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5 Joint Probability Distributions CHAPTER OUTLINE 5-1 Two or More Random Variables 5-1.1 Joint Probability Distributions 5-1.2 Marginal Probability Distributions 5-1.3 Conditional Probability Distributions 5-1.4 Independence 5-1.5 More Than Two Random Variables 5-2 Covariance and Correlation 5-3 Common Joint Distributions 5-3.1 Multinomial Probability Distribution 5-3.2 Bivariate Normal Distribution 5-4 Linear Functions of Random Variables 5-5 General Functions of Random Variables Chapter 4 Title and Outline 1 Learning Objective for Chapter 5 After careful study of this chapter, you should be able to do the following: 1. 2. 3. 4. 5. 6. 7. Use joint probability mass functions and joint probability density functions to calculate probabilities. Calculate marginal and conditional probability distributions from joint probability distributions. Interpret and calculate covariances and correlations between random variables. Use the multinomial distribution to determine probabilities. Understand properties of a bivariate normal distribution and be able to draw contour plots for the probability density function. Calculate means and variances for linear combinations of random variables, and calculate probabilities for linear combinations of normally distributed random variables. Determine the distribution of a general function of a random variable. Chapter 5 Learning Objectives © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 2 Concept of Joint Probabilities • Some random variables are not independent of each other, i.e., they tend to be related. – Urban atmospheric ozone and airborne particulate matter tend to vary together. – Urban vehicle speeds and fuel consumption rates tend to vary inversely. • The length (X) of a injection-molded part might not be independent of the width (Y). Individual parts will vary due to random variation in materials and pressure. • A joint probability distribution will describe the behavior of several random variables, say, X and Y. The graph of the distribution is 3-dimensional: x, y, and f(x,y). Chapter 5 Introduction 3 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Example 5-1: Signal Bars You use your cell phone to check your airline reservation. The airline system requires that you speak the name of your departure city to the voice recognition system. • Let Y denote the number of times that you have to state your departure city. • Let X denote the number of bars of signal strength on you cell phone. Figure 5-1 Joint probability distribution of X and Y. The table cells are the probabilities. Observe that more bars relate to less repeating. Bar Chart of Number of Repeats vs. Cell Phone Bars 0.25 Probability y = number of x = number of bars times city of signal strength name is stated 1 2 3 1 0.01 0.02 0.25 2 0.02 0.03 0.20 3 0.02 0.10 0.05 4 0.15 0.10 0.05 0.20 0.15 0.10 4 Times 3 Times Twice Once 0.05 0.00 1 2 3 Cell Phone Bars Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 4 Joint Probability Mass Function Defined T he joint probabilit y m ass function of the discrete random variables X and Y , denoted as f X Y x , y , satifies: (1) f X Y x , y 0 (2) x f XY x , y 1 A ll probabilities are non-negative T he sum of all probabilities is 1 y (3) f X Y x , y P X x , Y y Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. (5-1) 5 Joint Probability Density Function Defined The joint probability density function for the continuous random variables X and Y, denotes as fXY(x,y), satisfies the following properties: (1) f X Y x , y 0 for all x , y (2) f X Y x , y dxdy 1 (3) P X , Y R f X Y x , y dxdy (5-2) R Figure 5-2 Joint probability density function for the random variables X and Y. Probability that (X, Y) is in the region R is determined by the volume of fXY(x,y) over the region R. Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 6 Joint Probability Mass Function Graph Figure 5-3 Joint probability density function for the continuous random variables X and Y of different dimensions of an injection-molded part. Note the asymmetric, narrow ridge shape of the PDF – indicating that small values in the X dimension are more likely to occur when small values in the Y dimension occur. Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 7 Example 5-2: Server Access Time-1 Let the random variable X denote the time (msec’s) until a computer server connects to your machine. Let Y denote the time until the server authorizes you as a valid user. X and Y measure the wait from a common starting point (x < y). The range of x and y are shown here. Figure 5-4 The joint probability density function of X and Y is nonzero over the shaded region where x < y. Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 8 Example 5-2: Server Access Time-2 • The joint probability density function is: f X Y x , y ke 0.001 x 0.002 y for 0 x y and k 6 10 6 • We verify that it integrates to 1 as follows: f XY 0 .0 0 1 x 0 .0 0 2 y 0 .0 0 2 y 0 .0 0 1 x dy dx k e dy e dx x , y dxdy ke 0 x 0 x e 0 .0 0 2 x k 0.002 0 0 .0 0 1 x 0 .0 0 3 x e dx 0.003 e dx 0 1 0.003 1 0.003 Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 9 Example 5-2: Server Access Time-2 Now calculate a probability: 1000 2000 P X 1000, Y 2000 1000 k 0 1000 k 0 f X Y x , y dxdy x 2000 0.002 y 0.001 x dy e dx e x e 0.002 x e 4 0.002 0.001 x dx e 1000 0.003 e 0.003 x 4 e e 0.001 x dx 0 1 1 e 3 1 e 4 0.003 e 0.003 0.001 Figure 5-5 Region of integration for the probability that X < 1000 and Y < 2000 is darkly shaded. 0.003 316.738 11.578 0.915 Sec 5-1.1 Joint Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 10 Marginal Probability Distributions (discrete) For a discrete joint PDF, there are marginal distributions for each random variable, formed by summing the joint PMF over the other variable. fX x f xy y fY y f xy x y = number of x = number of bars times city of signal strength name is stated 1 2 3 f (y ) = 1 0.01 0.02 0.25 0.28 2 0.02 0.03 0.20 0.25 3 0.02 0.10 0.05 0.17 4 0.15 0.10 0.05 0.30 f (x ) = 0.20 0.25 0.55 1.00 Figure 5-6 From the prior example, the joint PMF is shown in green while the two marginal PMFs are shown in blue. Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 11 Marginal Probability Distributions (continuous) • Rather than summing a discrete joint PMF, we integrate a continuous joint PDF. • The marginal PDFs are used to make probability statements about one variable. • If the joint probability density function of random variables X and Y is fXY(x,y), the marginal probability density functions of X and Y are: fX x f X Y x , y dy y fY y f X Y x , y dx (5-3) x Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 12 Example 5-4: Server Access Time-1 For the random variables times in Example 5-2, find the probability that Y exceeds 2000. Integrate the joint PDF directly using the picture to determine the limits. 2000 P Y 2000 0 D ark region f X Y x , y dy dx 2000 left dark region f X Y x , y dy dx 2000 x right dark region Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 13 Example 5-4: Server Access Time-2 Alternatively, find the marginal PDF and then integrate that to find the desired probability. y fY y ke 0 .0 0 1 x 0 .0 0 2 y P Y 2000 0 fY y dy 2000 y ke 0 .0 0 2 y e 0 .0 0 1 x 6 10 dx 3 e 0.001 0 0 .0 0 1 x 0 .0 0 2 y y 0 .0 0 1 y ke 1 e 0.001 0 .0 0 2 y 3 6 10 e 0.002 y 1 e 0.001 y dy 2000 0 ke e 0 .0 0 2 y 1 e 6 10 0 .0 0 1 y 6 10 3 3 e 0.002 y 0.002 2000 e 0.003 y 0.003 2000 6 e 4 e 0.05 0.002 0.003 for y 0 Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 14 Mean & Variance of a Marginal Distribution Means E(X) and E(Y) are calculated from the discrete and continuous marginal distributions. D iscrete EX C ontinuous x fX x R E Y X y fY y X y f y dy Y Y R x fX x X 2 2 R V Y X R R V x f x dx x f X x dx X 2 2 R y fY 2 R y Y2 y fY 2 y dy Y2 R Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 15 Mean & Variance for Example 5-1 y = number of x = number of bars of signal strength times city name is stated 1 2 3 f (y ) = y *f (y ) = y 2*f (y ) = 1 0.01 0.02 0.25 0.28 0.28 0.28 2 0.02 0.03 0.20 0.25 0.50 1.00 3 0.02 0.10 0.05 0.17 0.51 1.53 4 0.15 0.10 0.05 0.30 1.20 4.80 f (x ) = 0.20 0.25 0.55 1.00 2.49 7.61 x *f (x ) = 0.20 0.50 1.65 2.35 x 2*f (x ) = 0.20 1.00 4.95 6.15 E(X) = 2.35 V(X) = 6.15 – 2.352 = 6.15 – 5.52 = 0.6275 E(Y) = 2.49 V(Y) = 7.61 – 2.492 = 7.61 – 16.20 = 1.4099 Sec 5-1.2 Marginal Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 16 Conditional Probability Distributions R ecall that P B A From Example 5-1 P(Y=1|X=3) = 0.25/0.55 = 0.455 P(Y=2|X=3) = 0.20/0.55 = 0.364 P(Y=3|X=3) = 0.05/0.55 = 0.091 P(Y=4|X=3) = 0.05/0.55 = 0.091 Sum = 1.001 PA B P A y = number of x = number of bars of signal strength times city name is stated 1 2 3 f (y ) = 1 0.01 0.02 0.25 0.28 2 0.02 0.03 0.20 0.25 3 0.02 0.10 0.05 0.17 4 0.15 0.10 0.05 0.30 f (x ) = 0.20 0.25 0.55 1.00 Note that there are 12 probabilities conditional on X, and 12 more probabilities conditional upon Y. Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 17 Conditional Probability Density Function Defined G iven continuous random variables X and Y w ith joint probability density function f X Y x , y , the conditional probability densiy function of Y given X = x is fY x y f XY x , y fX x for f X x 0 (5-4) w hich satifies the follow ing properties: (1) f Y (2) x fY y 0 x y dy 1 (3) P Y B X x fY x y dy for any set B in the range of Y B Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 18 Example 5-6: Conditional Probability-1 From Example 5-2, determine the conditional PDF for Y given X=x. fX x k e 0.001 x 0.002 y dy x ke ke 0.001 x 0.001 x 0.003 e fY x y e 0.002 y 0.002 0.003 x x 0.002 e x e 0.002 0.002 f XY x , y fX for x 0 ke 0.001 x 0.002 y 0.003 e 0.002 x 0.002 y 0.003 x for 0 x y Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 19 Example 5-6: Conditional Probability-2 Now find the probability that Y exceeds 2000 given that X=1500: P Y 2000 X 1500 fY 1500 y d y 2000 0 .0 0 2 e 0 .0 0 2 1 5 0 0 0 .0 0 2 y 2000 e 0 .0 0 2 y 3 0 .0 0 2 e 0 .0 0 2 2000 4 e 3 1 0 .0 0 2 e e 0 .3 6 8 0 .0 0 2 Figure 5-8 again The conditional PDF is nonzero on the solid line in the shaded region. Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 20 Example 5-7: Conditional Discrete PMFs Conditional discrete PMFs can be shown as tables. y = number of x = number of bars of signal strength f(x|y) for y = times city Sum of name is stated 1 2 3 f (y ) = 1 2 3 f(x|y) = 1 0.01 0.02 0.25 0.28 0.036 0.071 0.893 1.000 2 0.02 0.03 0.20 0.25 0.080 0.120 0.800 1.000 3 0.02 0.10 0.05 0.17 0.118 0.588 0.294 1.000 4 0.15 0.10 0.05 0.30 0.500 0.333 0.167 1.000 f (x ) = 0.20 0.25 0.55 1 0.050 0.080 0.455 2 3 4 Sum of f(y|x) = 0.100 0.100 0.750 1.000 0.120 0.400 0.400 1.000 0.364 0.091 0.091 1.000 Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 21 Mean & Variance of Conditional Random Variables • The conditional mean of Y given X = x, denoted as E(Y|x) or μY|x is: E Y x y f y dy (5-6) Y x y • The conditional variance of Y given X = x, denoted as V(Y|x) or σ2Y|x is: V Y x y Y y x 2 fY y dy x y fY 2 2 y x Y x y Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 22 Example 5-8: Conditional Mean & Variance From Example 5-2 & 6, what is the conditional mean for Y given that x = 1500? Integrate by parts. E Y X 1500 y 0.002 e 0.002 1500 0.002 y dy 0.002 e 3 1500 y e 0.002 y dy 1500 e 0.002 y 3 0.002 e y 0.002 1500 e 0.002 y dy 0.002 1500 0.002 y 1500 3 e 3 0.002 e e 0.002 0.002 0.002 1500 3 1500 3 e 0.002 e e 0.002 0.002 0.002 3 e 3 0.002 e 2000 2000 0.002 3 If the connect time is 1500 ms, then the expected time to be authorized is 2000 ms. Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 23 Example 5-9 For the discrete random variables in Exercise 5-1, what is the conditional mean of Y given X=1? y = number of x = number of bars of signal strength times city name is stated 1 2 3 1 0.01 0.02 0.25 2 0.02 0.03 0.20 3 0.02 0.10 0.05 4 0.15 0.10 0.05 f (x ) = 1 2 3 4 Sum of f(y|x) = f (y ) = 0.28 0.25 0.17 0.30 0.20 0.25 0.55 y*f(y|x=1) 0.050 0.080 0.455 0.05 0.100 0.100 0.750 1.000 0.120 0.400 0.400 1.000 0.364 0.091 0.091 1.000 0.20 0.30 3.00 3.55 y2*f(y|x=1) 0.05 0.40 0.90 12.00 13.35 12.6025 0.7475 The mean number of attempts given one bar is 3.55 with variance of 0.7475. Sec 5-1.3 Conditional Probability Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 24 Joint Random Variable Independence • Random variable independence means that knowledge of the values of X does not change any of the probabilities associated with the values of Y. • X and Y vary independently. • Dependence implies that the values of X are influenced by the values of Y. • Do you think that a person’s height and weight are independent? Sec 5-1.4 Independence 25 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Exercise 5-10: Independent Random Variables In a plastic molding operation, each part is classified as to whether it conforms to color and length specifications. 1 if the part conform s to length specs Y 0 otherw ise Figure 5-10(a) shows marginal & joint probabilities, fXY(x, y) = fX(x) * fY(y) Figure 5-10(b) show the conditional probabilities, fY|x(y) = fY(y) X 1 if the part conform s to color specs 0 otherw ise Sec 5-1.4 Independence 26 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Properties of Independence For random variables X and Y, if any one of the following properties is true, the others are also true. Then X and Y are independent. (1) f X Y x , y f X x f Y (2) f Y (3) f X x y y y y fY for all x and y w ith f X x 0 y f X x for all x and y w ith f Y y 0 (4) P X A , Y B P X A P Y B for any sets A and B in the range of X and Y , respectively. Sec 5-1.4 Independence (5-7) 27 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Rectangular Range for (X, Y) • A rectangular range for X and Y is a necessary, but not sufficient, condition for the independence of the variables. • If the range of X and Y is not rectangular, then the range of one variable is limited by the value of the other variable. • If the range of X and Y is rectangular, then one of the properties