Continuous Random Variables and Probability Distributions

Continuous Random Variables and Probability Distributions Instructor: Lingsong Zhang 1 4.1 Probability Density Functions Probability Density Functions Recall from Chapter 3 that a random variable X is continuous if 1. possible values comprise either a single interval on the number line or a union of disjoint intervals; 2. P (X = c) = 0 for any number c that is a possible value of X. Example 1. If in the study of the ecology of a lake, we make depth measurements at randomly chosen locations, then X = the depth at such a location is a continuous rv. Probability Density Functions One might argue that although in principle variables such as height, weight, and temperature are continuous, in practice the limitations of our measuring instruments restrict us to a discrete (though sometimes very finely subdivided) world. However, continuous models often approximate real-world situations very well, and continuous mathematics (the calculus) is frequently easier to work with than mathematics of discrete variables and distributions. Probability Distributions for Continuous Variables Suppose the variable X of interest is the depth of a lake at a randomly chosen point on the surface. Let M = the maximum depth (in meters), so that any number in the interval [0, M] is a possible value of X. If we discretize X by measuring depth to the nearest meter, then possible values are nonnegative integers less than or equal to M. The resulting discrete distribution of depth can be pictured using a probability histogram. 1

Probability Distributions for Continuous Variables If we draw the histogram so that the area of the rectangle above any possible integer k is the proportion of the lake whose depth is (to the nearest meter) k, then the total area of all rectangles is 1. If we continue in this way to measure depth more and more finely, the resulting sequence of histograms approaches a smooth curve. Probability Distributions for Continuous Variables Because for each histogram the total area of all rectangles equals 1, the total area under the smooth curve is also 1. The probability that the depth at a randomly chosen point is between a and b is just the area under the smooth curve between a and b. It is exactly a smooth curve of the type pictured earlier that specifies a continuous probability distribution. Probability Distributions for Continuous Variables Definition 2. Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a b, P (a X b) = b a f(x)dx The graph of f(x) is often referred to as the density curve Probability Distributions for Continuous Variables For f(x) to be a legitimate pdf, it must satisfy the following two conditions: 1. f(x) 0 for all x. 2. f(x)dx = area under the entire graph of f(x) =1 2

Example 3. The direction of an imperfection with respect to a reference line on a circular object such as a tire, brake rotor, or flywheel is, in general, subject to uncertainty. Consider the reference line connecting the valve stem on a tire to the center point, and let X be the angle measured clockwise to the location of an imperfection. One possible pdf for X is { 1 0 x < 360 f(x) = 360 0 otherwise What is the probability that 90 X 180? Example 180 P (90 X 180) = 90 = x 360 1 360 dx x=180 x=90 = 1 4 =.25 Probability Distributions for Continuous Variables Definition 4. A continuous rv X is said to have a uniform distribution on the interval [A, B] if the pdf of X is f(x; A, B) = { 1 B A A X B 0 otherwise Example 5. Time headway in traffic flow is the elapsed time between the time that one car finishes passing a fixed point and the instant that the next car begins to pass that point. Let X = the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy flow. The following pdf of X is essentially the one suggested in the Statistical Properties of Freeway Traffic (Transp. Res., vol. 11: 221-228): {.15e.15(x.5) x.5 f(x) = 0 otherwise 3

Example There is no density associated with headway times less than.5, and headway density decreases rapidly (exponentially fast) as x increases from.5. Example Clearly f(x) 0. Check whether Use Then a f(x)dx = f(x)dx = 1. e kx dx = 1 k e ka. 0.5.15e.15(x.5) dx =.15e.075 e.15x dx =.15e.075 = 1.5 1.15 e (.15)(.5) Example What is the probability that headway time is at most 5 sec is P (X 5) = = 5 5 =....5 =.491 f(x)dx.15e.15 (x.5)dx 4

