Random Variables

Random Variable

A random variable is a function from the sample space

Ω

ℝ

Random Variable Function Composition

Suppose that

X

is a random varaible and that

g : ℝ \to ℝ

is a function, then we define

g (X) : = g \circ X

where

g \circ X : Ω \to ℝ

Note that when we compose a real valued function with a random varaible, we obtain a new random varaible.

Random Variable and Binary Operation Syntax Sugar

Random Variable and Binary Operation Syntax Sugar: Suppose that

X : Ω \to ℝ

is a random variable and that

⋆

is a binary operation, then for any

a \in ℝ

we define

X ⋆ a : = {o \in Ω : X (o) ⋆ a}

The main thing to note with the above definition is that when we write something of the form $R = 3$ , then this is a subset of our sample space and therefore we can take the probability of it.

Discrete Random Variable

A random variable

X

is discrete if

\sum_{x \in ℝ} P (X = x) = 1

Probability Function

For a discrete random variable

X

, it's probability function is the function

p_{X} : ℝ \to [0, 1]

p_{X} (x) : = P (X = x)

Independent Random Variables

Given two random varaibles

X, Y

on the same sample space we say that

X

is independent of

Y

and write

X ⟂ Y

iff for any

A, B \subseteq ℝ

we have

P (X \in A | Y \in B) = P (X \in A)

Note that above simply says that given any $A, B \subseteq R$ the events $E_{A} : = X \in A$ and $E_{B} : = Y \in B$

Independence as a Product

A ⟂ B ⟺ P (X \in A \cap X \in B) = P (X \in A) P (X \in B)

Independence is Reflexive

A ⟂ B ⟺ B ⟂ A

Follows from the previous corollary.

Independence Practice

Suppose that

A, B

are two events such that

P (A ∖ B) = P (B) = \frac{2}{5}

and

A ⟂ B

Determine $P (A)$
Obtain the probabilities $P (A \cap B), P (A \cap B^{∁}), P (A^{∁} \cap B^{∁})$
Determine the conditional probability $P (A ∖ B | A \cap B)$

We start by determining $P (A)$ , firstly we know that $\begin{matrix} \frac{2}{5} & a m p; = P (B) = P (A ∖ B) \\ a m p; = P (A) - P (A \cap B) \\ a m p; = P (A) - P (A) P (B) \\ a m p; = P (A) (1 - P (B)) \\ a m p; = P (A) (1 - \frac{2}{5}) \\ a m p; = P (A) (\frac{3}{5}) \end{matrix}$ therefore we deduce that $P (A) = \frac{2}{3}$

Multivariate Random Variable

Suppose that we have a finite collection of random varaibles

X_{i}

each with their own sample space

Ω_{i}

then the tuple

𝐗 = (X_{1}, \dots X_{n})

is said to be a multivariate random variable and is a function from

Ω_{1} \times \dots \times Ω_{n} \to ℝ^{n}

Collection of Independent Events

Suppose we have some collection

A_{i}, i \in I

of events for some index set

I

, then we say that this collection of events are independent if