Ching-Han Hsu, BMES, National Tsing Hua University c 2015 by Ching-Han Hsu, Ph.D., BMIR Lab. = a + b 2. b a. x a b a = 12

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1 Lecture 5 Continuous Random Variables BMIR Lecture Series in Probability and Statistics Ching-Han Hsu, BMES, National Tsing Hua University c 215 by Ching-Han Hsu, Ph.D., BMIR Lab Uniform Distribution Continuous Uniform Distribution Definition 1 (Continuous Uniform Distribution). A continuous random variable X with probability density function is a continuous uniform random variable. The mean of X is E(X) f (x) 1 b a, a x b (1) b a x b a dx x 2 2(b a) b a a + b 2 The variance is V (X) b a ( ) x a+b 2 2 dx b a CDF: Continuous Uniform Distribution ) 3 ( x a+b 2 3(b a) b a (b a)2 12 The cumulative distribution function of a continuous uniform distribution X is x x 1 F(X) P(X x) f (t)dt dt. (2) b a If a x b, The complete form is F(X) F(X) x x a 1 dt (x a)/(b a). (3) b a 1 b a dt x < a x a b a a x < b 1 b x (4)

2 Probability and Statistics 2/19 Fall, 214 Example: Continuous Uniform Distribution Example 2. Let continuous random variable X denote the current measured in a thin copper in milliamperes (ma). Assume that the range of X is [,2mA], and the probability density function is f (x) 2 1.5, x 2. The mean of X is E(X) 2 x 2 dx 2 2 1mA The variance is V (X) 2 (x 1) 2 2 dx (2) mA2 What is the probability that a measurement of current is between 5 and 1 ma? P(5 < x 1)? P(5 < x 1) F(1) F(5) Normal Distribution Normal (Gaussian) Distribution Normal distribution is the most widely used model. For a repeated random experiment, the average of the outcomes tends to have a normal distribution (central limit theorem). The density function of a normal random variable is characterized by two parameters: mean µ and variance σ 2 as shown in Fig. 1. Each curve is symmetric and bell-shaped. µ determines the center and σ 2 determines the width. 5.5 Normal (Gaussian) Distribution Figure 1: Normal probability density functions for selected values of the parameters µ and σ

3 Probability and Statistics 3/19 Fall, 214 Normal (Gaussian) Distribution Definition 3 (Normal (Gaussian) Distribution). A continuous random variable X with probability density function f (x; µ,σ) 1 (x µ)2 e 2σ 2, < x < (5) 2πσ 2 1 σ (x µ) 2 2π e 2σ 2, < x < (6) is a normal (Gaussian) random variable with parameters µ, where < µ <, and σ >. Theorem 4. The distribution is denoted by N(µ,σ). And the mean and variance of X are equal to E(X) µ and V (X) σ 2, respectively. 5.7 Normal (Gaussian) Distribution Figure 2: Probability density function of a normal random variable with mean µ and σ 2. About 68% of the population is in the interval µ ± σ. About 95% of the population is in the interval µ ± 2σ. About 99.7% of the population is in the interval µ ± 3σ. 5.8 Show that I e y2 /2 dy 2π I 2 2π 2π 2π 2π e x2 /2 dx dθ I 2π e y2 /2 dy e (x2 +y 2 )/2 dxdy e r2 /2 rdrdθ e r2 /2 d r2 2 e r2 /2 rdr e s ds 2π e y2 /2 dy (dxdy rdrdθ) (r 2 /2 s) 5.9