of (5-7) must be demonstrated to prove independence. Sec 5-1.4 Independence 28 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Example 5-11: Independent Random Variables • Suppose the Example 5-2 is modified such that the joint PDF is: f X Y x , y 2 10 6 e 0 .0 0 1 x 0 .0 0 2 y for x 0 and y 0. • Are X and Y independent? Is the product of the marginal PDFs equal the joint PDF? Yes by inspection. fX x 6 2 10 e 0 .0 0 1 x 0 .0 0 2 y dy fY y 2 10 6 e 0.001 x 0.002 y dx 0 0 0.001e 0 .0 0 1 x for x 0 0.002 e 0.002 y for y 0 • Find this probability: P X 1000, Y 1000 P X 1000 P Y 1000 e 1 1 e 2 0.318 Sec 5-1.4 Independence 29 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Example 5-12: Machined Dimensions Let the random variables X and Y denote the lengths of 2 dimensions of a machined part. Assume that X and Y are independent and normally distributed. Find the desired probability. Normal Random Variables X Y Mean 10.5 3.2 Variance 0.0025 0.0036 P 10.4 X 10.6, 3.15 Y 3.25 P 10.4 X 10.6 P 3.15 Y 3.25 10.6 10.5 3.25 3.2 10.4 10.5 3.15 3.2 P Z Z P 0.05 0.05 0.06 0.06 P 2 Z 2 P 0.833 Z 0.833 0.568 Sec 5-1.4 Independence 30 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Example 5-13: More Than Two Random Variables • Many dimensions of a machined part are routinely measured during production. Let the random variables X1, X2, X3 and X4 denote the lengths of four dimensions of a part. • What we have learned about joint, marginal and conditional PDFs in two variables extends to many (p) random variables. Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 31 Joint Probability Density Function Redefined The joint probability density function for the continuous random variables X1, X2, X3, …Xp, denoted as f x x x ... x x1 , x 2 , x 3 , ..., x p satisfies the following properties: 1 2 3 p (1) f X 1 X 2 ... X p x1 , x 2 , ..., x p 0 (2) ... f X 1 X 2 ... X p x1 , x 2 , ..., x p dx1 dx 2 ...dx p 1 (3) For any region B of p-dim ensional s pace, P X 1 , X 2 ... X p B ... f X 1 X 2 ... X p x1 , x 2 , ..., x p dx1 dx 2 ...dx p (5-8) B Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 32 Example 5-14: Component Lifetimes In an electronic assembly, let X1, X2, X3, X4 denote the lifetimes of 4 components in hours. The joint PDF is: f X 1 X 2 X 3 X 4 x1 , x 2 , x 3 , x 4 9 10 12 e 0.001 x1 0.002 x 2 0.0015 x 3 0.003 x 4 for x i 0 What is the probability that the device operates more than 1000 hours? The joint PDF is a product of exponential PDFs. P(X1 > 1000, X2 > 1000, X3 > 1000, X4 > 1000) = e-1-2-1.5-3 = e-7.5 = 0.00055 Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 33 Example 5-15: Probability as a Ratio of Volumes • Suppose the joint PDF of the continuous random variables X and Y is constant over the region x2 + y2 =4. The graph is a round cake with radius of 2 and height of 1/4π. • A cake round of radius 1 is cut from the center of the cake, the region x2 + y2 =1. • What is the probability that a randomly selected bit of cake came from the center round? • Volume of the cake is 1. The volume of the round is π *1/4π = ¼. The desired probability is ¼. Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 34 Marginal Probability Density Function If the joint probability density functio n of continuous random variables X 1 , X 2 , ... X p is f X 1 X 2 ... X p x1 , x 2 , ... x p , the m arginal probability density functio n of X i is f X i xi ... f X 1 X 2 ... X p x1 , x 2 , ... x p dx1 dx 2 ...dx i 1 dx i 1 ...dx p (5-9) w here the integral is over all points in the range of X 1 , X 2 , ... X p for w hich X i x i . ( don't integrate out x i ) Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 35 Mean & Variance of a Joint PDF The mean and variance of Xi can be determined from either the marginal PDF, or the joint PDF as follows: E Xi ... x i f X 1 X 2 ... X p x1 , x 2 , ... x p dx1 dx 2 ...dx p and (5-10) V X i ... x i Xi 2 f X 1 X 2 ... X p x1 , x 2 , ... x p dx1 dx 2 ...dx p Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 36 Example 5-16 • There are 10 points in this discrete joint PDF. • Note that x1+x2+x3 = 3 • List the marginal PDF of X2 P(X2 = 0) = PXXX(0,0,3) + PXXX(1,0,2) + PXXX(2,0,1) + PXXX(3,0,0) P(X2 = 1) = PXXX(0,1,2) + PXXX(1,1,1) + PXXX(2,1,0) P(X2 = 2) = PXXX(0,2,1) + PXXX(1,2,0) P(X2 = 3) = PXXX(0,3,0) Note the index pattern Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 37 Reduced Dimensionality If the joint probability density functio n of continuous random variables X 1 , X 2 , ... X p is f X 1 X 2 ... X p x1 , x 2 , ... x p , then the probability density function of X 1 ,X 2 ,...X k , k p is f X 1 X 2 ... X k x1 , x 2 , ..., x k ... f X 1 X 2 ... X p x1 , x 2 , ... x p dx k 1 dx k 2 ...dx p (5-11) w here the integral is over all points in the range of X 1 , X 2 , ... X p for w hich X i x i for i 1 through k . ( integrate out p-k variables) Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 38 Conditional Probability Distributions • Conditional probability distributions can be developed for multiple random variables by extension of the ideas used for two random variables. • Suppose p = 5 and we wish to find the distribution conditional on X4 and X5. fX 1X 2 X3 X4 X5 x1 , x 2 , x 3 f X 1 X 2 X 3 X 4 X 5 x1 , x 2 , x 3 , x 4 , x 5 f X 4 X 5 x 4 , x5 for f X 4 X 5 x 4 , x 5 0. Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 39 Independence with Multiple Variables The concept of independence can be extended to multiple variables. R andom variables X 1 , X 2 , ..., X p are in depen d e nt if and only if f X 1 X 2 ... X p x1 , x 2 , ..., x p f X 1 x1 f X 2 x 2 ... f X p x p for all x 1 ,x 2 ,...,x p (5-12) (joint p df equals the product of all the m argina l P D Fs ) Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 40 Example 5-17 • In Chapter 3, we showed that a negative binomial random variable with parameters p and r can be represented as a sum of r geometric random variables X1, X2,…, Xr , each with parameter p. • Because the binomial trials are independent, X1, X2,…, Xr are independent random variables. Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 41 Example 5-18: Layer Thickness Suppose X1,X2,X3 represent the thickness in μm of a substrate, an active layer and a coating layer of a chemical product. Assume that these variables are independent and normally distributed with parameters and specified limits as tabled. What proportion of the product meets all specifications? Answer: 0.7783, 3 layer product. Which one of the three thicknesses has the least probability of meeting specs? Answer: Layer 3 has least prob. Note the index pattern Mean (μ) Std dev (σ) Lower limit Normal Random Variables X1 X2 X3 10,000 1,000 80 250 20 4 9,200 950 75 Upper limit 10,800 1,050 85 P(in limits) 0.99863 0.98758 0.78870 P(all in limits) = 0.77783 Sec 5-1.5 More Than Two Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 42 Covariance • Covariance is a measure of the relationship between two random variables. • First, we need to describe the expected value of a function of two random variables. Let h(X, Y) denote the function of interest. h x , y f X Y x , y for X , Y discrete E h X , Y (5-13) h x , y f X Y x , y dxdy for X , Y continuous Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 43 Example 5-19: E(Function of 2 Random Variables) Task: Calculate E[(X-μX)(Y-μY)] = covariance Mean Marginal Joint x y 1 1 3 3 3 1 3 1 2 1 2 3 1 2 3 μX = μY = f(x, y) x-μX y-μY Prod 0.1 -1.4 -1.0 0.14 0.2 -1.4 0.0 0.00 0.2 0.6 -1.0 -0.12 0.2 0.6 0.0 0.00 0.3 0.6 1.0 0.18 0.3 covariance = 0.20 0.7 0.3 0.4 0.3 2.4 2.0 Figure 5-12 Discrete joint distribution of X and Y. Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 44 Covariance Defined T he covariance betw een the random variab les X and Y , denoted as cov X , Y or XY is X Y E X X Y T he units of are units of X tim es units of Y . XY Y E X Y X Y (5 -1 4 ) For exam ple, if the units of X are feet and the units of Y are pounds, the units of the covariance are foot-pou nds. U nlike the range of variance, - XY . Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 45 Covariance and Scatter Patterns Figure 5-13 Joint probability distributions and the sign of cov(X, Y). Note that covariance is a measure of linear relationship. Variables with non-zero covariance are correlated. Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 46 Example 5-20: Intuitive Covariance y = number of x = number of bars of signal strength times city name is stated 1 2 3 1 0.01 0.02 0.25 2 0.02 0.03 0.20 3 0.02 0.10 0.05 4 0.15 0.10 0.05 The probability distribution of Example 5-1 is shown. By inspection, note that the larger probabilities occur as X and Y move in opposite directions. This indicates a negative covariance. Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 47 Correlation (ρ = rho) T he correlation betw een random variables X and Y , denoted as X Y , is XY cov X , Y V X V Y XY X Y (5-15) S ince X 0 and Y 0, X Y and cov X , Y have the sam e sign. W e say that X Y is norm alized, so 1 X Y 1 (5-16) N ote that X Y is dim ensionless. V ariables w ith non-zero correlation are correlated. Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 48 Example 5-21: Covariance & Correlation Joint x Figure 5-14 Discrete joint distribution, f(x, y). StDev Mean Marginal Determine the covariance and correlation. y 0 1 1 2 2 3 0 1 2 3 0 1 2 1 2 3 0 1 2 3 μX = μY = f(x, y) x-μX y-μY Prod 0.2 -1.8 -1.2 0.42 0.1 -0.8 -0.2 0.01 0.1 -0.8 0.8 -0.07 0.1 0.2 -0.2 0.00 0.1 0.2 0.8 0.02 0.4 1.2 1.8 0.88 0.2 covariance = 1.260 0.2 correlation = 0.926 0.2 0.4 Note the strong 0.2 positive correlation. 