2 4.2 Cumulative Distribution Functions and Expected Values The Cumulative Distribution Function The cumulative distribution function (cdf) F (x) for a discrete rv X gives, for any specified number x, the probability P (X x). It is obtained by summing the pmf p(y) over all possible values y satisfying y x. The cdf of a continuous rv gives the same probabilities P (X x) and is obtained by integrating the pdf f(y) between the limits and x. The Cumulative Distribution Function Definition 6. The cumulative distribution function F (x) for a continuous rv X is defined for every number x by F (x) = P (X x) = inf x f(y)dy. Example 7. Let X, the thickness of a certain metal sheet, have a uniform distribution on [A, B]. The density function is plotted in Calculate the cumulative density function. 5

Example (continued) For x < A, F (x) = 0; and for x B, F (x) = 1. Then for A x B, we have x F (x) = f(y)dy = y=x 1 = B A y = x A B A. y=a x A 1 B A dy Example (continued) The entire cdf is The graph of this cdf is F (x) = 0 x < A A x < B 1 x B x A B A Use F (x) to compute probabilities Proposition 1. Let X be a continuous rv with pdf f(x) and cdf F (x). Then for any number a, P (X > a) = 1 F (a) and for any two numbers a and b with a < b, P (a X b) = F (b) F (a) Example 8. Suppose the pdf of the magnitude X of a dynamic load on a bridge (in newtons) is { 1 f(x) = + 3x 0 x 2 8 8 0 otherwise For any number x between 0 and 2, Thus F (x) = x f(y)dy = F (x) = x 0 x + 3 8 ( 1 8 + 3 ) 8 y dy = x 8 + 3 16 x2 0 x < 0 16 x2 0 x 2 1 x > 2 6

Example What is the probability that the load is between 1 and 1.5? What is the probability that the load exceeds 1? Obtaining f(x) from F (x) Proposition 2. If X is a continuous rv with pdf f(x) and cdf F (x), then at every x at which the derivative F (x) exists, then F (x) = f(x). Example 9. When X has a uniform distribution, F (x) is differentiable except at x = A and x = B, where the graph of F (x) has sharp corners. 1. F (x) = 1 for x > B, and F (x) = 0 for x < A, thus, F (x) = 0 = f(x) for such x. 2. When A < x < B, we have F (x) = d dx Percentiles of a Continuous Distribution ( ) x A = 1 B A B A = f(x). Percentile: When we say that an individual s test score was at the 85th percentile of the population, we mean that 85% of all population scores were below that score and 15% were above. Definition 10. Let p be a number between 0 and 1. The (100p)th percentile of the distribution of a continuous rv X, denoted by η(p), is defined by Percentiles p = F (η(p)) = η(p) f(y)dy η(.75) of X, the 75th percentile, is such that the area under the graph of f(x) to the left of η(.75) is.75 7

Example 11. The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous rv X with pdf { 3 f(x) = (1 2 x2 ) 0 x 1 0 otherwise So what is the cdf for X? What is the 50% percentile of X? The cdf of sales for any x between 0 and 1 is F (x) = x 0 3 2 (1 y2 )dy = 3 2 (y y3 3 ) y=x y=0 = 3 2 ) (x x3. 3 Example (continued) The graph of both f(x) and F (x) are Example (continued) The 100pth percentile of this distribution will satisfy the equation p = F (η(p)) = 3 ) (η(p) η(p)3 2 3 Which leads to η(p) 3 3η(p) + 2p = 0 When p =.5, solve the equation will lead to η(.5) =.347. Thus, in the long run 50% of all weeks will result in sales of less than.347 ton and 50% in more than.347 ton. Percentiles Definition 12. The median of a continuous distribution, denoted by µ, is the 50th percentile, so µ satisfies.5 = F ( µ). That is, half the area under the density curve is to the left of µ and half is to the right of µ. Definition 13. A continuous distribution whose pdf is symmetric, if the graph of the pdf to the left of some point is a mirror image of the graph to the right of that point. Symmetric distribution will has median µ equal to the point of symmetry. 8

