probability models

Probability Model

A probability model consists of a non-empty set called the sample space

Ω

, a collection of events that are subsets of

S

and a probability measure

P

assigning a probability between 0 and 1 to each event with

P (\emptyset) = 0

and

P (S) = 1

and with

P

additive.

Finite Uniform Distribution

For any specific

N \in ℕ

, the random variable

X

is said to have a (finite discrete) uniform distribution on the sample space

Ω = {1, \dots, N}

- denoted:

X ~

unif

{1, \dots, N}

iff

P (X = k) = 1 / N, k = 1, \dots, N .

Conditional Probability

Let

A, B \subseteq Ω

, such that

P (B) \neq 0

then the conditional probability of

A

given

B

is defined as

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

P (B) = 0

then we define

P (A ∣ B) = 0

Intersection as Conditional Probability

Suppose that

P (A), P (B) \neq 0

then we have

P (A \cap B) = P (A ∣ B) P (B)

Finite Intersection as Conditional Probability

Suppose we have events

E_{1}, \dots, E_{n}

then

P (E_{1} \cap E_{2} \cap \dots \cap E_{n}) = P (E_{n} | (E_{1} \cap \dots \cap E_{n - 1})) P (E_{1} \cap \dots \cap E_{n - 1})

whenever

P (E_{1} \cap E_{2} \cap \dots \cap E_{n}) \neq 0

Let

A : = E_{1} \cap \dots \cap E_{n - 1}

and

B : = E_{n}

then from this we recall that:

P (B \cap A) = P (B ∣ A) P (B)

Since

P (B \cap A) = P (A \cap B)

we have that

P ((E_{1} \cap E_{2} \cap \dots \cap E_{n - 1}) \cap E_{n}) = P (E_{n} ∣ (E_{1} \cap E_{2} \cap \dots \cap E_{n - 1})) \cdot P (E_{1} \cap \dots \cap E_{n - 1})

Product Rule

Let

E_{1}, \dots, E_{n}

be events, then

P ((E_{1} \cap E_{2} \cap \dots \cap E_{n - 1}) \cap E_{n}) = P (E_{n} ∣ (E_{1} \cap E_{2} \cap \dots \cap E_{n - 1})) \cdot P (E_{n - 1} ∣ (E_{1} \cap \dots \cap E_{n - 2})) \dots P (E_{3} ∣ (E_{1} \cap E_{2})) \cdot P (E_{2} ∣ E_{1}) P (E_{1})

As stated it is notationally unwieldly we can clean it up using the following: $P ((E_{1} \cap E_{2} \cap \dots \cap E_{n - 1}) \cap E_{n}) = P (E_{1}) (\prod_{j = 2}^{n} P (E_{j} ∣ (E_{1} \cap \dots \cap E_{j - 1})))$ If you're confused by the usage of $\prod$ note that each factor uses exactly one more $E_{i}$ than the previous and moves in the reverse order of the of the original inequality

Symmetric Conditional Equation

P (A ∣ B) P (B) = P (B ∣ A) P (A)

Probability of Set Difference

Suppose that

A, B \subseteq Ω

then

P (A ⧵ B) = P (A) - P (A \cap B)

Probability of an Event through a Partition

Suppose that

X = {X_{1}, X_{2}, \dots}

is a partition of

Ω

then for any event

E

we have

P (E) = \sum_{i \in N_{1}} P (E \cap X_{i})

Observe that $E = E \cap Ω = E \cap (⊔_{i \in N_{1}} X_{i}) = ⊔_{i \in N_{1}} (E \cap X_{i})$ , therefore $P (E) = \sum_{i \in N_{1}} P (E \cap X_{i})$

It follows as an easy corollary that it holds for finite partitions by seting the rest of the partition to be the empty set.

Law of Total Probability

Suppose that

X = {X_{1}, X_{2}, \dots}

is a partition of

Ω

then for any event

E

we have

P (E) = \sum_{i \in N_{1}} P (E ∣ X_{i}) P (X_{i})

Independence of Events

Given two events

A, B \subseteq Ω

we say that

A

and

B

re independent iff

P (A | B) = P (A)

Independence is Symmetric

Suppose that

A

and

B

are independent, then

B

and

A

are independent

We know that $P (A ∣ B) = P (A)$ , we'd like to prove that $P (B ∣ A) = P (B)$ , $P (B ∣ A) & = \frac{P (A \cap B)}{P (A)} & = \frac{P (A ∣ B) P (B)}{P (A ∣ B)} & = P (B)$ on the second line we used the intersection as conditional formula

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