4 Probability and Statistics 4/19 Fall, 214 Mean and Variance of Normal Random Variable Show that E(X) µ. E(X) 1 x σ 2π e 1 (σy + µ)e y2 /2 dy 2π (x µ) 2 2σ 2 dx (y (x µ)/σ) σ ye y2 /2 dy + µ 1 e y2 /2 dy 2π 2π + µ 1 2π µ e y2 /2 dy Show that V (X) σ Standard Normal Random Variable Definition 5 (Standard Normal Random Variable). A normal random variable with µ and σ 1, N(,1), is called a standard normal random variable and is denoted as Z. Definition 6 (Standard Normal Random Variable). The cumulative distribution function of a standard normal random variable is denoted as Φ(z) P(Z z) Standardized Normal Random Variable Definition 7 (Standardized Normal Random Variable). If X is a normal random variable N(µ,σ), the random variable Z X µ (7) σ is a normal random variable N(,1). That is, Z is a standard normal random variable. Also, ( X µ P(X x) P σ where z x µ σ is the z value (or z score ) obtained by standardizing X. x µ ) P(Z z) (8) σ Example: Standard Normal Distribution Example 8. Aluminum sheets used to make beverage cans have thicknesses (in thousandths of an inch) that are normally distributed with mean 1 and standard deviation 1.3. A particular sheet is 1.8 thousandths of an inch thick. Find the z-score. The quantity 1.8 is an observation from a normal population with mean µ 1 and standard deviation σ 1.3. Therefore z x µ σ (9)

5 Probability and Statistics 5/19 Fall, 214 Example: Standard Normal Distribution Example 9. Referring to the previous example. The thickness of a certain sheet has a z-score of 1.7. Find the thickness of the sheet in the original units of thousandths of inches. We use Eq. (9), substituting 1.7 for z and solving for x. We obtain 1.7 x Solving for x yields x 7.8. The sheet is 7.8 thousandths of an inch thick Example: Standard Normal Distribution Example 1. Find the area under the normal curve to the left of z.47. From the z table, the area is Example: Standard Normal Distribution Example 11. Find the area under the normal curve to the right of z From the z table, the area to the left of z 1.38 is Therefore the area to the right is Example: Standard Normal Distribution Example 12. Find the area under the normal curve between z.71 and z From the z table, the area to the left of z 1.28 is The area to the left of z.71 is The area between z.71 and z 1.28 is therefore

6 Probability and Statistics 6/19 Fall, 214 Example: Grades and Normal Distribution Example 13. The grades in a large class (like Statistical Data Analysis) are approximately normal-distributed with mean 75 and standard deviation 6. The lowest D is 6, the lowest C is 68, the lowest B is 83, and the lowest A is 9. What proportion of the class will get A s, B s, C s, D s and F s? A s P(X 9) P(Z 15 6 ).62 B s P(83 X < 9) P(Z < 2.5) P(Z < 1.33) C s P(68 X < 83) P(Z < 1.33) P(Z < 1.17) Example: Grades and Normal Distribution D s P(6 X < 68) P(Z < 1.17) P(Z < 2.5) F s P(X < 6) P(Z < 2.5) Binomial vs Normal Example: Digital Communication Example 14. In a digital communication channel, assume that the number of bits received in error can be modelled by a binomial random variable, and assume that the probability that a bit received in error is If 16 million bits are transmitted, what is the probability that 15 or fewer errors occur?

7 Probability and Statistics 7/19 Fall, 214 Let X denote the number of errors. Then Xis a binomial random variable. And 15( ) 16 P(X 15) (1 5 ) x (1 1 5 ) 16 x x How to compute this equation?. 5.2 Approximation of Binomial Distribution Theorem 15. If X is a binomial random variable with parameters n and p Z X np np(1 p) (1) is approximately a standard normal random variable. To approximate a binomial probability with a normal distribution a continuity correction is applied as follows: ( ) x +.5 np P(X x) P(X x +.5) P Z (11) np(1 p) P(x X) P(x.5 X) P ( ) x.5 np Z np(1 p) (12) This approximation is good for np > 5 and n(1 p) > Binomial vs Normal Distribution Figure 3: Normal approximation to the binomial distribution with parameters n 1, and p