Examples Note that the error in a measurement of a physical quantity is often assumed to have a symmetric distribution. (Normal distribution, which will be discussed later) Expected values For discrete rv X, E(X) := x x p(x). For continuous rv X, we will replace summation by integration, pmf by pdf. Definition 14. The expected or mean value of a continuous rv X with pdf f(x) is µ X = E(X) = x f(x)dx. Example 15. The pdf of weekly gravel sales X was { 3 f(x) = (1 2 x2 ) 0 x 1 0 otherwise What is the expected value of X? E(X) = = 3 2 1 0 1 xf(x)dx = x 3 0 2 (1 x2 )dx (x x 3 )dx = 3 ( ) x 2 2 2 x4 x=1 = 3 4 x=0 8 Expected value It is similar to the discrete case, the E(h(X)) can be easily derived in the following proposition Proposition 3. If X is a continuous rv with pdf f(x), and h(x) is any function of X, then E[h(X)] = µ h(x) = h(x) f(x)dx. 9

Example 16. Two species are competing in a region for control of a limited amount of a certain resource. Let X = the proportion of the resource controlled by species 1 and suppose X has pdf { 1 0 x 1 f(x) = 0 otherwise which is a uniform distribution on [0, 1]. Then the species that controls the majority of this resource controls the amount { 1 X 0 X < 1/2 h(x) = max(x, 1 X) = X 1/2 X 1 What is the expected value of E(h(X))? Example E[h(X)] = = = = 3 4 1 0 1/2 0 max(x, 1 x) f(x)dx max(x, 1 x) 1dx (1 x) 1dx + 1 1/2 x 1dx Expected values Definition 17. The variance of a continuous random variable X with pdf f(x) and mean value µ is defined as σ 2 X = V (X) = The standard deviation (SD) of X is σ X = V (X). (x µ) 2 f(x)dx = E[(X µ) 2 ] Proposition 4. Calculation formula: V (X) = E(X 2 ) [E(X)] 2. Example 18. For the weekly gravel sales, we computed E(X) = 3, since 8 E(X 2 ) = x 2 f(x)dx = 0 1 x 2 3 2 (1 x2 )dx so = And thus σ X = V (X) =.244. 1 0 3 2 (x2 x 4 )dx = = 1 5 V (X) = 1 5 ( 3 8) 2 = =.059 10

Linear transformation Similarly as the discrete cases, if h(x) is a linear function E(h(X)) = E(aX + b) = ae(x) + b. For the variance and standard deviation, we have. V (ax + b) = a 2 V (X), σ ax+b = a σ X 3 4.3 The Normal Distribution The Normal Distribution The normal distribution is the most important one in all of probability and statistics. Many numerical populations have distributions that can be fit very closely by an appropriate normal curve. Examples include heights, weights, and other physical characteristics, measurement errors in scientific experiments, anthropometric measurements on fossils, reaction times in psychological experiments, measurements of intelligence and aptitude, scores on various tests, and numerous economic measures and indicators. The Normal Distribution Definition 19. A continuous rv X is said to have a normal distribution with parameters µ and σ (or µ and σ 2 ), where < µ < and 0 < σ, if the pdf of X is f(x; µ, σ) = 1 2πσ e (x µ)2 /(2σ 2), < x < e =the base of the natural logarithm system 2.71828. We typically write X N(µ, σ 2 ). It can be shown by some complicated calculus that f(x; µ, σ)dx = 1, E(X) = µ, V (X) = σ 2. 11

