CIS 2033 based on Dekking et al A

CIS 2033 based on Dekking et al. A Modern Introduction to Probability and Statistics. 2007 Instructor Longin Jan Latecki Chapter 5: Continuous Random Variables

Probability Density Function of X A random variable X is continuous if for some function ƒ: R R and for any numbers a and b with a ≤ b, P(a ≤ X ≤ b) = ∫ab ƒ(x) dx The function ƒ has to satisfy ƒ(x) ≥ 0 for all x and ∫-∞∞ ƒ(x) dx = 1. Whereas Discrete Random Variables {px(a) =P(X=a)} map: Ω -- (X) -> R -- (px) -> [0, 1] Continuous Random Variables {P(a ≤ X ≤ b)=∫ab ƒ(x) dx} map: Ω -- (X) -> R -- (ƒ) -> R

To approximate the probability density function at a point a, one must find an ε that is added and subtracted from a and then the area of the box under the curve is obtained by the following: (2ε) * ƒ(a). As ε approaches zero the area under the curve becomes more precise until one obtains an ε of zero where the area under the curve is that of a width-less box. This is shown through the following equation. P(a – ε ≤ X ≤ a +ε) = ∫a-εa+ε ƒ(x) dx ≈ 2ε*ƒ(a)

A few asides: DISCRETE NO DENSITY CONTINUOUS NO MASS BOTH CUMULATIVE DISTRIBUTION Ƒ(a) = P(X ≤ a) P(a < X ≤ b) = P(X ≤ b) – P(X ≤ a) = Ƒ(b) – Ƒ(a) Ƒ(b) = ∫-∞b ƒ(x) dx and ƒ(x) = (d/dx) Ƒ(x) *How the Distribution Function relates to the Density Function*

Uniform Distribution U(α, β) • A continuous random variable has a uniform distribution on the interval [α, β] if its probability density function ƒ is given by ƒ(x) = 0 if x is not in [α, β] and, ƒ(x) = 1/(β-α) for α ≤ x ≤ β This simply means that for any x in the interval of alpha to beta has the same probability and anything not in the interval is zero as shown in the figure below. • The cumulative distribution is given by F(x) = 0 if x < α, m. F(x) = 1 if x > β, and F(x) = (x − α)/(β − α) for α ≤ x ≤ β

Exponential Distribution Exp(λ) Intuitively: a continuous version of the geometric distribution. A continuous random variable has an exponential distribution with parameter λ if its probability density function ƒ is given by ƒ(x) = λe-λx for x ≥ 0 The Distribution function ƒ of an Exp(λ) distribution is given by Ƒ(a) = 1 – e-λa for a ≥ 0 P(X > s + t | x > s) = P(x > s + t)/P(x>s) = (e-λ(s+t))/(e-λs) =e-λt= P(X > t) This simply means that s becomes the origin where t increases therefore making s always less than t and the equation proven true.

ƒ(x) = λe-λx for x ≥ 0 Ƒ(a) = 1 – e-λa for a ≥ 0 How much time will elapse before an earthquake occurs in a given region? How long do we need to wait before a customer enters our shop? How long will it take before a call center receives the next phone call? How long will a piece of machinery work without breaking down? Questions such as these are often answered in probabilistic terms using the exponential distribution. All these questions concern the time we need to wait before a certain event occurs. If this waiting time is unknown, it is often appropriate to think of it as a random variable having an exponential distribution.

Roughly speaking, the time we need to wait before an event occurs has an exponential distribution if the probability that the event occurs during a certain time interval is proportional to the length of that time interval. More precisely, X has an exponential distribution if the conditional probability is approximately proportional to the length Δt of the time interval [t, t + Δt] for any time instant t. In most practical situations this property is very realistic and this is the reason why the exponential distribution is so widely used to model waiting times. From: http: //www. statlect. com/ucdexp 1. htm

Pareto Distribution Par(α) Used for estimating real-life situations such as • the number of people whose income exceeded level x, • city sizes, earthquake rupture areas, insurance claims, and sizes of commercial companies. A continuous random variable has a Pareto distribution with parameter α > 0 if its probability density function ƒ is given by ƒ(x) = 0 if x < 1 and for x ≥ 1

Normal Distribution N(μ, σ2) Normal Distribution (Gaussian Distribution) with parameters μ and σ2 > 0 if its probability density function ƒ is given by for -∞ < x < ∞ *Where μ = mean and σ2 = standard deviation* Distribution function is given by: for -∞ < a < ∞ However, since ƒ does not have an antiderivative there is no explicit expression for Ƒ. Therefore standard normal distribution where N(0, 1) is given as follows, and the distribution function is obtained similarly denoted by capital phi. for -∞ < x < ∞

Normal Distribution

Quantiles Portions of the whole which increase from left to right, meaning the 0 th percentile is on the left hand side and the 100 th percentile is on the right side. Let X be a continuous random variable and let p be a number between 0 and 1. The pth quantile or 100 pth percentile of the distribution of X is the smallest number qp such that Ƒ(qp) = P(X ≤ qp) = p The median is the 50 th percentile

Ƒ(qp) = P(X ≤ qp) = p For continuous random variables qp is often easy to determine. If F is strictly increasing from 0 to 1 on some interval (which may be infinite to one or both sides), then qp = Finv(p), where Finv is the inverse of F.

What is the median of the U(2, 7) distribution? The median is the number q 0. 5 = Finv(0. 5). You either see directly that you have got half of the mass to both sides of the middle of the interval, hence q 0. 5 = (2+7)/2 = 4. 5, or you solve with the distribution function: F(qp) = P(X ≤ qp) = p = 0. 5 F(q) = 0. 5 and F(x) = (x − α)/(β − α)

F(qp) = P(X ≤ qp) = p = 0. 9 F(a) = 0. 9 Matlab: a = norminv(0. 9, 0, 1) => a =1. 2816