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Chapter 5. Continuous Random Variables Uniform Random Variable • X is a uniform random variable on the interval (α, β) if its probability function is given by 1 if x f ( x) otherwise 0 0 x F ( x) 1 x x x 2 • Ex 3a. Let X be uniformly distributed over (α, β). Find (a) E[X] and (b) Var(X). (a ) E[ X ] xf ( x)dx x dx 2 2 2( ) 2 • The expected value of a random variable uniformly distributed over some interval is equal to the midpoint of that interval. 3 (b) To find Var( X ), we first calculate E[ X 2 ] E[ X ] 2 1 2 x dx 3 3 3( ) 2 2 3 2 2 ( ) 2 Var ( X ) 3 4 4 • Ex 3b. If X is uniformly distributed over (0, 10), calculate the probability that (a) X < 3, (b) X > 6, and (c) 3 < X < 8. • f(x) = 1/10, 0 ≤ x ≤ 10. 3 (a ) P( X 3) (1 / 10)dx 3 / 10 0 10 (b) P( X 6) (1 / 10)dx 4 / 10 6 8 (c) P(3 X 8) (1 / 10)dx 1 / 2 3 5 • Ex 3c. Buses arrive at a specified stop at 15-minute intervals starting at 7AM. That is, they arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time that is uniformly distributed between 7 and 7:30, find the probability that he waits (a) less than 5 minutes for a bus; (b) more than 10 minutes for a bus. f(x) = 1/30, 0 ≤ x ≤ 30. 30 1 1 (a ) P(10 X 15) P(25 X 30) dx dx 1 / 3 10 30 25 30 15 (b) P(0 X 5) P(15 X 20) 5 0 20 1 1 dx dx 1 / 3 15 30 30 6 Normal Random Variables 1 ( x )2 / 2 2 f ( x) e 2 • The distribution associated with Normal random variable is called Normal distribution. • It was first studied by French mathematician Abraham DeMoivre. • Carl Friedrich Gauss analyzed astronomical data using Normal distribution and defined the equation of its probability density function. – The distribution is also called Gaussian distribution. 7 Importance of Normal Distribution • Describes many random processes of continuous phenomena – Height of a man, velocity of a molecule in gas, error made in measuring a physical quantity. • Can be used to approximate discrete probability distributions – Example: binomial • Basis for classical statistical inference – Central limit theorem. 8 Normal Distribution • ‘Bell-shaped & symmetrical f(X) • Random variable has infinite range • Mean measures the center of the distribution • Standard deviation measures the spread of the distribution X 1 ( x )2 / 2 2 f ( x) e 2 = Mean of random variable x = Standard deviation = 3.14159… e = 2.71828… 9 Effect of Varying Parameters ( & ) f(X) B A C X A: = 3.7 =1 B: = 3.7 =0.5 C: = 5.3 =1 10 • If X is normally distributed with parameter μ and σ2, then Y = aX + b is normally distributed with parameters aμ + b and (aσ)2. FY ( x) P(Y x) P(aX b x) x b P( X ) a xb FX ( ) a 1 x b fX ( ) a a 1 x b exp[ ( ) 2 / 2 2 ] a 2 a 1 exp[ ( x b a ) 2 / 2(a ) 2 ] 2 a 1 exp[ ( x (a b)) 2 / 2(a ) 2 ] 2 a fY ( x) 11 Standard Normal • Normal distribution with parameter (0, 1), N(0,1), is also called standard normal. 1 x2 / 2 f ( x) e 2 • If X is N(μ, σ2), then Z = (X - μ)/σ is standard normal. 12 • Ex 4a. Find E[X] and Var[X] when X is a normal random variable with parameters μ and σ2. For standard normal, we have E[ Z ] xfZ ( x)dx 1 x2 / 2 xe dx 2 1 x2 / 2 e | 2 0 13 Var ( Z ) E[ Z 2 ] 1 2 x2 / 2 xe dx 2 Integratio n by parts (u x,dv xe ) gives 1 x2 / 2 x2 / 2 Var ( Z ) ( xe | e dx) 2 1 x2 / 2 e dx 2 1 -x 2 / 2 Integratio n by parts : b a b f ( x) g ( x)dx [ f ( x) g ( x)] | f ' ( x) g ( x)dx ' b a a 14 Because X = μ + σZ, E[X] = μ + σE[Z] = μ Var(X) = σ2Var(Z) = σ2. 15 The cumulative distributi on function of a standard normal random variable is denoted by Φ( x). 1 x y2 / 2 ( x) e dy 2 For negative values of x: ( x) 1 ( x) - x For standard normal: P(Z ≤ -x) = P(Z > x) For X ~ N(μ, σ) FX (a) P( X a) P( X a ) ( a ) 16 • Φ(x) can be found from the table for standard normal distribution. 17 • Ex 4b. If X is a normal random variable with parameter μ = 3 and σ2 = 9, find (a) P(2 < X < 5); (b) P(X > 0); (c) P(|X-3| > 6). (a ) P(2 X 5) P( 23 X 3 53 ) 3 3 3 1 2 P( Z ) 3 3 2 1 3 3 2 1 1 3 3 .7486 (1 .6293) .3779 18 X 3 03 (b) P( X 0) P( ) 3 3 P( Z 1) 1 (1) (1) .8413 ( c) P(| X 3 | 6) P( X 9) P( X 3) X 3 93 X 3 33 ) P( ) 3 3 3 3 P( Z 2) P( Z 2) P( 1 (2) (2) 2[1 (2)] .0456 19 • Ex 4c. An examination is often regarded as being good if the test scores of those taking the examination can be approximated by a normal density function. • The instructor often uses the test scores to estimate the normal parameters μ, and σ2 and then assign the letter grade A to those whose test score is greater than μ + σ, B to those whose score is between μ and μ + σ, C to those whose score is between μ - σ and μ, D to those whose score is between μ - 2σ and μ - σ, and F to those getting a score below μ - 2σ. This is sometimes referred to as grading “on the curve”. 