Visual impression Each density curve is symmetric about µ and bell-shaped. µ - both mean and median σ: the distance from µ to the inflection points of the curve (the points at which the curve changes from turning downward to turning upward) The standard Normal distribution Computation of P (a X b) for X N(µ, σ 2 ) is P (a X b) = Use some transformation, we will have P (a X b) = = b b a a (b µ)/σ 1 2πσ e (x µ)2 /(2σ 2) dx ( ) 1 x µ e ((x µ)/σ)2 /2 d 2π σ (a µ)/σ 1 2π e z2 /2 dz Here we replace (x µ)/σ = z, and the domain of a x b, becomes to (a µ)/σ < z < (b µ)/σ Standard Normal Distribution Definition 20. The normal distribution with parameter values µ = 0 and σ = 1 is called the standard normal distribution. A random variable having a standard normal distribution is called a standard normal random variable and will be denoted by Z. The pdf of Z is f(z; 0, 1) = 1 2π e z2 /2, < z < The graph of f(z; 0, 1) is called the standard normal (or z) curve. Its inflection points are at 1 and -1. The cdf of Z is P (Z z) = z f(y; 0, 1)dy, which we will denote by Φ(z). 12

The Standard Normal Distribution The standard normal distribution almost never serves as a model for a naturally arising population. Instead, it is a reference distribution from which information about other normal distributions can be obtained. Appendix Table A.3 gives Φ(z) = P (Z z), the area under the standard normal density curve to the left of z, for z = 3.49, 3.48,, 3.48, 3.49. The Standard Normal Distribution Figure 4.14 illustrates the type of cumulative area (probability) tabulated in Table A.3. From this table, various other probabilities involving Z can be calculated. Example 21. Let s determine the following standard normal probabilities: 1. P (Z 1.25), 2. P (Z > 1.25), 3. P (Z 1.25), 4. P (.38 Z 1.25). Percentiles of the Standard Normal Distribution For any p between 0 and 1, Appendix Table A.3 can be used to obtain the (100p)th percentile of the standard normal distribution. Example 22. Find the 99th percentile of the standard normal distribution. This is to identify z-value, such that P (Z z) =.9900 13

Percentiles Symmetry of N(0, 1) is helpful. What if p is not in the table? Use the nearest p in the table (use this one in our class) Use linear interpolation R can help us to find the percentile (will be discussed later) Upper α point for Standard Normal Distribution In statistical inference, we will need the values on the horizontal z axis that capture certain small tail areas under the standard normal curve. Notation 1. Z α will denote the value on the z axis for which α of the area under the z curve lies to the right of z α. z critical values Since α of the area under the z curve lies to the right of z α, 1 α of the area lies to its left. Thus z α is the 100(1 α)thpercentile of the standard normal distribution. By symmetry the area under the standard normal curve to the left of z α is also α. The z α s are usually referred to as z critical values. Useful z critical values 14

Nonstandard Normal Distributions When X N(µ, σ 2 ), probabilities involving X are computed by standardizing. The standardized variable is (X µ)/σ. Subtracting µ shifts the mean from µ to zero, and then dividing by σ scales the variable so that the standard deviation is 1 rather than σ. Proposition 5. If X has a normal distribution with mean µ and standard deviation σ, then has a standard normal distribution Z = X µ σ Example 23. The time that it takes a driver to react to the brake lights on a decelerating vehicle is critical in helping to avoid rear-end collisions. The article Fast-Rise Brake Lamp as a Collision-Prevention Device (Ergonomics, 1993: 391-395) suggests that reaction time for an in-traffic response to a brake signal from standard brake lights can be modeled with a normal distribution having mean value 1.25 sec and standard deviation of.46 sec. What is the probability that reaction time is between 1.00 sec and 1.75 sec? Percentiles of an Arbitrary Normal Distribution The (100p)th percentile of a normal distribution with mean µ and standard deviation σ is easily related to the (100p)th percentile of the standard normal distribution. Proposition 6. (100p)th percentile for N(µ, σ 2 ) is µ + Cσ where C is the 100pth percentile for standard normal distribution. Example 24. The amount of distilled water dispensed by a certain machine is normally distributed with mean value 64 oz and standard deviation.78 oz. What container size c will ensure that overflow occurs only.5% of the time? The Normal Distribution and Discrete Populations The normal distribution is often used as an approximation to the distribution of values in a discrete population. In such situations, extra care should be taken to ensure that probabilities are computed in an accurate manner. Example 25. IQ in a particular population (as measured by a standard test) is known to be approximately normally distributed with µ = 100 and σ = 15. What is the probability that a randomly selected individual has an IQ of at least 125? We need a new technique called continuity correction, when we use continuous random variable to approximate a discrete random variable. P (X 125) P (X > 124.5) 15

