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_________________________________________________________________________________ Chapter 7 Special Continuous Distributions § 7.1 Uniform Random Variable • Definition: A random variable is said to be uniformly distributed over an interval (a, b) if its probability density function is given by f(x) = = 1/(b-a) 0 a<x<b otherwise. * We can easily show that for the above random variable, E[X] = (a+b)/2 and Var(X) = (b-a)2/12. * Example 7.3: What is the probability that a random chord of a circle is longer than a side of an equilateral triangle inscribed into that circle. Solution: There are three possible ways of randomly choosing a chord in a circle. All are uniform except that the sample space is different, thus giving different answers. Fig. 7.2, Fig. 7.3, Fig. 7.4 __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ § 7.2 Normal Random Variable • Theorem 7.1 (De Moivre-Laplace Theorem): Let X be a binomial random variable with parameters n and p. Then for any numbers a and b, a < b, lim P( a < [X-np]/[np(1-p)]0.5 < b ) = 1/(2)0.5 ab e-t 2/2 dt n Note that E(X) = np and X = [np(1-p)]0.5 . * Note that as n approaches infinity, the standardized binomial random variable has a distribution function FX*(t) = 1/(2)0.5 - t e-x 2/2 dx = (t). * It can be shown that (see text) (-) = 0 and () = 1. • Definition: A random variable X is called standard normal if its distribution function is FX(t) = (t) = 1/(2)0.5 - t e-x 2/2 * The density function is f(x) = 1/(2)0.5e-x 2/2 dx . Fig. 7.5 __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ • To approximate a probability based on a discrete random variable by a probability based on a continuous random variable, we have P(i X j) ~ i-0.5j+0.5 f(x) dx ; P(X = k) ~ k -0.5k+0.5 f(x) dx ; P(X i) ~ i -0.5 f(x) dx ; P(X j) ~ - j +0.5 f(x) dx . * (x) can not be calculated directly, often we look up its value in a table (see Table 1 of the appendix in the textbook). There is also an approximation formula (x) ~ 0.5 + x(4.4-x)/10. * see Example 7.4 * It can be shown (see text) that the expectation and variance of the standard normal random variable are 0 and 1, respectively. • Definition: A random variable X is called normal, with parameters and , if its density function is given by f(x) = 1/[(2)0.5]e-(x-) 2/22 and we write X ~ N(,2). The normal random variable is sometimes called the Gaussian random variable. • Lemma: If X ~ N(,2), then X* = (X-)/ is N(0,1). That is if X ~ N(,2), then the standardized X is N(0,1). __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ Fig. 7.8 Fig. 7.9 * Example 7.5: Suppose that a man's chest size is N(39.8,2.052). Randomly choose 20 men, what is the probability that five have a chest size larger than 40. Solution: First find the probability of a men's chest size is larger than 40. p = P(X 40) = P{(X-39.8)/2.05 0.2/2.05} = P(X* 0.1) = 1 - (0.1) = 1- 0.5398 0.46. Then C(20,5) (0.46)5 (0.54)15 * Example 7.7: The score on a test is ~ N(500,1002). What should the score of a student be to place him on the top 10%? Solution: P{(X-500)/100 < (k-500)/100} = 0.9. From the appendix, we have (1.28) = 0.8997 ~ 0.9. So (k-500)/100 = 1.28 and k = 628. __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ § 7.3 Exponential Random Variable • Let {N(t), t0} be a Poisson process and let Xi be the time between the (i-1)th event and the ith event. We have P(X1 > t) = P(N(t)=0) = e-t(t)0/0! = e-t and P(X1 t) = 1 - e-t Since the Poisson process is stationary, we have P(Xn t) = 1 - e-t =0 t0 t<0 We have the cdf of Xn being FXn(t) and is given by FXn(t) = 1 - e-t =0 t0 t<0 and fXn(t) = F'Xn(t) = e-t =0 It can also e shown that -fXn(t) = 1. t0 t<0 • Definition: A continuous random variable X is called exponential with parameter > 0 if its probability density function is as given above. * Example of exponential random variables 1. The inter-arrival time between two customers. 2. The duration between two phone calls. 3. The time between two consecutive earthquakes. 4. The time between two accidents in an intersection. __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ • We know that is the average number of events in a unit time interval. So we expect the average time between two events is 1/. E(X) = 0 x e-xdx = [-xe-x]0 + 0 e-xdx = 0 - (1/)[e-x]0 = 1/ • It can also be shown that for an exponential random variable X, Var(X) = (1/)2 and X = 1/ Fig. 7.10 Fig. 7.11 __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ • Memoryless Property: A nonnegative random variable is called memoryless if for all s, t 0, P(X > s+t|X > t) = P(X > s) * It can be shown that the exponential random variable is the only continuous random variable that is memoryless. It can also be shown that the geometric random variable is the only discrete memoryless random variable. * Example 7.10: Suppose that on average two earthquakes occur in SF and two in LA every year. If the last earthquake in SF is 10 months ago and the last earthquake in LA is 2 months ago. What is the probability that the next earthquake in SF occurs after the next earthquake in LA? Solution: Due to memoryless property, the times before the next earthquake in SF and LA are both exponential with parameter 2. So the probability is 0.5 due to symmetry. __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ Q: How to generate an exponential function from U[0,1]? __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ § 7.4 Gamma Distribution • Given X1, X2, X3 .... as exponential random variables in the previous section. Define a new random variable Yn = X1 + X2 + .... + Xn In other words, Yn is the time it takes before the nth event occurs. We have e-t(t)i/i! FYn (t) = P(Yn t) = P(N(t) n) = i=n, * Differentiating FYn with respect to t, we have fYn (t) = e-t(t)n-1/(n-1)! * The density function f(t) = = e-t(t)n-1/(n-1)! 0 t0 elsewhere is called the n-Erlang density or the gamma density with parameter (n,). * We can extend the gamma density to parameters (r,), where r is a positive real number. Let the gamma function be defined as (r) = tr-1e-t dt r>0 It can then be shown that for r=n, where n is an integer, then (n+1) = n!. (p. 293) __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___ _________________________________________________________________________________ • Definition: A random variable with probability density function f(x) = e-x(x)r-1/(r) =0 x0 elsewhere is said to have a gamma distribution with parameters (r,), r, >0. * Note that the exponential random variable is a special case of the gamma random variable with r=1. Fig. 7.12, Fig. 7.13 * Finding the expectation and variance of a gamma random variable is rather tedious and it can be shown that for a gamma random variable with parameters (r,) E(X) = r/ and Var(X) = r/2. * Example 7.15 (p. 296) __________________________________________________________ © Shi-Chung Chang, Tzi-Dar Chiueh ___