20 P( X ) P( X 1) 1 (1) .1587 P ( X ) P (0 X P ( X ) P (1 X P ( 2 X ) P(2 P ( X 2 ) P( X 1) (1) (0) .3413 0) (0) (1) .3413 X 1) (2) (1) .1359 2) (2) .0228 • Approximately 16 percent of the class will receive an A grade on the examination, 34 percent a B grade, 34 percent a C grade, and 14 percent a D grade; 2 percent will fail. 21 • Ex 4e. Suppose that a binary message – either 0 or 1 must be transmitted by wire from location A to location B. However, the data sent over the wire are subject to a channel noise disturbance, so to reduce the possibility of error, the value 2 is sent over the wire when the message is 1 and the value -2 is sent when the message is 0. If x, x = ±2, is the value sent at location A, then R, the value received at location B is given by R = x + N, where N is the channel noise disturbance. When the message is received at location B the receiver decodes it according to the following rule: If R ≥ .5, then 1 is concluded If R < .5, then 0 is concluded. Assuming the channel noise is normally distributed, determine the error probabilities when N is a standard normal random variable. 22 • There are two types of errors: one is that the message 1 can be incorrectly concluded to be 0, and the other that 0 is concluded to be 1. The first type of error will occur if the message is 1 and 2 + N<.5, whereas the second will occur if the message is 0 and -2 + N ≥ .5. P(error | message is 1) P( N -1.5) 1 - (1.5) .0668 P(error | message is 0) P( N 2.5) 1 - (2.5) .0062 23 Normal Approximation to the Binomial Distribution • When n is large, a binomial random variable with parameter n and p will have approximately the same distribution as a normal random variable with the same mean and variance. 24 • Ex 4f. Let X be the number of times that a fair coin, flipped 40 times, lands heads. Find the probability that X = 20. Use the normal approximation and then compare it to the exact solution. • Since binomial is discrete and normal is continuous, we need to apply a continuity correction – we compute P(i-1/2 < X < i+1/2) instead of P(X = i). P ( X 20) P (19.5 X 20.5) 19.5 20 X 20 20.5 20 ) 10 10 10 X 20 P (.16 .16) 10 (.16) (.16) .1272 P( Exact result using Binomial is : 40 1 P ( X 20) .1254 20 2 40 25 • Ex 4g. The ideal size of a first-year class at a particular college is 150 students. The college, knowing from past experience that on the average only 30 percent of those accepted for admission will actually attend, uses a policy of approving the applications of 450 students. Compute the probability that more than 150 first-year students attend this college. X 450(.3) 150.5 450(.3) P( X 150.5) P 450(.3)(.7) 450(.3)(.7) 1 (1.59) 0.0559 26 • Ex 4h. To determine the effectiveness of a certain diet in reducing the amount of cholesterol in the bloodstream, 100 people are put on the diet. After they have been on the diet for a sufficient length of time, their cholesterol count will be taken. The nutritionist running this experiment has decided to endorse the diet if at least 65 percent of the people have a lower cholesterol count after going on the diet. What is the probability that the nutritionist endorses the new diet if, in fact, it has no effect on the cholesterol level? 100 1 i 65 i 2 100 100 P( X 64.5) X (100)(1 / 2) P 2.9 100(1 / 2)(1 / 2) 1 (2.9) .0019 27 Exponential Random Variables • pdf of exponential random variable e x if x 0 f ( x) if x 0 0 • The distribution of the amount of time until some specific event occurs. – The amount of time until an earthquake occurs. 28 Exponential Distribution 29 Distributi on function : F (a) P( X a) a e x dx 0 e x |0a 1 e a • Ex 5a. Let X be an exponential random variable with parameter λ. Calculate (a) E[X] and (b) Var(X). 