Normal approximation to Binomial Recall that the mean value and standard deviation of a binomial random variable X are µ X = np and σ X = np(1 p) respectively. Binomial probability histogram for the binomial distribution with n = 20, p =.6, i.e., µ X = 12 and σ = 20(.6)(.4) = 2.19 Normal approximation to Binomial Distribution A normal curve with this µ and σ has been superimposed on the probability histogram. Although the probability histogram is a bit skewed (because p.5), the normal curve gives a very good approximation, especially in the middle part of the picture. The area of any rectangle (probability of any particular X value) except those in the extreme tails can be accurately approximated by the corresponding normal curve area. Approximating the Binomial Distribution For example where by using normal curve, we have P (X = 10) = B(10; 20,.6) B(9; 20,.6) =.117 P (9.5 X 10.5) = P ( 1.14 Z.68) =.1212 More generally, as long as the binomial probability histogram is not too skewed, binomial probabilities can be well approximated by normal curve areas. Approximating the Binomial Distribution Proposition 7. Let X be a binomial rv based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, X has approximately a normal distribution with µ = np and σ = np(1 p). In practice, the approximation is adequate provided that both np 10 and n(1 p) 10, since there is then enough symmetry in the underlying binomial distribution. 16

Example 26. Suppose that 25% of all students at a large public university receive financial aid. Let X be the number of students in a random sample of size 50 who receive financial aid, so that p =.25. Then µ = 12.5 and σ = 3.06. Since np = 50(.25) = 12.5 10 and n(1 p) = 37.5 10, the approximation can safely be applied. What is the probability that at most 10 students receive aid? What is the probability that between 5 and 15 (both inclusive) of the selected students receive aid? 4 4.4 Other distributions Other important distributions for continuous random variables Exponential distribution Gamma distribution Chi-square distribution t-distribution Beta distribution Weibull distribution Lognormal distribution Exponential Distribution The family of exponential distributions provides probability models that are very widely used in engineering and science disciplines. Definition 27. X is said to have an exponential distribution with parameter λ (λ > 0) if the pdf of X is { λe λx λ > 0 f(x; λ) = 0 otherwise By using integration by parts, we will get the the following useful properties. E(X) = 0 xλe λx = 1 λ, V (X) = 1 λ 2 Note that Both mean and standard deviation are 1/λ. 17

PDF of exponential distribution The CDF can be calculated easily as F (x; λ) = { 1 e λx x 0 0 x < 0 Example 28. The article Probabilistic Fatigue Evaluation of Riveted Railway Bridges (J. of Bridge Engr., 2008: 237-244) suggested the exponential distribution with mean value 6 MPa as a model for the distribution of stress range in certain bridge connections. Let s assume that this is in fact the true model. Then E(X) = 1/λ = 6 implies that λ =.1667. What is the probability that stress range is at most 10 MPa? What is the probability that stress range is between 5 and 10? P (X 10) = F (10;.1667) = 1 e (.1667)(10) = P (5 X < 10) = F (10;.1667) F (5;.1667) = More on Exponential distribution The exponential distribution is frequently used as a model for the distribution of times between the occurrence of successive events, such as customers arriving at a service facility or calls coming in to a switchboard. It has a strong connection with Poisson distribution and Poisson process. For a given time period t, if the interarrival time between two consecutive occurrences is from an exponential distribution, then the number of occurrences within this period t is from a Poisson distribution. 18

