<a href="/keyword/continuous-random-variable/" >continuous random variable</a> and Their Probability Distributions: Part I Cyr Emile M LAN, Ph.D. [e-mail] <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 1/29 Introduction Text Reference: Mathematical Statistics with Application, Chapter 4. Reading Assignment: Sections 4.1-4.3, 4.9, 4.10, October 16 - October 21 As mentioned before, there are two types of random variables: discrete and continuous. In this set of notes, we study the second types of random variable that arises in many applied problems. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 2/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s A random variable is continuous if the values it can assumed cannot be enumerated (can be represented by an interval). Example 4.1: If a chemical compound is randomly selected and its pH X is determine, then X is a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> because any pH value between 0 and 14 is possible. If in the study of the ecology of a lake, we make depth measurements at randomly chosen locations, then X = the depth at such location is a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 3/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Unlike discrete random variables, it is impossible to assign nonzero probabilities to all the possible values of a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> . Question: How do deal with this type of random variable? The idea is to nd an alternative representation to the probability mass function that will characterize continuous distributions. For, &quot;discretize&quot; the random variable. The resulting discrete distribution can be pictured using a probability histogram. As one re nes the discretization, a much smoother curve appears on top of the histogram plot. If we continue in this way, the sequence of histograms Continuous p. 4/29 approaches a smooth curve. Random Variables: Part I <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Histogram of x Histogram of x 0.3 Density Density 4 2 0 x 2 4 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 4 2 0 x 2 4 Histogram of x Histogram of x 0.4 0.3 Density Density 0.2 0.1 0.0 4 2 0 x 2 4 4 2 2 4 Continuous Random0Variables: Part I x 0.0 0.1 0.2 0.3 0.4 p. 5/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s In any of these histograms, the area of all rectangles is 1. Thus, the total area under the smooth curve should also be 1. The probability that the random variable, Y , falls between two values a and b, P a Y b , is approximately equal to the sum of all rectangles that fall in the interval [a, b]. Thus, the total area under the smooth curve between a and b is just the probability that the random variable falls in the interval [a, b]. The smooth curve is called a probability density function and it is the quantity that is used to characterize continuous distribution. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 6/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s De nition 4.1: Let Y be a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> . Then there exists a mathematical function f (y) called the the probability density function of the random variable. Y satisfying 1. f (y) 0 for all &lt; y &lt; . 2. f (x) dx = 1. For any two numbers a and b such that a b, b P a Y b = f (x) dx . a That is, the probability that Y takes on a value in the interval [a, b] is the area above this interval and under the graph of the density function curve. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 7/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s All probability statements about Y can be answered with f (y). Let b = a. Then a P (Y = a) = P a Y a = f (x) dx = 0 . a In words, the probability that a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> will assume any particular value is zero. (Note that there is no area). An important practical implication is: The probability that Y lies in some interval between a and b does not depend on whether the lower limit a or the upper limit b is included in the probability calculation: P a Y b =P a&lt;Y &lt;b =P a&lt;Y b =P a Y &lt;b . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 8/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Let a = and b = y . y P (Y y) = P ( Y y) = f (x) dx . De nition 4.2: Let Y be a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> . Then the cumulative distribution function F (x) is de ned for every number x by y F (y) = P (Y y) = f (x) dx That is, for each y, F (y) is the area under the probability density curve to the left of y. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 9/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s P (Y &gt; a) = 1 F (a) For any two numbers a and b with a &lt; b, P (a Y b) = F (b) F (a) Theorem 4.1: Properties of a Distribution Function Let F (y) be a distribution function, then 1. F ( ) = lim F (y) = 0 . y y + 2. F (+ ) = lim F (y) = 1 . 3. F (y) is a nondecreasing function of y. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 10/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s De nition 4.3: A random variable Y with probability distribution function F (y) is said to be continuous if F (y) is continuous, for &lt; y &lt; . If one differentiates both sides of the equation in de nition 3.6, one obtains d F (y) = F (y) = f (y) . dy Theorem 4.2: Let Y be a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> with probability density function f (y) and cumulative distribution function F (y), then at every y at which the derivative F (y) exists, F (y) = f (y) . