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Lesson 5: Continuous Probability Distributions Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-1 Outline Continuous probability distributions Features of univariate probability distribution Features of bivariate probability distribution Marginal density and Conditional density Expectation, Variance, Covariance and Correlation Coefficient Importance of normal distribution The normal approximation to the binomial Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-2 Types of Probability Distributions Probability distribution may be classified according to the number of random variables it describes. Number of random variables Ka-fu Wong © 2007 Joint distribution 1 Univariate probability distribution 2 Bivariate probability distribution 3 Trivariate probability distribution … … n Multivariate probability distribution ECON1003: Analysis of Economic Data Lesson5-3 Continuous Probability Distributions A continuous random variable is a variable that can assume any value in an interval thickness of an item time required to complete a task temperature of a solution height, in inches These can potentially take on any value, depending only on the ability to measure accurately. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-4 Cumulative Distribution Function The cumulative distribution function, F(x), for a continuous random variable X expresses the probability that X does not exceed the value of x F(x) P(X x) Let a and b be two possible values of X, with a < b. The probability that X lies between a and b is P(a X b) F(b) F(a) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-5 Probability Density Function Let X be a random variable that takes any real values in an interval between a and b. The number of possible outcomes are by definition infinite. The main features of a probability density function f(x) are: P(X (-, +)) = P(X (a,b)) = 1. P(X = x) = 0. f(x) 0 for all x and may be larger than 1. The probability that X falls into an subinterval (c,d) is d P ( X (c, d )) f ( x )dx c and lies between 0 and 1. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-6 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 1. b a A Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-7 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0.5. 1/2 a b A Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-8 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0.25. 1/4 a 1/2 b A Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-9 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0.125. 1/8 a 1/4 1/2 b A Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-10 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly at a point A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0!! a 1/8 1/4 1/2 b A Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-11 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly at a point A or a point B (which lie on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0!! Since A & B are mutually exclusive, P(A or B) = P(A) + P(B) =0. a A Ka-fu Wong © 2007 1/8 1/4 1/2 b B ECON1003: Analysis of Economic Data Lesson5-12 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly at one of the single point on the line? Suppose the dart has equal chance to land on any point of the line. 0!! Since for distinct points A & B are mutually exclusive, P(A or B) = P(A) + P(B) =0. P (one of the single point on the line) = 0 ????? a A Ka-fu Wong © 2007 1/8 1/4 1/2 b B ECON1003: Analysis of Economic Data Lesson5-13 The probability concepts for the discrete case is not readily applicable…. Would like to modify the concept of discrete probability to fit into the case of continuous random variable! What do we know? P(X<k) should be proportional to (k-0) Try P(X<k) = (k-a)*c P(X<b) = (b-a)*c Ka-fu Wong © 2007 where c is a constant. implies c= 1/(b-a) ECON1003: Analysis of Economic Data Lesson5-14 The probability concepts for the discrete case is not readily applicable…. What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)? Suppose the dart has equal chance to land on any point of the line. 0.5 = (b-a)/2 * c = 1/2. Probability is simply the area C=1/(b-a) 1/2 a b A c is called the probability density. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-15 The probability concepts for the discrete case is not readily applicable…. In discrete case, E(X) = ∑X X P(X). In continuous case, P(X) =0 for any point of X. How do we compute E(X) then? Split the domain into n equal parts, so that the width of these n interval is dx=(b-a)/n. In each of these n intervals, the probability is well define. Then, take the left boundary value of each of this n intervals, multiply by the probability for the interval. Then, we have a weighted average similar to the discrete case. E(X)= a*dx*c + (a+dx)*dx*c+ (a+2dx)*dx*c+ … +(a+(n-1)dx)*dx*c c a Ka-fu Wong © 2007 b dx ECON1003: Analysis of Economic Data Lesson5-16 The probability concepts for the discrete case is not readily applicable…. E(X)= a*dx*c + (a+dx)*dx*c+ (a+2dx)*dx*c+ … +(a+(n-1)dx)*dx*c However, it is an approximation because a is only an approximate of the points within the interval (a, a+dx). Approximation improves if dx is made smaller, or n larger. That is, when dx is very very very closed to zero (but still positive), E(X)= limn-> [a*dx*c + (a+dx)*dx*c +… +(a+(n-1)dx)*dx*c] = limn-> ∑i [xi*dx*c ] b =limdx->0 ∑i [xi*dx*c ] E(X) x[cdx] a c a Ka-fu Wong © 2007 xi dx xi+dx ECON1003: Analysis of Economic Data b Lesson5-17 The probability concepts for the discrete case is not readily applicable…. What if we want to allow it more likely for the dart to end up in some segment of the line (say, the middle segment)? c2 c1 c1 a 1/4 3/4 b We can do it as long as we have the areas added up to 1: ¼*c1 + ½*c2 + ¼*c1 = 1. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-18 The probability concepts for the discrete case is not readily applicable…. What is we want to allow it is more likely for the dart to end up in some segment of the line (say, the middle segment)? c2 c1 c1 b a 3/4 1/4 It is better to define the density “function” to allow the density to vary with x in a general way. b f(x) = c1 if x in (a, a+1/4) E(X) x[ f ( x)dx] = c2 if x in (a+1/4, a+3/4) a = c3 if x in (a+3/4, b) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-19 Probability as an area Shaded area under the curve is the probability that X is between c and d d P(c x d ) f(x)dx f(x) c E( X ) xf(x)dx c Ka-fu Wong © 2007 d ECON1003: Analysis of Economic Data x Lesson5-20 Probability Density Function The cumulative density function F(x0) is the area under the probability density function f(x) from the minimum x value (a) up to x0 F(x0 ) Ka-fu Wong © 2007 x0 x0 a f(x)dx f(x)dx ECON1003: Analysis of Economic Data Lesson5-21 Expectations for Continuous Random Variables The mean of X, denoted μX , is defined as the expected value of X μ X E(X) xf(x)dx - The variance of X, denoted σX2 , is defined as the expectation of the squared deviation, (X - μX)2, of a random variable from its mean σ X2 E[(X μ X )2 ] (x μ X )2 f(x)dx Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-22 Linear Functions of Variables Let W = a + bX , where X has mean μX and variance σX2 , and a and b are constants Then the mean of W is E(W) = E(a+bX) = a + bE(X) = a + b μX the variance is Var(W) = Var(a+bX) = b2Var(X) = b2σX2 the standard deviation of W is |b|σX Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-23 Linear Functions of Variables An important special case of the previous results is the standardized random variable Z =( X- μX ) /σX which has a mean 0 and variance 1 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-24 The Univariate Uniform Distribution If c and d are numbers on the real line, the random variable X ~ U(c,d), i.e., has a univariate uniform distribution if 1 f(x) = d - c 0 for c x d otherwise The mean and standard deviation of a uniform random variable x are X Ka-fu Wong © 2007 cd 2 and X d c 12 ECON1003: Analysis of Economic Data Lesson5-25 The Uniform Density Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-26 Learning exercise 4: Part-time Work on Campus 8 A student has been offered part-time work in a laboratory. The professor says that the work will vary from week to week. The number of hours will be between 10 and 20 with a uniform probability density function, represented as follows: How tall is the rectangle? What is the probability of getting less than 15 hours in a week? Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will 10 12 14 16 18 20 22 be available? Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-27 Learning exercise 4: Part-time Work on Campus 8 10 12 14 16 18 20 P(hour>17.5)/P(hour>15) Ka-fu Wong © 2007 22 How tall is the rectangle? (20-10)*h = 1 h=0.1 What is the probability of getting less than 15 hours in a week? 0.1*(15-10) = 0.5 Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will be available? 0.1*(20-17.5) = 0.25 0.25/0.5 = 0.5 ECON1003: Analysis of Economic Data Lesson5-28 The Normal Distribution Probability Distributions Continuous Probability Distributions Uniform Normal Exponential Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-29 Normal Distribution N(,2) ‘Bell Shaped’ f(x) Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, μ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: + to σ x μ Mean = Median = Mode The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-30 The Normal Distribution N(,2) The normal distribution closely approximates the probability distributions of a wide range of random variables Distributions of sample means approach a normal distribution given a “large” sample size Computations of probabilities are direct and elegant The normal probability distribution has led to good business decisions for a number of applications Sum of normal random variables remain normal. Normal distribution is completely characterized by two parameters, mean and variance. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-31 N(,2) Changing shifts the location of the distribution. Changing 2 changes the dispersion. x (a) x (b) Ka-fu Wong © 2007 (c) ECON1003: Analysis of Economic Data x Lesson5-32 The Normal Probability Distribution The random variable X ~ N(,2), i.e., has a univariate normal distribution if for all x on the real line (-,+ ) 1 f(x) = e 2 1 x - 2 2 and are the mean and standard deviation, = 3.14159 … and e = 2.71828 is the base of natural or Naperian logarithms. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-33 Normal Distribution Probability Probability is the area under the curve! A table may be constructed to help us find the probability f(X) c Ka-fu Wong © 2007 d X ECON1003: Analysis of Economic Data Lesson5-34 Moments of Standard Normal Random Variables N(0, 1) Mean=0 Variance =1 Skewness = 0 Kurtosis = 3 Excess kurtosis =0 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-35 Infinite Number of Normal Distribution Tables Normal distributions differ by mean & standard deviation. f(X) Each distribution would require its own table. X Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-36 The Standard Normal Probability Distribution -- N(0,1) The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is also called the z distribution. A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is: Z X 1 X 1 E (Z ) E E( X ) [E( X ) ] 0 1 X 1 Var ( Z ) Var 2 Var ( X ) 2 Var ( X ) 1 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data The Standard Normal Probability Distribution Any normal random variable can be transformed to a standard normal random variable Suppose X ~ N(µ, 2) Z=(X-µ)/ ~ N(0,1) P(X<k) = P [(X-µ)/ < (k-µ)/ ] Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Standardize the Normal Distribution Z X Normal Distribution Standardized Normal Distribution X =1 0 Z Because we can transform any normal random variable into standard normal random variable, we need only one table! Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-39 Standardizing Example Normal distribution N(5,100) = 5, = 10 5 6.2 Standardized Normal Distribution N(0,1) = 0, = 1 X Z X 5 5 0 10 0 .12 Z Z X 6.25 0.12 10 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-40 Obtaining the Probability Standardized Normal Probability Table (Portion) Z .00 .01 =1 .02 0.0 .0000 .0040 .0080 0.0478 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871 0 0.3 .1179 .1217 .1255 Probabilities Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data 0.12 Z Shaded Area Exaggerated Lesson5-41 Example P(3.8 X 5) Z X 3.85 0.12 10 Normal Distribution Standardized Normal Distribution = 10 0.0478 3.8 =5 X -0.12 0 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-42 Example (2.9 X 7.1) Z X 2.95 0.21 10 Z X 7.15 0.21 10 Normal Distribution Standardized Normal Distribution = 10 .1664 .0832 .0832 2.9 5 7.1 X -.21 0 .21 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-43 Example P(X 8) Normal Distribution Z X 85 0.30 10 Standardized Normal Distribution = 10 .5000 .3821 .1179 =5 8 X 0 .30 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-44 Example P(7.1 X 8) Normal Distribution Z X 7.15 0.21 10 Z X 85 0.3 10 Standardized Normal Distribution = 10 .1179 .0347 .0832 =5 7.1 8 X 0 .21 .30 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-45 Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with µ= 2000 hours & = 200 hours. What’s the probability that a bulb will last between 2000 & 2400 hours? less than 1470 hours? Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-46 Solution P(2000 X 2400) P(2000<X<2400) = P [(2000-µ)/ <(X-µ)/ < (2400-µ)/ ] = P[(X-µ)/ < (2400-µ)/ ] – P [(X-µ)/ < (2000-µ)/ ] = P[(X-µ)/ < (2400-µ)/ ] – 0.5 Normal Distribution Z X μ 2400 20002.0 σ 200 Standardized Normal Distribution = 200 .4772 = 2000 2400 X 0 2.0 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-47 Solution P(X 1470) P(X<1470) = P [(X-µ)/ < (1470-µ)/ ] Normal Distribution Z X μ 1470 2000 2.65 σ Standardized Normal 200 Distribution = 200 .5000 .4960 .0040 1470 = 2000 X -2.65 0 Z Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-48 Finding Z Values for Known Probabilities What Is Z Given P(Z) = 0.1217? .1217 Z =0 Z .31 Standardized Normal Probability Table (Portion) =1 Z Z .00 .01 .02 0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871 .1179 .1217 .1255 0.3 Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-49 Finding X Values for Known Probabilities Normal Distribution Standardized Normal Distribution = 10 Z =1 .1217 =5 ? X .1217 Z =0 .31 Z X Z 5(0.31)108.1 Shaded Area Exaggerated Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-50 EXAMPLE 1 The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $20,000 and a standard deviation of $2,000. What is the z-value for a salary of $24,000? z Ka-fu Wong © 2007 X $24,000 $20,000 2.00 $2000 ECON1003: Analysis of Economic Data Lesson5-51 EXAMPLE 1 continued What is the z-value of $17,000 ? z X $17,000 $20,000 1.50 $2000 A z-value of 2 indicates that the value of $24,000 is 2 standard deviation above the mean of $20,000. A z-value of –1.50 indicates that $17,000 is 1.5 standard deviation below the mean of $20,000. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-52 Areas Under the Normal Curve About 68 percent of the area under the normal curve is within one standard deviation of the mean. ± P( - < X < + ) = 0.6826 About 95 percent is within two standard deviations of the mean. ±2 P( - 2 < X < + 2 ) = 0.9544 Practically all is within three standard deviations of the mean. ±3 P( - 3 < X < + 3 ) = 0.9974 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-53 EXAMPLE 2 The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water? About 68% of the daily water usage will lie between 15 and 25 gallons. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-54 EXAMPLE 2 continued What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day? X 20 20 z 0.00 5 z X 24 20 0.80 5 P(20<X<24) =P[(20-20)/5 < (X-20)/5 < (24-20)/5 ] =P[ 0<Z<0.8 ] The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881. We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-55 How do we find P(0<z<0.8) P(0<z<c) P(z<c) c P(0<z<0.8) = P(z<0.8) – P(z<0) =0.7881 – 0.5 =0.2881 Ka-fu Wong © 2007 0 c P(0<z<0.8) = 0.2881 ECON1003: Analysis of Economic Data Lesson5-56 EXAMPLE 2 continued What percent of the population use between 18 and 26 gallons of water per day? z z X 18 20 0.40 5 X 26 20 1.20 5 Suppose X ~ N(µ, 2) Z=(X-µ)/ ~ N(0,1) P(X<k) = P [(X-µ)/ < (k-µ)/ ] Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-57 How do we find P(-0.4<z<1.2) P(0<z<c) P(z<c) c P(-0.4<z<1.2) = P(z<1.2) - P(z<-0.4) = P(z<1.2) - P(z>0.4) = P(z<1.2) – [1- P(z<0.4)] =0.8849 – [1- 0.6554] =0.5403 0 c P(-0.4<z<1.2) = P(-0.4<z<0) + P(0<z<1.2) =P(0<z<0.4) + P(0<z<1.2) =0.1554+0.3849 =0.5403 P(-0.4<z<0) =P(0<z<0.4) because of symmetry of the z distribution. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-58 Finding the X value for a Known Probability Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: X μ Zσ Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-59 Finding the X value for a Known Probability Example: Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X .20 ? ? Ka-fu Wong © 2007 8.0 0 ECON1003: Analysis of Economic Data X Z Lesson5-60 Find the Z value for 20% in the Lower Tail 1. Find the Z value for the known probability Standardized Normal Probability 20% area in the lower tail is consistent with a Z value of Table (Portion) 0.84 z F(z) .82 .7939 .83 .7967 .84 .7995 .85 Ka-fu Wong © 2007 .8023 .80 .20 ? 