8 Probability and Statistics 8/19 Fall, 214 Example: Digital Communication (cont) Since np ( )(1 1 5 ) 16 > 5 and n(1 p) > 5, we can use the normal distribution to approximate the original binomial distribution as: P(X 15) P(X ) ( ) X P 16(1 1 5 ) 16(1 1 5 ) P(Z.75) Non-symmetric Binomial vs Normal Distribution Figure 4: Binomial distribution is not symmetric if p is close to or 1. (If np or n(1 p) is small, the binomial is quite skewed and the symmetric normal distribution is not a good approximation. ) Poisson vs Normal Approximation of Poisson Distribution Theorem 16. If X is a Poisson random variable with E(X) λ and V (X) λ, Z X λ λ (13) is approximately a standard normal distribution. This approximation is good for λ > Poisson Distribution with Small λ 5.26 Poisson Distribution with Large λ 5.27

9 Probability and Statistics 9/19 Fall, 214 Figure 5: Poisson distributions for small values of the parameter λ. Example: Normal Approximation to Poisson Example 17. Assume that the number particles in a squared meter of dust on a surface follows a Poisson distribution with a mean of 1. If a squared meter of dust is analyzed, what is the probability that 95 or fewer particles are found? The probability can be expressed as P(X 95) The probability can be approximated as P(X x) P(Z 95 e 1 1 x x! ) P(Z 1.58) Exponential Distribution Exponential Distribution Recall that the distribution of the number of trials needed for the first success in a sequence of Bernoulli trials is geometric. Consider a sequence of events that occur randomly in time according to the Poisson distribution at rate λ >. The distribution of the number of events N(t) in the interval [,t] is λt (λt)k P(N(t) k) e. k! Suppose that we are interested in the distribution of the waiting time for the first event. Let T denote this random variable. Then P(T > t) P(no event in [, t]) P(N(t) ) e λt. 5.29

10 Probability and Statistics 1/19 Fall, 214 Figure 6: Poisson distributions for selected large values of the parameter λ. Exponential Distribution Since the cumulative distribution function of T is F(t) P(T t) 1 P(T > t) 1 e λt, the density of T is given by f (t) d dt F(t) d P(T > t) { dt λe λt, fort, fort <. (14) 5.3 Exponential Distribution Definition 18. The random variable X that equals the distance (time or length) of a Poisson process with the rate λ > is an exponential random variable with parameter λ. The probability density function of X is f (x) λe λx, x < (15) The cumulative distribution function is F(x) 1 e λx, x < (16) It is important to use consistent units in calculation of probabilities, means and variances involving exponential random variables Exponential Distribution 5.32

11 Probability and Statistics 11/19 Fall, 214 Figure 7: Probability density functions of exponential random variables for selected values of the parameter λ. Exponential Distribution Theorem 19. If random variable X has an exponential distribution with parameter λ, µ E(X) 1 and σ 2 V (X) 1 λ λ 2 (17) µ E(X) xλe λx dx xde λx xe λx dλx ( xe λx) + e λx dx 1 λ e λx 1 λ 1 λ 5.33 Example: Computer Network Usage Example 2. In a large corporate computer network, user log-ons to the system can be modeled as a Poisson process with mean of 25 log-ons per hour. What is probability that there are no log-ons in an interval of 6 minutes. Let the random variable X denote the time from the start of the interval until the first log-on. X has an exponential distribution with λ 25 log-ons per hour. In addition, 6 minutes is equal to.1 hour. The probability of no log-ons in an interval of 6 minutes is P(X >.1) 1 F(.1) e 25(.1)

