Rules of Probability
Probabilities must be between zero and one, i.e., $0≤P(A)≤1$ for any event A.
Probabilities add to one, i.e., $\sum{P(X_i)} = 1$
The complement of an event, $A^c$, denotes that the event did not happen. Since probabilities must add to one, $P(A^c) = 1 - P(A)$
If A and B are two events, the probability that A or B happens (this is an inclusive or) is the probability of the union of the events: $$ P(A \cup B) = P(A) + P(B) - P(A\cap B) $$ where $\cup$ represents union (“or”) and $\cap$ represents intersection (“and”). If a set of events $A_i$ are mutually exclusive (only one event may happen), then $$ P(\cup_{i=1}^n{A_i}) = \sum_{i=1}^n{P(A_i)} $$
Odds
The odds for event A, denoted $\mathcal{O}(A)$ is defined as $\mathcal{O}(A) = P(A)/P(A^c)$
This is the probability for divided by probability against the event
From odds, we can also compute back probabilities $$ \frac{P(A)}{P(A^c)} = \mathcal{O}(A) $$
$$ \frac{P(A)}{1-P(A)} = \mathcal{O}(A) $$
$$ \frac{1 -P(A)}{P(A)} = \frac{1}{\mathcal{O}(A)} $$
$$ \frac{1}{P(A)} - 1 = \frac{1}{\mathcal{O}(A)} $$
$$ \frac{1}{P(A)} = \frac{1}{\mathcal{O}(A)} + 1 $$
$$ \frac{1}{P(A)} = \frac{1 + \mathcal{O}(A)}{\mathcal{O}(A)} $$
$$ P(A) = \frac{\mathcal{O}(A)}{1 + \mathcal{O}(A)} $$
Expectation
The expected value of a random variable X is a weighted average of values X can take, with weights given by the probabilities of those values. $$ E(X) = \sum_{i=1}^n{x_i * P(X=x_i)} $$
Frameworks of probability
Classical – Outcomes that are equally likely have equal probabilities
Frequentist – In an infinite sequence of events, what is the relative frequency
Bayesian – Personal perspective (your own measure of uncertainty)
In betting, one must make sure that all the rules of probability are followed. That the events are “coherent”, otherwise one might construct a series of bets where you’re guaranteed to lose money. This is referred to as a Dutch book.
Conditional probability
$$ P(A|B) = \frac{P(A\cup B)}{P(B)} $$
Where $A|B$ denotes “A given B”
Example from lecture:
Suppose there are 30 students, 9 of which are female. From the 30 students, 12 are computer science majors. 4 of those 12 computer science majors are female $$ P(Female) = \frac{9}{30} = \frac{3}{10} $$
$$ P(CS) = \frac{12}{30} = \frac{2}{5} $$
$$ P(F\cap CS) = \frac{4}{30} = \frac{2}{15} $$
$$ P(F|CS) = \frac{P(F \cap CS)}{P(CS)} = \frac{2/15}{2/5} = \frac{1}{3} $$
An intuitive way to think about a conditional probability is that we’re looking at a subsegment of the original population, and asking a probability question within that segment $$ P(F|CS^c) = \frac{P(F\cap CS^c)}{PS(CS^c)} = \frac{5/30}{18/30} = \frac{5}{18} $$ The concept of independence is when one event does not depend on another. $$ P(A|B) = P(A) $$ It doesn’t matter that B occurred.
If two events are independent then the following is true $$ P(A\cap B) = P(A)P(B) $$ This can be derived from the conditional probability equation.
Conditional Probabilities in terms of other conditional
Suppose we don’t know what $P(A|B)$ is but we do know what $P(B|A)$ is. We can then rewrite $P(A|B)$ in terms of $P(B|A)$ $$ P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A^c)P(A^c)} $$ Let’s look at an example of an early test for HIV antibodies known as the ELISA test. $$ P(+ | HIV) = 0.977 $$
$$ P(- | NO_HIV) = 0.926 $$
As you can see over 90% of the time, this test was accurate.