More on Exponential distribution Another reason that leads to the popularity of exponential distribution is the memoryless property of the exponential distribution. That is why the exponential distribution is commonly used to model the distribution of component lifetime. Example 29. Suppose component lifetime is exponentially distributed with parameter λ. After putting the component into service, we leave for a period of t 0 hours and then return to find the component still working; what now is the probability that it lasts at least an additional t hours? In symbols, we wish to find P (X t + t 0 X t 0 ). Memoryless property P (X t + t 0 X t 0 ) = P [(X t + t 0) (X t 0 )] P (X t 0 ) = P [(X t + t 0)] P (X t 0 ) = 1 F (t + t 0; λ) = e λt 1 F (t 0 ; λ) Which is the same as P (X t) = e λt. Thus the distribution of additional lifetime is exactly the same as the original distribution of lifetime, so at each point in time the component shows no effect of wear. In other words, the distribution of remaining lifetime is independent of current age. Gamma Function To define the family of gamma distributions, we first need to introduce a function that plays an important role in many branches of mathematics. Definition 30. For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx. Important properties of Gamma Function The most important properties of the gamma function are the following: 1. For any α > 1, Γ(α) = (α 1) Γ(α 1). 2. For any integer n, Γ(n) = (n 1)!. 3. Γ(1/2) = π. Take Γ(a) = 0 x α 1 e x dx, Let f(x; α) = { x α 1 e x Γ(α) x 0 0 otherwise Then f(x; α) 0, and 0 f(x; α)dx = 1. So f(x; α) is a pdf. 19

Gamma Distribution Definition 31. A continuous random variable X is said to have a gamma distribution if the pdf of X is { 1 β f(x; α, β) = α Γ(α) xα 1 e x/β x 0 0 otherwise where the parameters α and β satisfy α > 0, β > 0. The standard gamma distribution has β = 1, so the pdf of a standard gamma rv is given in the former slide. Visual impression The exponential distribution results from taking α = 1 and β = 1/λ. (a) several Gamma distribution, (b) standard Gamma distribution β = 1. Gamma distribution For the standard pdf, when α 1, f(x; α), is strictly decreasing as x increases from 0; when α > 1, f(x; α) rises from 0 at x = 0 to a maximum and then decreases. The parameter β is called the scale parameter because values other than 1 either stretch or compress the pdf in the x direction. The mean and variance of a random variable X having the gamma distribution f(x; α, β) are E(X) = µ = αβ, V (X) = σ 2 = αβ 2 There are extensive tables of available; in Appendix Table A.4, we present a small tabulation for α = 1, 2,, 10 and x = 1, 2,, 15. Example 32. Suppose the reaction time X of a randomly selected individual to a certain stimulus has a standard gamma distribution with α = 2. What are the probabilities of P (3 X 5), P (X > 4)? Arbitrary Gamma Distributions Definition 33. When X is a standard gamma rv, the cdf of X, F (x; α) = x is called the incomplete gamma function. 0 y α 1 e y dy, x > 0 Γ(α) 20

Arbitrary Gamma Distributions The incomplete gamma function can also be used to compute probabilities involving nonstandard gamma distributions. These probabilities can also be obtained almost instantaneously from various software packages. Proposition 8. Let X have a Gamma distribution with parameters α and β. Then for any x > 0, the CDF of X is given by ( ) x P (X x) = F (x; α, β) = F β ; α where F ( ; α) is the incomplete gamma function. Example 34. Suppose the survival time X in weeks of a randomly selected male mouse exposed to 240 rads of gamma radiation has a gamma distribution with α = 8 and β = 15. What is the probability that a mouse survives between 60 and 120 weeks? P (60 X 120) = P (X 120) P (X < 60) = F (120/15; 8) F (60/15; 8) = F (8; 8) F (4; 8) =.547.051 =.496 Chi-square distribution The chi-squared distribution is important because it is the basis for a number of procedures in statistical inference. The central role played by the chi-squared distribution in inference springs from its relationship to normal distributions. Definition 35. Let v be a positive integer. Then a random variable X is said to have a chi-squared distribution with parameter v if the pdf of X is the gamma density with α = v/2 and β = 2. The pdf of a chi-squared rv is thus { 1 f(x; v) = 2 v/2 Γ(v/2) x(v/2) 1 e x/2 x 0 0 x < 0 The parameter v is called the number of degrees of freedom (df) of X. The symbol χ 2 is often used in place of Chi-squared.. 21