8.0 -0.84 0 ECON1003: Analysis of Economic Data X Z Lesson5-61 Finding the X value 2. Convert to X units using the formula: X μ Zσ 8.0 ( 0.84 )5.0 3.80 So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-62 EXAMPLE 3 Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A? Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-63 Example 3 continued To begin let k be the score that separates an A from a B. 15 percent of the students score more than k, then 35 percent must score between the mean of 72 and k. Write down the relation between k and the probability: P(X>k) = 0.15 and P(X<k) =1-P(X>k) = 0.85 Transform X into z: P[(X-72)/5) < (k-72)/5 ] = P[z < (k-72)/5] P[0<z < s] =0.85 -0.5 = 0.35 0.35 0.15 Ka-fu Wong © 2007 72 0 ECON1003: Analysis of Economic Data k X ? Z Lesson5-64 Example 3 continued Find s from table: P[0<z<1.04]=0.35 Compute k: (k-72)/5=1.04 implies K=77.2 0.35 72 0 0.15 77.2 X 1.04 Z Those with a score of 77.2 or more earn an A. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-65 How do we know that the data are likely drawn from normal? 1. Check the moments Skewness =0 Excess Kurtosis = 0 (For instance, refer to section 16.7 “Tests for Skewness and Excess Kurtosis”, p.567 of Estimation and Inference in Econometrics by Davidson and MacKinnon) 2. Normal probability plot 1. Suppose we have n observations in the sample. Sort them in ascending order. Compute the empirical z value (i.e., (x-mx)/sx) 2. Generate a column 0.5, 1.5, …..,[ 0.5+(n-1)]. Call this column U. 3. Generate another column p(z) = U/n. 4. Generate another column theoretical z = NORMSINV(p(z)) 5. Plot empirical z against the theoretical z. 6. If the data has normal distribution, the plot should be a straight line. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-66 Normal Probability Plot (The data are generated from a normal distribution.) 3.5 3 2.5 2 z value from data 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3 -3.5 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 Theoretical z value Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-67 Normal Probability Plot (The data are generated from a uniform distribution.) 3.5 3 2.5 2 z value from data 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3 -3.5 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 Theoretical z value Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-68 The Normal Approximation to the Binomial The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n (number of trials). The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5. Why can we approximate binomial by normal? Because of the Central Limit Theorem. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-69 Normal Distribution Approximation for Binomial Distribution Recall the binomial distribution: n independent trials probability of success on any given trial = P Random variable X: Xi =1 if the ith trial is “success” Xi =0 if the ith trial is “failure” E(X) μ nP Var(X) σ 2 nP(1- P) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-70 Normal Distribution Approximation for Binomial Distribution The shape of the binomial distribution is approximately normal if n is large The normal is a good approximation to the binomial when nP(1 – P) > 9 Standardize to Z from a binomial distribution: Z Ka-fu Wong © 2007 X E(X) X np Var(X) nP(1 P) ECON1003: Analysis of Economic Data Lesson5-71 Normal Distribution Approximation for Binomial Distribution Let X be the number of successes from n independent trials, each with probability of success P. If nP(1 - P) > 9, a nP b nP P(a X b) P Z nP(1 P) nP(1 P) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-72 Binomial Approximation Example 40% of all voters support ballot proposition A. What is the probability that between 76 and 80 voters indicate support in a sample of n = 200 ? E(X) = µ = nP = 200(0.40) = 80 Var(X) = σ2 = nP(1 – P) = 200(0.40)(1 – 0.40) = 48 ( note: nP(1 – P) = 48 > 9 ) 76 80 80 80 P(76 X 80) P Z 200(0.4)(1 0.4) 200(0.4)(1 0.4) P(0.58 Z 0) F(0) F(0.58) 0.5000 0.2810 0.2190 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-73 