30 e x if x 0 f ( x) if x 0 0 E[ X n ] x n e x dx 0 Integratin g by parts (dv e x , u x n ) yields E[ X ] x e 0 n n | nx n 1e x dx n x 0 n 0 0 x n 1e x dx Integratio n by parts : b udvdx uv | a b a b a duvdx E[ X n 1 ] Letting n 1, and then n 2, gives E[ X ] 1 / E[ X 2 ] (2 / ) E[ X ] 2 / 2 2 1 Var ( X ) 2 1 / 2 2 31 • Ex 5b. Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1/10. If someone arrives immediately ahead of you at a public telephone booth, find the probability that you will have to wait (a) more than 10 minutes; (b) between 10 and 20 minutes. (a ) P( X 10) 1 F (10) 1 (1 e 10(1/10) ) e 1 F (a) 1 e a (b) P(10 X 20) P( X 20) P( X 10) F (20) F (10) e 1 e 2 32 Memory of A Random Variable • We say a nonnegative random variable X is memoryless if P( X s t | X t ) P( X s) for all s, t 0 P( X s t , X t ) P( X s) P( X t ) or P( X s t ) P( X s) P( X t ) Since e ( s t ) es et , exponentia l random variables are memoryless . 33 • Ex 5c. Consider a post office that is staffed by two clerks. Suppose that Ms. Jones is being served by one of the clerks and Mr. Brown by the other. Suppose also that Mr. Smith is told that his service will begin as soon as either Jones or Brown leaves. If the amount of time that a clerk spends with a customer is exponentially distributed with parameter λ. what is the probability that, of the three customers, Mr. Smith is the last to leave the post office? • Because exponential random variable is memoryless, the probability of Mr. Smith is the last to leave the post office is 1/2. 34 • Ex 5d. Suppose that the number of miles that a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5000-mile trip, what is the probability that he or she will be able to complete the trip without having to replace the battery? What can be said when the distribution is not exponential? The remaining lifetime of the battery is exponential with parameter 1/10 (in thousands of miles). P(remaining lifetime 5) 1 F (5) e 5 e 1/ 2 .604 If the lifetime distributi on is not exponentia l, P(lifetime t 5, lifetime t ) P(lifetime t 5 | lifetime t ) P(lifetime t ) 1 F (t 5) 1 F (t ) 35 Gamma Distribution e x (x) 1 f ( x) ( ) 0 x0 x0 ( ) e y y 1dy 0 (n) (n 1)! • Gamma distribution is often used as the distribution of the amount of time one has to wait until a total of n events has occurred. – A generalization of exponential distribution 36 Gamma Distribution 37 • Ex 6a. Let X be a gamma variable with parameters α and λ. Calculate (a) E[X] and Var(X). 1 k x 1 x e ( x ) dx 0 ( ) 1 k e x ( x) k 1dx ( ) 0 ( k ) k ( ) E[ X k ] E[ X ] E[ X 2 ] ( 1) ( ) ( 2) ( 1) 2( ) 2 Var ( X ) E[ X 2 ] ( E[ X ]) 2 2 38 Gamma Distribution – Special Cases • When α = 1, the distribution reduce to exponential. ex (x) 1 f ( x) f ( x) e x , 1 ( ) • When λ = 1, α = n/2, the distribution is also called chi-square distribution with n degree of freedom. – Model errors involved in attempting to hit a target in n dimensional space when each coordinate error is normally distributed. – Comparison of categorical data. 39 Beta Distribution 1 x a 1 (1 x)b 1 0 x 1 f ( x ) B ( a, b) 0 otherwise (a)(b) B ( a, b) ( a b) • Used to model a random phenomenon whose set of possible values is some finite interval [c, d] – which can be transformed into the interval [0, 1]. • Frequently used in Bayesian inference. 40 Beta Distribution 41 Distribution of a Function of a Random Variable • Sometimes we would like to know the distribution function of g(X) when the pdf of X is known. • Ex 7a. Let X be uniformly distributed over (0,1). What is the distribution function of the random variable Y, defined by Y = Xn? 42 • Distribution function FY ( y ) P(Y y ) P( X n y ) P ( X y1 / n ) FX ( y ) 1/ n y1 / n 0 x F ( x) 1 x x x • Probability density function fY ( y) (1 / n) y1/ n1 0 y 1 43 • Ex 7b. If X is a continuous random variable with probability density fX, then the distribution of Y = X2 is: FY ( y ) P(Y y ) P( X 2 y ) P( y X y ) FX ( y ) FX ( y ) Differenti ation yields 1 fY ( y ) [ f X ( y ) f X ( y )] 2 y 44 Summary of Chapter 5 • • • • • Continuous random variables Expectation and expectation of function of random variables Variance Uniform random variable Normal random variable – Standard normal • Other random variables – Exponential – Gamma – Beta • Distribution function of function of random variables 45