> anishgoyal █


Notes › EENG 3421: Probability and Statistics for Engineers Lecture 4

Independence and Tree Diagrams

calendar_today   article 466 words   access_time 4 min   replay Modified

Table of Contents

Comments on Bayes Theorem

Definition 04.1 (Prior and Posterior Probabilities).
  • $P[A_i]$ prior probability of cause $A_i$
  • $P[A_i \mid B]$ posterior probability of cause $A_i$ given effect $B$
Theorem 04.2 (Bayes Theorem for Causes).

Let $\{A_1, \dots, A_n\}$ be mutually exclusive and collectively exhaustive

$$ P[A_i \mid B] = \frac{P[A_i] P[B \mid A_i]}{\sum_{j=1}^n P[A_j] P[B \mid A_j]} $$

Independence

Definition 04.3 (Two Independent Events).

Events $A$ and $B$ are independent if and only if

$$ P[A \cap B] = P[A] P[B] $$

Independent Vs Mutually Exclusive

Comment

Independent and mutually exclusive are not synonymous

Example

Example 04.4.

Draw one card from a standard deck

  • $A$ event card is a spade
  • $B$ event card is a queen
$$ P[A] = \frac{13}{52}, \quad P[B] = \frac{4}{52} $$$$ P[A \cap B] = \frac{1}{52} $$

Since $P[A \cap B] = P[A] P[B]$, the events are independent

Three Independent Events

Definition 04.5 (Three Independent Events).

Events $A_1, A_2, A_3$ are mutually independent if and only if

  • $A_1$ and $A_2$ are independent
  • $A_1$ and $A_3$ are independent
  • $A_2$ and $A_3$ are independent
  • $$ P[A_1 \cap A_2 \cap A_3] = P[A_1] P[A_2] P[A_3] $$

Example

Example 04.6.

Let $S = \{1,2,3,4\}$ with equiprobable outcomes

  • $A_1 = \{1,3,4\}$
  • $A_2 = \{2,3,4\}$
  • $A_3 = \emptyset$

Although

$$ P[A_1 \cap A_2 \cap A_3] = P[A_1] P[A_2] P[A_3] $$

the events are not mutually independent since $A_1$ and $A_2$ are not independent

More Than Three Events

Definition 04.7 (Mutual Independence).

For $n \ge 3$, events $A_1, \dots, A_n$ are mutually independent if and only if

  • Every collection of $n-1$ events is mutually independent
  • $$ P\!\left[\bigcap_{i=1}^n A_i\right] = \prod_{i=1}^n P[A_i] $$

Tree Diagrams

Definition 04.8 (Tree Diagram).

A tree diagram represents a sequential experiment

  • Branch labels are probabilities or conditional probabilities
  • Each path corresponds to a joint event
  • Path probability is the product of branch probabilities
$$ P[\text{path}] = \prod \text{branch probabilities} $$

Example

Example 04.9.

Radar detection

  • $A$ aircraft present
  • $B$ alarm generated
$$ P[A] = 0.05 $$$$ P[B \mid A] = 0.99, \quad P[B \mid A^c] = 0.10 $$

Joint probabilities computed along tree paths

Example

Example 04.10.

False positive test

  • $A$ person has disease
  • $B$ test positive
$$ P[A] = 0.001 $$$$ P[B \mid A] = 0.95, \quad P[B^c \mid A^c] = 0.95 $$

Using Bayes theorem

$$ P[A \mid B] = \frac{P[B \mid A] P[A]}{P[B \mid A] P[A] + P[B \mid A^c] P[A^c]} $$

References

Sources

Graph

mindmap ↗