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 11/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Example 4.2: An accounting rm that does not have its own computing facilities rents time from a consulting company. The rm must plan its computing budget carefully and hence has studied the weekly use of CPU time quite thoroughly. The weekly use of CPU time in hours, X, approximately follows the probability density function given by K x2 (4 x) , 0 x 4 f (x) = 0, elsewhere a). Find the constant K. b). Find the cumulative distribution function F (x) for weekly CPU time X. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 12/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Solution: 4 a). To get K, solve the equation 1 = K 0 4 0 x2 (4 x) dx. We have x=4 x=0 x (4 x) dx = K 2 4 3 1 4 x x 3 4 = 64 K. 3 3 Hence, K = . 64 b). The cumulative distribution function F (y) for weekly CPU time y between 0 and 4 is y 3 y 2 1 3 x (4 x) dx = y (16 3y) . F (y) = f (x) dx = 64 0 256 0 Thus, F (x) = 0, x&lt;0 x3 (16 3x) , 0 x 4 4&lt;x 1 256 1, <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 13/29 <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s Example 4.3: Suppose that the random variable Y has a distribution function 0, y 0 F (y) = y 2 , y &gt; 0 1 e What is the probability that Y exceeds 1? b). Find the probability density function of Y . a). Solution: a). P Y &gt; 1 = 1 F (1) = e 1 = .368 . b). f (y) = F (y) = y 2 2y e . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 14/29 Joke on statistics What does a politician and a drunken man has in common? A Politician uses statistics as a drunken man uses the lamp-posts for support rather than illumination. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 15/29 Quantiles (or Percentiles) of a Continuous Distribution When we say that an individual s test score was the .85th quantile or 85th percentile of the population, we mean that 85% of all population scores were below that score and 15% were above. De nition 4.4: Let p be a number between 0 and 1. The the pth quantile or the (100p)th percentile of the distribution of a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> , Y with distribution function F (y), denoted by p , is the smallest value such that p p = F ( p ) = P (Y p ) = f (x) dx . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 16/29 Percentiles of a Continuous Distribution Example 4.4: Time headway&quot; in traf c ow is the elapsed time between the time that one car nishes passing a xed point and the instant that the next car begins to pass that point. Let X be the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy ow. The following probability density is essentially the one suggested in the Statistical Properties of Freeway Traf c&quot; (Transp. Res., vol. 11: 221-228): .15 e .15(y .5) , y .5 f (y) = 0, elsewhere a). Find the 50th percentile of this distribution. Find the 15th percentile of this distribution. Find the 85th percentile of this distribution. b). c). <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 17/29 Percentiles of a Continuous Distribution Solution: The cumulative distribution function F (x) for any number x larger or equal to .5 is given by y y F (y) Thus = 0 f (x) dx = 5 .15 e .15(x .5) dx = 1 e .15(y .5) . The (100p)th percentile satis es the equation: p = F ( p ) = 1 e .15( p .5) . Hence, p = .5 In particular, a). b). c). 1 e .15(y .5) , y .5 F (y) = 0, elsewhere ln(1 p) . .15 The median of this distribution is 5.12. The 15th percentile of this distribution is 1.58. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 18/29 The 85th percentile of this distribution is 13.15. Expected Value of a Continuous Distribution As with discrete random variables, the expected value of a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> is a measure of location. That is, it gives information about what a typical&quot; value is. It is also regarded as the balancing point of the distribution. De nition 4.5: If a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> Y has for density f (y), then the expected value is de ned as: E(Y ) = y f (y) dy . |y| f (y) dy &lt; (absolute con- provided that the integral vergence). <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 19/29 Expected Value of a Continuous Distribution Example 4.5: The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> Y with probability density function 3 (1 y 2 ) , 0 y 1 2 f (y) = 0, elsewhere Find the expected value of Y . Solution: E(Y ) = 3 3 2 y (1 y ) dy = 2 2 3 (y y ) dy = . 8 3 Hence, if gravel sales are determined week after week according to the given probability density function, then the long run average values of sales per week will be .375 ton. <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 20/29 Expected Value of a Continuous Distribution Theorem 4.4: Let Y be a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> with probability density function f (y) and let h(Y ) be any function of Y . Then, distribution function F (y). E h(Y ) = h(y) f (y) dy . In particular, E Y2 = y 2 f (y) dy . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 21/29 Variance of a Continuous Distribution De nition 4.6: Let Y be a <a href="/keyword/continuous-random-variable/" >continuous random variable</a> with the probability density function f (y) and mean E(Y ) = . The variance, 2 , of the random variable Y is de ned as 2 = V (X) = E (Y )2 = (y )2 f (y) dy . 2 In other words, V (Y ) = E(Y 2 ) E(Y ) . The standard deviation, , of the random variable Y is de ned as = 2 . <a href="/keyword/continuous-random-variable/" >continuous random variable</a> s: Part I p. 22/29 Variance of a Continuous Distribution Example 4.6: Example 4.5 revisited. Calculate the standard deviation of Y . Solution: We have E(Y ) = 2 3 2 2 3 y (1 y ) dy = 2 2 1 (y y ) dy = . 5 2 4 2 Thus, V (Y ) = E(Y 2 ) E(Y ) and = .24...