The probability of someone in North America having this disease was $P(HIV) = .0026$
Now let’s consider the following problem: the probability of having the disease given that they tested positive $P(HIV | +)$ $$ P(HIV|+) = \frac{P(+|HIV)P(HIV)}{P(+|HIV)P(HIV) + P(+|NO_HIV){P(NO_HIV)}} $$
$$ P(HIV|+) = \frac{(.977)(.0026)}{(.977)(.0026) + (1-.977)(1-.0026)} $$
$$ P(HIV|+) = 0.033 $$
This example looked at Bayes Theorem for the two event case. We can generalize it to n events through the following formula $$ P(A|B) = \frac{P(B|A_1){(A_1)}}{\sum_{i=1}^{n}{P(B|A_i)}P(A_i)} $$
Bernoulli Distribution
~ means ‘is distributed as’
We’ll be first studying the Bernoulli Distribution. This is when your event has two outcomes, which is commonly referred to as a success outcome and a failure outcome. The probability of success is $p$ which means the probability of failure is $(1-p)$ $$ X \sim B(p) $$
$$ P(X = 1) = p $$
$$ P(X = 0) = 1-p $$
The probability of a random variable $X$ taking some value $x$ given $p$ is $$ f(X = x | p) = f(x|p) = p^x(1-p)^{1 - x}I $$ Where $I$ is the Heavenside function
Recall the expected value $$ E(X) = \sum_{x_i}{x_iP(X=x_i)} = (1)p + (0)(1-p) = p $$ We can also define the variance of Bernoulli $$ Var(X) = p(1-p) $$
Binomial Distribution
The binomial distribution is the sum of n independent Bernoulli trials $$ X \sim Bin(n, p) $$
$$ P(X=x|p) = f(x|p) = {n \choose x} p^x (1-p)^{n-x} $$
$n\choose x$ is the combinatoric term which is defined as $$ {n \choose x} = \frac{n!}{x! (n - x)!} $$
$$ E(X) = np $$
$$ Var(X) = np(1-p) $$
Uniform distribution
Let’s say X is uniformally distributed $$ X \sim U[0,1] $$
$$ f(x) = \left{ \begin{array}{lr} 1 & : x \in [0,1]\ 0 & : otherwise \end{array} \right. $$
$$ P(0 < x < \frac{1}{2}) = \int_0^\frac{1}{2}{f(x)dx} = \int_0^\frac{1}{2}{dx} = \frac{1}{2} $$
$$ P(0 \leq x \leq \frac{1}{2}) = \int_0^\frac{1}{2}{f(x)dx} = \int_0^\frac{1}{2}{dx} = \frac{1}{2} $$
$$ P(x = \frac{1}{2}) = 0 $$
Rules of probability density functions
$$ \int_{-\infty}^\infty{f(x)dx} = 1 $$
$$ f(x) \ge 0 $$
$$ E(X) = \int_{-\infty}^\infty{xf(x)dx} $$
$$ E(g(X)) = \int{g(x)f(x)dx} $$
$$ E(aX) = aE(X) $$
$$ E(X + Y) = E(X) + E(Y) $$
If X & Y are independent $$ E(XY) = E(X)E(Y) $$
Exponential Distribution
$$ X \sim Exp(\lambda) $$
Where $\lambda$ is the average unit between observations $$ f(x|\lambda) = \lambda e^{-\lambda x} $$
$$ E(X) = \frac{1}{\lambda} $$
$$ Var(X) = \frac{1}{\lambda^2} $$
Uniform (Continuous) Distribution
$$ X \sim [\theta_1, \theta_2] $$
$$ f(x|\theta_1,\theta_2) = \frac{1}{\theta_2 - \theta_1}I_{\theta_1 \le x \le \theta_2} $$
Normal Distribution
$$ X \sim N(\mu, \sigma^2) $$
$$ f(x|\mu,\sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(x-\mu)^2} $$
$$ E(X) = \mu $$
$$ Var(X) = \sigma^2 $$
Variance
Variance is the squared distance from the mean $$ Var(X) = \int_{-\infty}^\infty {(x - \mu)^2f(x)dx} $$
Geometric Distribution (Discrete)
The geometric distribution is the number of trails needed to get the first success, i.e, the number of Bernoulli events until a success is observed. $$ X \sim Geo(p) $$
$$ P(X = x|p) = p(1-p)^{x-1} $$
$$ E(X) = \frac{1}{p} $$
Multinomial Distribution (Discrete)
Multinomial is like a binomial when there are more than two possible outcomes.
$$ f(x_1,…,x_k|p_1,…,p_k) = \frac{n!}{x_1! … x_k!}p_1^{x_1}…p_k^{x_k} $$
Poisson Distribution (Discrete)
The Poisson distribution is used for counts. The parameter $\lambda > 0$ is the rate at which we expect to observe the thing we are counting. $$ X \sim Pois(\lambda) $$
$$ P(X=x|\lambda) = \frac{\lambda^xe^{-\lambda}}{x!} $$
$$ E(X) = \lambda $$
$$ Var(X) = \lambda $$
Gamma Distribution (Continuous)
If $X_1, X_2, …, X_n$ are independent and identically distributed Exponentials,waiting time between success events, then the total waiting time for all $n$ events to occur will follow a gamma distribution with shape parameter $\alpha = n$ and rate parameter $\beta = \lambda$ $$ Y \sim Gamma(\alpha, \beta) $$
$$ f(y|\alpha,\beta) = \frac{\beta^n}{\Gamma(\alpha)}y^{n-1}e^{-\beta y}I_{y\ge0}(y) $$
$$ E(Y) = \frac{\alpha}{\beta} $$
$$ Var(Y) = \frac{\alpha}{\beta^2} $$
Where $\Gamma(x)$ is the gamma function. The exponential distribution is a special case of the gamma distribution with $\alpha = 1$. As $\alpha$ increases, the gamma distribution more closely resembles the normal distribution.
Beta Distribution (Continuous)
The beta distribution is used for random variables which take on values between 0 and 1. For this reason, the beta distribution is commonly used to model probabilities. $$ X \sim Beta(\alpha, \beta) $$
$$ f(x|\alpha,\beta) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{n -1}(1 - x)^{\beta - 1}I_{{0 < x < 1}} $$
$$ E(X) = \frac{\alpha}{\alpha + \beta} $$
$$ Var(X) = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha+\beta+1)} $$
The standard uniform distribution is a special case of the beta distribution with $\alpha = \beta = 1$
Bayes Theorem for continuous distribution
$$ f(\theta|y) = \frac{f(y|\theta)f(\theta)}{\int{f(y|\theta)f(\theta)d\theta}} $$