0.2 0.2 0.4 1.8 1.8 σX = 1.1662 σY = 1.1662 Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 49 Example 5-21 Calculate the correlation. Figure 5-15 Discrete joint distribution. Steepness of line connecting points is immaterial. StDev Mean Marginals Joint x y 1 2 3 1 2 3 7 9 11 7 9 11 μX = μY = f(x, y) x*y*f 0.2 1.4 0.6 10.8 0.2 6.6 0.2 18.8 = E(XY) 0.6 0.80 = cov(X ,Y ) 0.2 1.00 = ρXY 0.2 Using Equations 0.6 5-14 & 5-15 0.2 2 9.0 σX = 0.6325 σY = 1.2649 Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 50 Independence Implies ρ = 0 • If X and Y are independent random variables, σXY = ρXY = 0 (5-17) • ρXY = 0 is necessary, but not a sufficient condition for independence. – Figure 5-13d (x, y plots as a circle) provides an example. – Figure 5-13b (x, y plots as a square) indicates independence, but a non-rectangular pattern would indicate dependence. Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 51 Example 5-23: Independence Implies Zero Covariance Let f X Y xy x y 16 for 0 x 2 and 0 y 4 S how that X Y E X Y E X E Y 0 EX 2 x yd x d y 16 0 0 1 1 4 4 16 0 2 x y 3 3 2 1 y 16 2 E Y d y 0 2 8 1 16 4 3 6 2 3 0 4 2 xy d x d y 16 0 0 1 1 4 4 16 0 2 x2 y 2 2 3 2 y 16 3 d y 0 2 1 64 8 8 3 3 0 4 2 2 2 E XY x y dx dy 16 0 0 1 1 16 1 16 4 4 0 4 0 x3 y 3 2 dy 0 2 2 8 y dy 3 3 1y 6 3 1 64 32 6 3 9 0 4 XY E X Y E X E Y 32 9 Figure 5-15 A planar joint distribution. 4 8 0 3 3 Sec 5-2 Covariance & Correlation © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 52 Common Joint Distributions • There are two common joint distributions – Multinomial probability distribution (discrete), an extension of the binomial distribution – Bivariate normal probability distribution (continuous), a two-variable extension of the normal distribution. Although they exist, we do not deal with more than two random variables. • There are many lesser known and custom joint probability distributions as you have already seen. Sec 5-3 Common Joint Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 53 Multinomial Probability Distribution • Suppose a random experiment consists of a series of n trials. Assume that: 1) 2) 3) • The outcome of each trial can be classifies into one of k classes. The probability of a trial resulting in one of the k outcomes is constant, denoted as p1, p2, …, pk. The trials are independent. The random variables X1, X2,…, Xk denote the number of outcomes in each class and have a multinomial distribution and probability mass function: P X 1 x1 , X 2 x 2 , ..., X k x k n! x1 ! x 2 !... x k ! x x x p1 1 p 2 2 ... p k k (5-18) for x1 x 2 ... x k n and p1 p 2 ... p k 1. Sec 5-3.1 Multinomial Probability Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 54 Example 5-24: Digital Channel Of the 20 bits received over a digital channel, 14 are of excellent quality, 3 are good, 2 are fair, 1 is poor. The sequence received was EEEEEEEEEEEEEEGGGFFP. The probability of that sequence is 0.6140.330.0820.021 = 2.708*10-9 However, the number of different ways of receiving those bits is a lot! 20 ! x E G F P P(x) 0.60 0.30 0.08 0.02 2, 325, 600 14 !3!2 !1! The combined result is a multinomial distribution. P x1 14, x 2 3, x 3 2, x 4 1 20 ! 0.6 0.3 0.08 0.02 0.0063 14 3 2 1 14 !3!2 !1! Sec 5-3.1 Multinomial Probability Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 55 Example 5-25: Digital Channel Refer again to the prior Example 5-24. What is the probability that 12 bits are E, 6 bits are G, 2 are F, and 0 are P? P x1 12, x 2 6, x 3 2, x 4 0 20 ! 0.6 0.3 0.08 0.02 0.0358 12 6 2 0 12 !6 !2 !0 ! Using Excel 0.03582 = (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2 Sec 5-3.1 Multinomial Probability Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 56 Multinomial Means and Variances The marginal distributions of the multinomial are binomial. If X1, X2,…, Xk have a multinomial distribution, the marginal probability distributions of Xi is binomial with: E(Xi) = npi and V(Xi) = npi(1-pi) (5-19) Sec 5-3.1 Multinomial Probability Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 57 Example 5-26: Marginal Probability Distributions Refer again to the prior Example 5-25. The classes are now {G}, {F}, and {E, P}. Now the multinomial changes to: PX 2 X 3 x 2 , x 3 n! x 2 ! x3 ! n x 2 x3 ! p x2 2 p x3 3 1 p2 p3 n x 2 x3 for x 2 0 to n x 3 and x 3 0 to n x 2 . Sec 5-3.1 Multinomial Probability Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 58 Example 5-27: Bivariate Normal Distribution Earlier, we discussed the two dimensions of an