The Weibull distribution The family of Weibull distributions was introduced by the Swedish physicist Waloddi Weibull in 1939; his 1951 article A Statistical Distribution Function of Wide Applicability (J. of Applied Mechanics, vol. 18: 293-297) discusses a number of applications. Definition 36. A random variable X is said to have a Weibull distribution with parameters α and β (α > 0, β > 0) if the pdf of X is { α x α 1 e (x/β)α x 0 f(x; α, β) = β α 0 x < 0 Weibull distribution When α = 1, the pdf reduces to the exponential distribution (with λ = 1/β), so the exponential distribution is a special case of both the gamma and Weibull distributions. Weibull distribution β is called a scale parameter, and α is called the shape parameter. The mean and variance will be µ = E(X) = βγ ( 1 + 1 α ), ( σ 2 = V (X) = β {Γ 2 1 + 2 ) [ ( Γ 1 + 1 )] } 2 α α The cdf is { F (x; α, β) = 0 x < 0 1 e (x/β)α x 0 In recent years the Weibull distribution has been used to model engine emissions of various pollutants. Example 37. Let X denote the amount of NO x emission (g/gal) from a randomly selected four-stroke engine of a certain type, and suppose that X has a Weibull distribution with α = 2 and β = 10 (suggested by information in the article Quantification of Variability and Uncertainty in Lawn and Garden Equipment NO x and Total Hydrocarbon Emission Factors, J. of the Air and Waste Management Assoc., 2002: 435-448). What are the probabilities of P (X 10) and P (X 25)? What is the 95th percentile of the emission distribution? 22

Lognormal distribution Definition 38. A nonnegative rv X is said to have a lognormal distribution if the rv Y = ln(x) has a normal distribution. The pdf for lognormal is f(x; µ, σ) = { 1 2πσx e [ln(x) µ]2 /(2σ 2 ) x 0 0 x < 0 Note that µ and σ here are not the mean and standard deviation of X, but ln(x). E(X) = e µ+σ2 /2 V (X) = e 2µ+σ2 (e σ2 1) Lognormal distribution It is skewed right (positive skew)! Lognormal distribution The cdf of X can be written as a function of the cdf of standard normal distribution Φ(x) for x > 0. F (x; µ, σ) = P (X x) = P (ln(x) ln(x)) = P (Z ln(x) µ ) ( σ ) ln(x) µ = Φ σ 23

Example 39. According to the article Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines (Corrosion, 2009: 332-342), the lognormal distribution has been reported as the best option for describing the distribution of maximum pit depth data from cast iron pipes in soil. The authors suggest that a lognormal distribution with µ =.353 and σ =.754 is appropriate for maximum pit depth (mm) of buried pipelines. What are the mean and variance of pit depth? What are the probability of P (1 X 2)? What value c is such that only 1% of all specimens have a maximum pit depth exceeding c? The Beta Distribution Definition 40. A random variable X is said to have a beta distribution with parameters α, β (both positive), A, and B if the pdf of X is { 1 Γ(α+β) ( x A ) α 1 ( B x β 1 f(x; α, β, A, B) = B A Γ(α)Γ(β) B A B A) A x B 0 otherwise. The case A = 0, B = 1 gives the standard beta distribution. Beta Distribution Visual impression of several standard beta distribution Beta Distribution Mean and variance of Beta distribution µ = A + (B A) α α + β, σ2 = (B A) 2 αβ (α + β) 2 (α + β + 1) Example 41. Project managers often use a method labeled PERT: for program evaluation and review technique: to coordinate the various activities making up a large project. 24

