website/content/notes/bayesianstatistics/week1.md

Bayesian Statistics

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}}