$ : B has happend, the prob of A.
$P [A]$ and $P [A^{C}]$ : “The prori” a priori info about the events you are interested in, regard of the observation.
$P [B | A]$ and $P [B | A^{C}]$ : “Likelihood” likelihoods that eventsthat you observe B actually happen given the events of interest to you.