NotesAtHKU

A continuous random variable has probabilities the area under its curve. Hence, $P(X=n)$ for any outcome $n$ is $0$ . A discrete random variables have specific probabilities assigned to an outcome.

Operation on continous ranges

$P(X<x)=P(X\leq x)$

$P(X>x)=1 - P(X<x)$

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

Operation on discrete ranges

$P(X<x)=P(X\leq x-1)$

$P(X>x)=1-P(X\leq x)$

$P(a\leq X\leq b)=P(X\leq b)-P(X\leq a-1)$

Probability distribution functions

A p.d.f is a continuous function that returns the probability of the given outcome. The following is an example of a p.d.f:

X\sim f(x) = \begin{cases} Kx^2 & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}

Using the function

$P(X=x)=0$

$P(a<X<b)=P(a\leq X\leq b)=\int_{a}^{b}f(x)dx$

Note that $\int_{\infty}^{-\infty}f(x)dx = 1$ , as the total probability of any event is 1. This condition must be true for $f(x)$ to be a valid p.d.f. Hence for the example $K=3$

Expected value

$E(X^n)=\int_{\infty}^{-\infty}x^nf(x)dx$

Cumulative distribution function

To obtain a c.d.f $X'\sim F(x)$ where $P(X'=x) = P(X < x)$ , all we have to do is integrate the p.d.f:

F(X)=\int f(x)dx

Using our example:

F(X)=\begin{cases} 1 & x > 1 \\ x^3 & 0 < x < 1 \\ 0 & x < 0 \\ \end{cases}

And to convert a c.d.f back to it's p.d.f, all we have to do is differentiate $F(x)$ :

f(x)=\frac{d}{dx}F(x)

Statistical distributions

Common statistical distribution


Continuous	Discrete	Discrete	Discrete	Discrete	Discrete
Exponential	Bernoulli	Binomial	Geometric	Negative Binomial	Possion
$Expo(\lambda)$	$Ber(p)$	$B(n, p)$	$Geo(p)$	$NB(n, p)$	$Po(\lambda)$
$\lambda e^{-\lambda x}$	$p$ if true, else $(1-p)$	$\binom{n}{x} p^x(1-p)^{n-x}$	$p(1-p)^{x-1}$	$\binom{x-1}{n-1} p^n(1-p)^{x-n}$	$\frac{\lambda^x}{x!} e^{-\lambda}$
$E=\lambda^{-1}$	$E=p$	$E=np$	$E=\frac{1}{p}$	$E=\frac{n}{p}$	$E=\lambda$
$Var=\lambda^{-2}$	$Var=p(1-p)$	$Var=np(1-p)$	$Var=\frac{1-p}{p2}$	$Var=\frac{n(1-p)}{p2}$	$Var=\lambda$

Usage cases

Ber: Outcomes only $True$ or $False$
B: $x$ successes in $n$
Geo: $1$ st success at $x$ tries
NB: $n$ th success at $x$ tries
Po: $x$ successes in interval $\lambda$

Normal distribution

The Normal distribution is given by the following formula (which you don't have to memorize):

X\sim N(\mu, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}},\quad\quad\text{where }\underbrace{\mu}_{\text{mean}},\underbrace{\sigma^2}_{\text{variance}}

To calculate $P(X<x)$ , we have to standardize our $N$ by $Z\sim N(0, 1)$ , $P(X<x)=P(Z<\frac{x-\mu}{\sigma})$ . Here $Z$ is the standard normal variable.

Reading the Z-table

For $P(Z<z)=p$ , to find $p$ , locate the header and leftmost column in the z-table such that their sum is $z$ . The corresponding intersecting cell is $p$ .

$Z_p$ gives the value $z$ in $P(Z<z)=p$

Critical interval

The critical interval is given by:

C.I.=[\bar{X}\pm Z_{\frac{C.L.+1}{2}}\times\frac{\sigma}{\sqrt{n}}],\quad\quad\text{where }\underbrace{C.L.}_{\text{Confidence level}},\ \underbrace{\bar{X}}_{\text{Mean value}}

Hence, we can derive that the critical interval width is:

2\times Z_{\frac{C.L.+1}{2}}\times\frac{\sigma}{\sqrt{n}}

Probability distributions