-1.91-1.25 -.75 -.53.20.35.72.87 1.40 1.56 (One successful application was in the construction of the Apollo spacecraft.) A standard assumption in PERT analysis is that the time necessary to complete any particular activity once it has been started has a beta distribution with A = the optimistic time (if everything goes well) and B = the pessimistic time (if everything goes badly). Example 42. Suppose that in constructing a single-family house, the time X (in days) necessary for laying the foundation has a beta distribution with A = 2, B = 5, α = 2, and β = 3. What is the mean of this distribution? What is the probability of P (X 3)? 5 4.5 Probability Plots Probability plots We have assumed X f(x), i.e., from some specific continuous distribution. Given now we have a data with sample size n, how can we check whether the data is from this distribution or not. Example 43. The value of a certain physical constant is known to an experimenter. The experimenter makes n = 10 independent measurements of this value using a particular measurement device and records the resulting measurement errors. These observations appear in the accompanying table. Is it plausible that the random variable measurement error has a standard normal distribution? Scenario Let us assume we have n observations Calculate the population percentile for f(x) at p i = i/n. Denote them as {x i } s. Calculate the sample percentile for the data we observed at p i = i/n. Denote them as {y i } s. It is expected that when the data is from f(x), y i x i. We will plot (x i, y i ) in a 2-d plane, and check whether the dots lie closed to the 45 line. Sample percentiles Recall: We know that the (100p)th percentile of a continuous distribution with cdf F ( ) is the number η(p) that satisfies F (η(p)) = p. That is, η(p) is the number on the measurement scale such that the area under the density curve to the left of η(p) is p. 25

-1.91-1.25 -.75 -.53.20.35.72.87 1.40 1.56 Roughly speaking, sample percentiles are defined in the same way that percentiles of a population distribution are defined. However, we may not be able to calculate all p s for the sample percentiles. Let us assume that we have n observations. (e.g. n = 30). Sample percentile Definition 44. Order the n sample observations from smallest to largest. Then the ith smallest observation in the list is taken to be the [100(i.5)/n]th sample percentile. Once the percentage values 100(i.5)/n (i = 1, 2,, n) have been calculated, sample percentiles corresponding to intermediate percentages can be obtained by linear interpolation. In R, we can use routine quantile to calculate any sample percentiles. A probability plot Suppose now that for percentages 100(i.5)/n (i = 1, 2,, n) the percentiles are determined for a specified population distribution whose plausibility is being investigated. If the sample was actually selected from the specified distribution, the sample percentiles (ordered sample observations) should be reasonably close to the corresponding population distribution percentiles. We will draw x i = [100(i.5)/n]th percentile of the distribution y i = ith smallest sample observation in the data on a two-dimensional coordinate system, i = 1,, n. Probability plot If the sample percentiles are close to the corresponding population distribution percentiles, the first number in each pair will be roughly equal to the second number. The plotted points will then fall close to a 45 line. Substantial deviations of the plotted points from a 45 line cast doubt on the assumption that the distribution under consideration is the correct one. Example 45. The n = 10 measurement errors appear in the accompanying table. plausible that the measurement error is from N(0, 1)? Is it 26

Example (continued) Calculate the needed standard normal (z) percentiles, will lead to the following points in the probability plot: (-1.645, -1.91), (-1.037, -1.25), and (1.645, 1.56). The 45 line gives a good fit to the middle part of the sample but not to the extremes. (Previous plot) Probably a well-defined S-shaped appearance. Typical Probability plots Note that probability plot is not a formal test. Kolmogorov-Smirnov test in Statistics. Refer to goodness-of-fit test, such as 27