The Exponential Distribution Used to model the length of time between two occurrences of an event (the time between arrivals) Examples: Time between trucks arriving at an unloading dock Time between transactions at an ATM Machine Time between phone calls to the main operator Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-74 The Exponential Distribution The exponential random variable T (t>0) has a probability density function f(t) λe λt for t 0 Where is the mean number of occurrences per unit time t is the length of time until the next occurrence e = 2.71828 T is said to follow an exponential probability distribution Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-75 The Exponential Distribution Defined by a single parameter, its mean (lambda) The cumulative distribution function (the probability that an arrival time is less than some specified time t) is F(t) 1 e λt where e = mathematical constant approximated by 2.71828 = the population mean number of arrivals per unit t = any value of the continuous variable where t > 0 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-76 Exponential Distribution Example Example: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes? The mean number of arrivals per hour is 15, so = 15 Three minutes is .05 hours P(arrival time < .05) = 1 – e- X = 1 – e-(15)(.05) = 0.5276 So there is a 52.76% probability that the arrival time between successive customers is less than three minutes Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-77 Joint Cumulative Distribution Functions Let X1, X2, . . .Xk be continuous random variables Their joint cumulative distribution function, F(x1, x2, . . .xk) defines the probability that simultaneously X1 is less than x1, X2 is less than x2, and so on; that is F(x1 , x 2 ,..., x k ) P(X1 x1, X2 x 2 ,..., Xk x k ) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-78 Joint Cumulative Distribution Functions The cumulative distribution functions F(x1), F(x2), . . .,F(xk) of the individual random variables are called their marginal distribution functions The random variables are independent if and only if F(x1 , x 2 ,..., x k ) F(x1 )F(x 2 )...F(x k ) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-79 Features of a Bivariate Continuous Distribution Let X1 and X2 be a random variables that takes any real values in a region (rectangle) of (a,b,c,d). The number of possible outcomes are by definition infinite. The main features of a probability density function f(x1,x2) are: f(x1,x2) 0 for all (x1,x2) and may be larger than 1. The probability that (X1,X2) falls into a region (rectangle) or (p,q,r,s) is s q r p P (( X 1, X 2 ) ( p, q, r , s )) f ( x1, x 2 )dx1dx2 and lies between 0 and 1. P((X1,X2) (a,b,c,d)) = 1. P((X1,X2) = (x1,x2) ) = 0. Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-80 The Bivariate Uniform Distribution If a, b, c and d are numbers on the real line, , the random variable (X1,X2) ~ U(a,b,c,d), i.e., has a bivariate uniform distribution if 1 for a x1 b and c x 2 d f(x1, x 2 ) = (b - a)(d - c) 0 otherwise Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-81 The Marginal Density The marginal density functions are: f(y) f(x,y)dx f(x) f(x,y)dy Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-82 The Conditional Density The conditional density functions are: f(x|y) f(x,y)/f(y) f(y|x) f(x,y)/f(x) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-83 The Expectation (Mean) of Continuous Probability Distribution For univariate probability distribution, the expectation or mean E(X) is computed by the formula: E(X) xf(x)dx For bivariate probability distribution, the the expectation or mean E(X) is computed by the formula: E(X) Ka-fu Wong © 2007 xf(x,y)dx dy ECON1003: Analysis of Economic Data Lesson5-84 Conditional Mean of Bivariate Discrete Probability Distribution For bivariate probability distribution, the conditional expectation or conditional mean E(X|Y) is computed by the formula: E(X|Y y) xf(x|y)dx Unconditional expectation or mean of X, E(X) E(X) xf(x|y)dx f(y)dy E[E(X|Y)] E[μ X ] Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-85 Expectation of a linear transformed random variable If a and b are constants and X is a random variable, then E(a) = a E(bX) = bE(X) E(a+bX) = a+bE(X) E(a bx) (a bx) f(a bx) dx (a bx) f(x) dx a