Discrete Random Variables - lec. 7

ucla | MATH 170E | 2023-02-01T10:23

Definitions
Big Ideas
Resources

Definitions

Big Ideas

Random Variable (r.v.)

random variables are a function given a set $S$ and prob. space $(Ω, F, P)$ :

$X : Ω \to S$

if $x \in S$ and $A \sube S$

$\mathbb P(X=x):=\mathbb P({\omega\in\Omega:X(\omega)=x})$

the function’s diagram can be depicted as

Discrete Random Variables

a random variable is discrete if the range (output) is finite or countable i.e. one-to-one

$X : Ω \to S \sube \N$

Probability Mass Function (PMF) ( or Discrete Density)

PMF of r.v.s. function $X$ :

$p_X:S\to[0,1]$

Cumulative Distribution Function (CDF)

oof $X$

$F_{X} : \R \to [0, 1]$

two r.v.s are identically distributed if they have the same CDF

$X \sim Y$

Visual for PMF vs CDF

Propositions

$P (X \in A \sube \N) = \sum_{x \in A \cap S} p_{X} (x) ⟹ A \cap S \sube S$

$F_{X} (x) = \sum_{y \in S, y \leq x} p_{X} (y) ⟹ y \in A \cap S$

$P (a < X \leq b) = F_{X} (b) - F_{X} (a)$

Uniform Random Variables

r.v.s. $X$ is uniformly distributed on $1, \dots, m$ s.t.

$X \sim Uniform (1, \dots, m)$

i.e. if it has PMF

$p_{X} (x) = \frac{1}{m} x \in 1, \dots, m$

this uniform random variable has CDF

$F_{X} (x) =$

Geometric Random Variable

Number of trials until FIRST success

independent, identical Bernoulli trials with $p \in 0, 1$ probability
$X$ is an r.v. taking values from $S = 1, \dots, n$

$X \sim Geometric(p)$

$p_{X} (x) = p \cdot (1 - p)^{x - 1} x \in 1, \dots$

$F_{X} (x) = P (X \leq x) = 1 - (1 - p)^{x}$

$M_{X} (t) = \frac{p e^{t}}{1 - (1 - p) e^{t}} t < - \log (1 - p)$

Mean

$E [X] = \frac{1}{p}$

Variance

$\text{var( $X$ )}=\frac{1-p}{p^2}\space$

Binomial Distribution

Number of success in fixed number of trials

A Binomial distribution is the distribution of $n \geq 1$ independent, identical Bernoulli trials that we write $X \sim Binomial(n,p)$ where $n$ is the number of trials and $p$ is the probability of each success (prob. of success of 1 trial)
then the PMF is

$p_{X} (x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x} x \in 0, 1, \dots, n$

$Missing \end{cases}$

then the MGF is

$M_{X} (t) = E [e^{t X}] = (1 - p + p e^{t})^{n} \forall t \in \R$

then the expected value (mean) is

$E [X] = n p$

then the variance is

$var (X) = n p (1 - p)$

Negative Binomial R.V.

Probability of $n^{t h}$ success on $r^{t h}$ trial

independent, identical Bernoulli trials with probability $p \in 0, 1$ of success
let $r \geq 1$ and let $X$ be the d.r.v. the first trial on which we first achieve the $r^{t h}$ success takes values from $S = r, r + 1, r + 2, \dots$

$X\sim \text{Negative Binomial( $r, p$ )}$

$p_{X} (x) = (\binom{x - 1}{r - 1}) p^{r} (1 - p)^{x - r} x \in S$

$Missing \end{cases}$

Lemma for finding sum

${(\frac{1}{1 - s})}^{r} = \sum_{x = r}^{\infin} (\binom{x - 1}{r - 1}) s^{x - r}$

$M_{X} (t) = {(\frac{p e^{t}}{1 - (1 - p) e^{t}})}^{r} t < - \log (1 - p)$

mean and variance

$\mathbb E[X]=\frac rp\space$

Poisson R.V.

there are $λ > 0$ occurrences
let $X$ be the number of occurrences in some time span and takes values from $S = 1, 2, \dots$ (assuming population is infinite)
Assuming time intervals are disjoint: $(t_{1}, t_{2}], (t_{2}, t_{3}], \dots, (t_{n}, t_{n + 1}] ⟹$ then occurrences on each time interval are independent
If $h = t_{2} - t_{1} > 0$ is sufficiently small $⟹ P (X = 1 in (t_{1} t_{2}]) = λ h$ and converges rapidly to zero as $h \to 0$ i.e. an approximate Poisson r.v.

$X \sim Poisson(λ)$

$p_{X} (x) = \frac{e^{- λ} λ^{x}}{x!}$

$F_{X} (x) = e^{- λ} \sum_{k = 0}^{x} \frac{λ^{k}}{k!}$

Mean and Variance

$\mathbb E[X]=\lambda$

Resources

https://youtu.be/YXLVjCKVP7U

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