injection-molded part as two random variables (X and Y). Let each dimension be modeled as a normal random variable. Since the dimensions are from the same part, they are typically not independent and hence correlated. Now we have five parameters to describe the bivariate normal distribution: μX, σX, μY, σY, ρXY Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 59 Bivariate Normal Distribution Defined f XY x , y ; X , X , Y , Y , u 1 2 1 2 x X 2 X 2 1 2 X Y 1 2 x X e u 2 y Y y Y X Y Y 2 2 for x and y . x 0, P aram eter lim its: y 0, x , y , 1 1 Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 60 Role of Correlation Figure 5-17 These illustrations show the shapes and contour lines of two bivariate normal distributions. The left distribution has independent X, Y random variables (ρ = 0). The right distribution has dependent X, Y random variables with positive correlation (ρ > 0, actually 0.9). The center of the contour ellipses is the point (μX, μY). Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 61 Example 5-28: Standard Bivariate Normal Distribution Figure 5-18 This is a standard bivariate normal because its means are zero, its standard deviations are one, and its correlation is zero since X and Y are independent. The density function is: f XY x , y 1 2 e 2 0.5 x y 2 Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 62 Marginal Distributions of the Bivariate Normal If X and Y have a bivariate normal distribution with joint probability density function fXY(x,y;σX,σY,μX,μY,ρ), the marginal probability distributions of X and Y are normal with means μX and μY and σX and σY, respectively. (5-21) Figure 5-19 The marginal probability density functions of a bivariate normal distribution are simply projections of the joint onto each of the axis planes. Note that the correlation (ρ) has no effect on the marginal distributions. Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 63 Conditional Distributions of the Joint Normal If X and Y have a bivariate normal distribution with joint probability density fXY(x,y;σX,σY,μX,μY,ρ), the conditional probability distribution of Y given X = x is normal with mean and variance as follows: Y x Y Y x 2 2 Y Y X x X 1 2 Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 64 Correlation of Bivariate Normal Random Variables If X and Y have a bivariate normal distribution with joint probability density function fXY(x,y;σX,σY,μX,μY,ρ), the correlation between X and Y is ρ. Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. (5-22) 65 Bivariate Normal Correlation & Independence • In general, zero correlation does not imply independence. • But in the special case that X and Y have a bivariate normal distribution, if ρ = 0, then X and Y are independent. (5-23) Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 66 Example 5-29: Injection-Molded Part The injection- molded part dimensions has parameters as tabled and is graphed as shown. The probability of X and Y being within limits is the volume within the PDF between the limit values. This volume is determined by numerical integration – beyond the scope of this text. Figure 5-3 Mean Std Dev Correlation Upper Limit Lower Limit Sec 5-3.2 Bivariate Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. Bivariate X Y 3 7.7 0.04 0.08 0.8 3.05 7.80 2.95 7.60 67 Linear Functions of Random Variables • A function of random variables is itself a random variable. • A function of random variables can be formed by either linear or nonlinear relationships. We limit our discussion here to linear functions. • Given random variables X1, X2,…,Xp and constants c1, c2, …, cp Y= c1X1 + c2X2 + … + cpXp (5-24) is a linear combination of X1, X2,…,Xp. Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 68 Mean & Variance of a Linear Function Let Y= c1X1 + c2X2 + … + cpXp and use Equation 5-10: E Y c1 E X 1 c 2 E X 2 ... c p E X V Y c1 V 2 p (5-25) X 1 c 22V X 2 ... c 2pV X p 2 c i c j cov X i X j (5-26) i j If X 1 , X 2 , ..., X V Y c1 V 2 p are independent , then cov X i X j 0, X 1 c 22V X 2 .. . c 2pV X p Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. (5-27) 69 Example 5-30: Negative Binomial Distribution Let Xi be a geometric random variable with parameter p with μ = 1/p and σ2 = (1-p)/p2 Let Y = X1 + X2 +…+Xr, a linear combination of r independent geometric random variables. Then Y is a negative binomial random variable with μ = r/p and σ2 = r(1-p)/p2 by Equations 525 and 5-27. Thus, a negative binomial random variable is a sum of r identically distributed and independent geometric random variables. Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 70 Example 5-31: Error Propagation A semiconductor product consists of three layers. The variances of the thickness of each layer is 25, 40 and 30 nm. What is the variance of the finished product? Answer: X X1 X 2 X 3 3 V X V X i 25 40 30 95 nm 2 i 1 SD X 95 9.747 nm Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 71 Mean & Variance of an Average If X X 1 X 2 ... X p p T hen E X p and E X i (5 -28a) p If the X i are independent w ith V T hen V X p p 2 2 Xi 2 2 (5-28b) p Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 72 Reproductive Property of the Normal Distribution If X 1 , X 2 , ..., X p are independent , norm al random variables w ith E X i , and V X i = 2 , for i 1, 2, ..., p , then Y c1 X 1 c 2 X 2 ... c p X p is a norm al random variable w ith E Y c1 1 c 2 2 ... c p p and V Y c1 1 c 2 2 ... c p 2 2 2 2 2 2 p Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. (5-29) 73 Example 5-32: Linear Function of Independent Normals Let the random variables X1 and X2 denote the independent length and width of a rectangular manufactured part. Their parameters are shown in the table. What is the probability that the perimeter exceeds 14.5 cm? Parameters of X1 X2 Mean 2 5 Std Dev 0.1 0.2 Let Y 2 X 1 2 X 2 perim eter E Y 2 E X 1 2 E X 2 2 2 2 5 14 cm V Y 2 V 2 SD Y X 1 2 2 V X 2 4 0.1 2 4 0.2 0.04 0.16 0.20 2 0.20 0.4472 cm 14.5 14 P Y 14.5 1 1 1.1180 0.1318 .4472 Using Excel 0.1318 = 1 - NORMDIST(14.5, 14, SQRT(0.2), TRUE) Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 74 Example 5-33: Beverage Volume Soft drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid ounces, and the standard deviation is 0.1 fl oz. Assume that the fill volumes are independent, normal random variables. What is the probability that the average volume of 10 cans is less than 12 fl oz? Let X i denote the fill volum e of the i th can 10 Let X X i 10 i 1 EX n EX i 10 10 12.1 10 i 1 X 11 0 V X 2 V 12.1 fl oz 10 10 0.1 i i 1 2 0.001 fl oz 2 100 12 12.1 P X 12 3.16 0.00079 0.001 Using Excel 0.000783 = NORMDIST(12, 12.1, SQRT(0.001), TRUE) Sec 5-4 Linear Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 75 General Functions of Random Variables • A linear combination is not the only way that we can form a new random variable (Y) from one or more of random variables (Xi) that we already know. We will focus on single variables, i.e., Y = h(X), the transform function. • The transform function must be monotone: – Each value of x produces exactly one value of y. – Each value of y translates to only one value of x. • This methodology produces the probability mass or density function of Y from the function of X. Sec 5-5 General Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 76 General Function of a Discrete Random Variable Suppose that X is a discrete random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability mass function of the random variable Y is fY(y) = fX[u(y)] (5-30) Sec 5-5 General Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 77 Example 5-34: Function of a Discrete Random Variable Let X be a geometric random variable with PMF: fX(x) = p(1-p)x-1 for x = 1, 2, … Find the probability distribution of Y = X2. Solution: – Since X > 0, the transformation is one-to-one. – The inverse transform function is X = sqrt(Y). – fY(y) = p(1-p)sqrt(y)-1 for y = 1, 4, 9, 16,… Sec 5-5 General Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 78 General Function of a Continuous Random Variable Suppose that X is a continuous random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability density function of the random variable Y is fY(y) = fX[u(y)]∙|J| (5-31) where J = u’(y) is called the Jacobian of the transformation and the absolute value is used. Sec 5-5 General Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 79 Example 5-35: Function of a Continuous Random Variable Let X be a continuous random variable with probability distribution: x fX for 0 x 4 8 Find the probability distribution of Y = h(X) = 2X + 4 N ote that Y has a one-to-one relationship to X . x u y fY y y4 2 y 4 8 and the Jacobian is J u ' y 1 2 2 1 y4 for 4 y 12. 2 32 Sec 5-5 General Functions of Random Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger. 80 Important Terms & Concepts for Chapter 5 Bivariate distribution Bivariate normal distribution Conditional mean Conditional probability density function Conditional probability mass function Conditional variance Contour plots Correlation Covariance Error propagation General functions of random variables Independence Joint probability density function Joint probability mass function Linear functions of random variables Marginal probability distribution Multinomial distribution Reproductive property of the normal distribution Chapter 5 Summary 81 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.