12 Probability and Statistics 12/19 Fall, 214 Example: Computer Network Usage What is probability that the time until next log-on is between 2 and 3 minutes ( 2 P 6.33 < X < 3 ) 6.5 F(.5) F(.33) e 25(.5) e 25(.33).152. The mean time until the next log-on is µ 1 λ 1.4 hr 2.4 min 25 The standard deviation of the time until the next log-on is mean time until the next log-on is σ 1 25 hr 2.4 min 5.35 Example: Lack of Memory Example 21. Let X denote the time between detections of a particle with a Geiger counter. Assume that X has an exponential distribution with E(X) 1.4 minutes. The probability that we detect a particle within 3 seconds of starting the counter is P(X <.5 min) F(.5) 1 e.5/ Suppose that the counter has been on for 3 minutes without detecting a particle. What is the probability that we detect a particle in next 3 seconds: P(X < 3.5 X > 3) P(3 < X < 3.5)/P(X > 3) 5.36 Example: Lack of Memory We have P(3 < X < 3.5) F(3.5) F(3).35 P(X > 3) 1 F(3).117 P(X < 3.5 X > 3) P(3 < X < 3.5)/P(X > 3).35/ After waiting for 3 minutes without a detection, the probability of a detection in next 3 seconds is the same as the probability of a detection in the 3 seconds immediately after starting the counter. Theorem 22 (Lack of Memory). For an exponential random variable X P(X < (t 1 +t 2 ) X > t 1 ) P(X < t 2 ) (18) 5.37

13 Probability and Statistics 13/19 Fall, Erlang and Gamma Distributions Example: Erlang Distribution Example 23 (CPU Failure). The failure of the CPUs of large computer systems are often modeled as a Poisson process. Assume that the units that fail are immediately repaired, and assume that the mean number of failure per hour is.1. Let X denote the time until four failures occur in a system. Determine the probability that X exceeds 4, hours. Let the random variable N denote the number of failures in 4, hours. The time until four failures occur exceeds 4, hours if the number of failures in 4, hours in three or less: Example: Erlang Distribution N has a Poisson distribution with Therefore, P(X > 4,) P(N 3) E(N) 4, (.1) 4 failures in 4, hours P(X > 4,) P(N 3) 3 e 4 4 k k k! Erlang Distribution If X is the time until the rth event in a Poisson process then P(X > x) r 1 e λx (λx) k k k! Since P(X > x) 1 F(X), the probability density function of X equals f (x) d dx P(X > x) λ r x r 1 e λx (r 1)! for x > and r 1,2,... This probability density function defines an Erlang distribution. With r 1, an Erlang RV becomes an exponential RV. Gamma Function Definition 24 (Gamma Function). The gamma function of γ is Γ(γ) x γ 1 e x dx, for γ >. (19) The value of the integral is a positive finite number. Using the integral by parts, it can be shown that Γ(γ) (γ 1)Γ(γ 1) If γ is a positive integer, (as in Erlang distribution), Γ(γ) (γ 1)!, given that γ(1)! 1. β γ Γ(γ) y γ 1 e y/β dy

14 Probability and Statistics 14/19 Fall, 214 Gamma Distribution Definition 25 (Gamma Distribution). A random variable X that has a pdf { λ γ x γ 1 e λx f (x;γ,λ) Γ(γ) λ Γ(γ) (λx)γ 1 e λx, < x <, elsewhere. (2) is said to have a gamma distribution with parameters γ > and λ >. The parameters γ and λ are called the scale and shape parameters, respectively. If γ is a positive integer r, X becomes an Erlang distribution Gamma Distribution Figure 8: Gamma probability density functions for selected values of the parameters γ (scale) and λ (shape) Gamma Distribution Theorem 26 (Mean and Variance of Gamma Distribution). If X is a gamma random variable with parameters λ and γ, µ E(X) γ λ (21) and σ 2 V (X) γ λ 2 (22) Definition 27 (Chi-Squared Distribution). The chi-squared distribution is a special case of gamma distribution in which λ 1/2 and γ 1/2, 1, 3/2, 2,... The chi-squared distribution is used extensively in interval estimation and hypothesis testing. 5.44

