180 likes | 443 Views
Relation between Binomial and Poisson Distributions. Binomial distribution Model for number of success in n trails where P (success in any one trail) = p .
E N D
Relation between Binomial and Poisson Distributions • Binomial distribution Model for number of success in n trails where P(success in any one trail) = p. • Poisson distribution is used to model rare occurrences that occur on average at rate λ per time interval. Can think of “rare” occurrence in terms of p 0 and n ∞. Take these limits so that λ = np. • So we have that week 5
Continuous Probability Spaces • Ω is not countable. • Outcomes can be any real number or part of an interval of R, e.g. heights, weights and lifetimes. • Can not assign probabilities to each outcome and add them for events. • Define Ω as an interval that is a subset of R. • F – the event space elements are formed by taking a (countable) number of intersections, unions and complements of sub-intervals of Ω. • Example: Ω = [0,1] and F = {A = [0,1/2), B = [1/2, 1], Φ, Ω} week 5
How to define P ? • Idea - P should be weighted by the length of the intervals. - must have P(Ω) = 1 - assign 0 probability to intervals not of interest. • For Ω the real line, define P by a (cumulative) distribution function as follows: F(x) = P((- ∞, x]). • Distribution functions (cdf) are usually discussed in terms of random variables. week 5
Recalls week 5
Cdf for Continuous Probability Space • For continuous probability space, the probability of any unique outcome is 0. Because, P({ω}) = P((ω, ω]) = F(ω) - F(ω) = 0. • The intervals (a, b), [a, b), (a, b], [a, b] all have the same probability in continuous probability space. • Generally speaking, • discrete random variable have cdfs that are step functions. • continuous random variables have continuous cdfs. week 5
Examples (a) X is a random variable with a uniform[0,1] distribution. The probability of any sub-interval of [0,1] is proportional to the interval’s length. The cdf of X is given by: (b) Uniform[a, b] distribution, b > a. The cdf of X is given by: week 5
Formal Definition of continuous random variable • A random variable X is continuous if its distribution function may be written in the form for some non-negative function f. • fX(x)is the (Probability) Density Function of X. • Examples are in the next few slides…. week 5
The Uniform distribution (a) X has a uniform[0,1] distribution. The pdf of X is given by: (b) Uniform[a, b] distribution, b > a. The pdf of X is given by: week 5
Facts and Properties of Pdf • If X is a continuous random variable with a well-behaved cdf F then • Properties of Probability Density Function (pdf) Any function satisfying these two properties is a probability density function (pdf) for some random variable X. • Note:fX (x) does not give a probability. • For continuous random variable X with density f week 5
The Exponential Distribution • A random variable X that counts the waiting time for rare phenomena has Exponential(λ) distribution. The parameter of the distribution λ = average number of occurrences per unit of time (space etc.). The pdf of X is given by: • Questions: Is this a valid pdf? What is the cdf of X? • Note: The textbook uses different parameterization λ = 1/θ. • Memoryless property of exponential random variable: week 5
The Gamma distribution • A random variable X is said to have a gamma distribution with parameters α > 0 and λ > 0 if and only if the density function of X is where • Note: the quantity г(α) is known as the gamma function. It has the following properties: • г(1) = 1 • г(α + 1) = α г(α) • г(n) = (n – 1)! if n is an integer. week 5
The Beta Distribution • A random variable X is said to have a beta distribution with parameters α > 0 and β > 0 if and only if the density function of X is week 5
The Normal Distribution • A random variable X is said to have a normal distribution if and only if, for σ > 0 and -∞ < μ < ∞, the density function of X is • The normal distribution is a symmetric distribution and has two parameters μ and σ. • A very famous normal distribution is the Standard Normal distribution with parameters μ = 0 and σ = 1. • Probabilities under the standard normal density curve can be done using Table III on 574 in the text book. • Example: week 5
Example • Kerosene tank holds 200 gallons; The model for X the weekly demand is given by the following density function • Check if this is a valid pdf. • Find the cdf of X. week 5
Summary of Discrete vs. Continuous Probability Spaces • All probability spaces have 3 ingredients: (Ω, F, P) week 5