f(x) dx bx f(x) dx a f(x) dx b x f(x) dx a bE(x) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-86 The Variance of a Continuous Probability Distribution For univariate continuous probability distribution Var(X) E[(X μ)2 ] 2 (X μ) f(x)dx - If a and b are constants and X is a random variable, then Var(a) = 0 Var(bX) = b2Var(X) Var(a+bX) = b2Var(X) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-87 The Covariance of a Bivariate Discrete Probability Distribution Covariance measures how two random variables co-vary. Cov(X,Y) E[(X μ X )(Y μY )] (X μ X )(Y μY )f ( x, y )dxdy If a and b are constants and X is a random variable, then Cov(a,b) = 0 Cov(a,bX) = 0 Cov(a+bX,Y) = bCov(X,Y) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-88 Correlation coefficient The strength of the dependence between X and Y is measured by the correlation coefficient: Cov(X,Y) Corr(X,Y) Var(X)Var(Y) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-89 Variance of a sum of random variables If a and b are constants and X and Y are random variables, then Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y) Var(X Y) E[ X Y (μ X μY ) ] 2 E[ (X μ X ) (Y μY )] 2 E[ (X μ X ) 2 (Y μY )2 2(X μ X )(Y μY )] E[ (X μ X ) 2 ] E[(Y μY )2 2 E[(X μ X )(Y μY )] Var(X) Var(Y) 2Cov(X,Y) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-90 Variance of a sum of random variables If a and b are constants and X and Y are random variables, then Var(aX+bY) =a2Var(X) + b2Var(Y) + 2abCov(X,Y) Var(aX bY) E[ aX bY (aμ X bμY ) ] 2 E[ (aX aμ X ) (bY bμY )] 2 E[ a 2 (X μ X ) 2 b 2 (Y μY )2 2(aX aμ X )(bY bμY )] a 2 E[ (X μ X ) 2 ] b 2 E[(Y μY )2 2abE[(X μ X )(Y μY )] a 2Var(X) b 2Var(Y) 2abCov(X,Y) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-91 Sums of Random Variables Let X1, X2, . . .Xk be k random variables with means μ1, μ2,. . . μk and variances σ12, σ22,. . ., σk2. Then, the mean of their sum is the sum of their means E(X 1 X 2 ... X k ) μ1 μ2 ... μk Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-92 Sums of Random Variables Let X1, X2, . . .Xk be k random variables with means μ1, μ2,. . . μk and variances σ12, σ22,. . ., σk2. Then: If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances Var(X 1 X 2 ... X k ) σ12 σ 22 ... σ 2k However, if the covariances between pairs of random variables are not 0, the variance of their sum is K 1 K Var(X 1 X 2 ... X k ) σ σ ... σ 2 Cov(X i , X j ) 2 1 Ka-fu Wong © 2007 2 2 ECON1003: Analysis of Economic Data 2 k i 1 ji 1 Lesson5-93 Differences Between Two Random Variables For two random variables, X and Y The mean of their difference is the difference of their means; that is E(X Y) μX μY If the covariance between X and Y is 0, then the variance of their difference is Var(X Y) σ 2X σ 2Y If the covariance between X and Y is not 0, then the variance of their difference is Var(X Y) σ 2X σ 2Y 2Cov(X, Y) Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-94 Linear Combinations of Random Variables A linear combination of two random variables, X and Y, (where a and b are constants) is W aX bY The mean of W is μW E[W] E[aX bY] aμX bμY Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-95 Linear Combinations of Random Variables The variance of W is σ 2W a2σ 2X b2σ 2Y 2abCov(X, Y) Or using the correlation, σ 2W a2σ 2X b2σ 2Y 2abCorr(X, Y)σ Xσ Y If both X and Y are joint normally distributed random variables then the linear combination, W, is also normally distributed Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-96 Example Two tasks must be performed by the same worker. X = minutes to complete task 1; μx = 20, σx = 5 Y = minutes to complete task 2; μy = 20, σy = 5 X and Y are normally distributed and independent What is the mean and standard deviation of the time to complete both tasks? Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-97 Example X = minutes to complete task 1; μx = 20, σx = 5 Y = minutes to complete task 2; μy = 30, σy = 8 What are the mean and standard deviation for the time to complete both tasks? W XY μW μX μY 20 30 50 Since X and Y are independent, Cov(X,Y) = 0, so σ 2W σ 2X σ 2Y 2Cov(X, Y) (5)2 (8)2 89 The standard deviation is σW Ka-fu Wong © 2007 89 9.434 ECON1003: Analysis of Economic Data Lesson5-98 Lesson 5: Continuous Probability Distributions - END - Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-99