15 Probability and Statistics 15/19 Fall, 214 Example: Gamma Distribution Example 28. The time to prepare a micro-array slide for high throughput genomics is a Poisson process with a mean of two hours per slide. What is the probability that 1 slides require more than 25 hours to prepare? Let X denote the time to prepare 1 slides. X has gamma distribution with λ 1/2 (slide/hour) and γ 1. Th requested probability P(X > 25) P(X > 25) 9 e 12.5 (12.5) k k k!.214 The mean time to prepare 1 slides is E(X) γ/λ 1/.5 2 hours. And the variance time is V (X) γ/λ Weibull Distribution Weibull Distribution Definition 29 (Weibull Distribution). The random variable X with the probability density function f (x) β ( x ) [ β 1 ( x ) ] β exp, x > (23) δ δ δ is a Weibull random variable with scale parameter δ > and shape parameter β >. The Weibull distribution is often used to model the time until failure of many different physical systems. When β 1, the Weibull distribution is identical to the exponential. The Raleigh distribution is a special case when the shape parameter β Weibull Distribution 5.47 Weibull Distribution Theorem 3. If X has a Weibull distribution with parameters δ and β, then the cumulative distribution function of X is [ ( x ) ] β F(x) 1 exp (24) δ Theorem 31. If X has a Weibull distribution with parameters δ and β, ( µ E(X) δγ ) β (25) ( σ 2 V (X) δ 2 Γ ) [ ( δ 2 Γ )] 2 (26) β β 5.48

16 Probability and Statistics 16/19 Fall, 214 Figure 9: Weibull probability density functions for selected values of the parameters δ (scale) and β (shape). Example: Weibull Distribution Example 32. The time to failure (in hours) of bearing a mechanical shaft is modeled as a Weibull random variable with δ 5 hours and β 1/2. Determine the mean time until failure E(X) 5Γ[1 + (1/.5)] 5Γ[3] 5 2! 1, hours Determine the probability that a bearing last at least 6 hours [ ( ) ] 6 1/2 P(X > 6) 1 F(6) exp 5 e Lognormal Distribution Lognormal Distribution Let W be a normal distribution. X exp(w) is also an random variable. Since log(x) is normally distributed, X is called a lognormal distribution. The cumulative distribution function for X is F(x) P(X x) P(exp(W) x) P(W log(x)) ( P Z log(x) θ ) ( ) log(x) θ Φ ω ω for x >, where Z is a standard normal random variable. F(x) for x. 5.5

17 Probability and Statistics 17/19 Fall, 214 Lognormal Distribution Definition 33 (Lognormal Distribution). Let W have a normal distribution with mean θ and variance ω 2 ; then X exp(w) is a lognormal random variable with probability density function ] 1 f (x) [ xω 2π exp (log(x) θ)2 2ω 2 (27) for < x <. The mean and variance of X are E(X) exp(θ + ω 2 /2) V (X) e 2θ+ω2 (e ω2 1) The lifetime of a product that degrades over time is often modeled by a lognormal distribution random variable Lognormal Distribution Figure 1: Normal probability density functions with θ for selected values of σ Lognormal Distribution 5.53 Lognormal Distribution 5.54 Example: Lognormal Distribution Example 34. The lifetime of a semiconductor laser has a lognormal distribution with θ 1 hours and ω 1.5 hours.

18 Probability and Statistics 18/19 Fall, 214 Figure 11: Lognormal probability density functions with θ for selected values of ω 2. What is the probability the lifetime exceeds 1, hours? P(X > 1) 1 P(exp(W) 1) Example: Lognormal Distribution What lifetime is exceeded by 99% of lasers? 1 P(W log(1)) ( ) log(1) 1 1 Φ Φ(.52) P(X > x) P(exp(W) > x) P(W > log(x)) ( ) log(x) 1 1 Φ Φ(z).99 when z Therefore, log(x) x exp(6.55) hours Example: Lognormal Distribution Determine the mean and variance of lifetime. E(X) exp(θ + ω 2 /2) exp( ) 67, V (X) e 2θ+ω2 (e ω2 1) exp( )[exp(2.25) 1] 39, 7, 59, σ V (x) 197,661.5 E(X) Notice that the standard deviation of life time is much larger to the mean. 5.57

19 Probability and Statistics 19/19 Fall, 214 Figure 12: Lognormal probability density functions with selected values of θ and ω 2.