Joint Distributions. Tieming Ji. Fall 2012

Transcription

1 Joint Distributions Tieming Ji Fall / 33

2 X : univariate random variable. (X, Y ): bivariate random variable. In this chapter, we are going to study the distributions of bivariate random variables joint distributions, both in discrete cases and continuous cases. Motivation: Many problems require the study of two random variables at the same time. For example, studying the effect of pesticide application time for the soybean yield; the population size and the crime rate relationship; the phenotype and genotype in agronomy studies; the yield of a chemical reaction in conjunction with the temperature at which the reaction is run, etc. 2 / 33

3 Discrete Cases Definition: Let X and Y be discrete random variables. The ordered pair (X, Y ) is called a two-dimensional discrete random variable. The joint (probability) density function f for (X, Y ) is defined as f (x, y) = P(X = x, Y = y). Definition: Function f is a valid discrete joint density function if and only if f (x, y) 0 for x, y R, and x R y R f (x, y) = 1. 3 / 33

4 Example (book page 157): In an automobile plant two tasks are performed by robots. The first entails welding two joints; the second, tightening three bolts. Let X denote the number of defective welds and Y the number of improperly tightened bolts produced per car. Past data indicates that the joint density for (X, Y ) is shown in the table below. x/y Is f (x, y) a proper joint density distribution? f (x, y) is a valid joint discrete density for (X, Y ) because (1) for any x and y, f (x, y) 0; and (2) x R y R f (x, y) = 2 3 x=0 y=0 f (x, y) = = 1. 4 / 33

5 2. What is P(X = 1, Y = 0)? P(X = 1, Y = 0) = What is the probability that there is no improperly tightened bolts? Solution: When there is no improperly tightened bolts, Y =0. So, we want to compute P(Y = 0). We have P(Y = 0) = P(X = 0, Y = 0) + P(X = 1, Y = 0) + P(X = 2, Y = 0) = = / 33

6 Definition: The marginal probability density functions f X (x) for X given the joint probability density function f X,Y (x, y) is given by f X (x) = f X,Y (x, y). y R Similarly, the marginal probability density function f Y (y) for Y is given by f Y (y) = f X,Y (x, y). x R 6 / 33

7 Example: Compute the marginal distributions of X and Y given the table in the last example. x/y f X (x) f Y (y) / 33

8 Continuous Cases Definition: Let X and Y be continuous random variables. The ordered pair (X, Y ) is called a two-dimensional continuous random variable. The probability that X [a, b] and Y [c, d] given the joint continuous probability density f X,Y (x, y) is computed by P(a X b, c Y d) = b d a c f X,Y (x, y)dydx. Definition: Function f is proper continuous joint density function if and only if f (x, y) 0 for x, y R, and f X,Y (x, y)dydx = 1. 8 / 33

9 Example: In a healthy individual age 20 to 29 years, the calcium level in the blood, X, is usually between 8.5 and 10.5 milligrams per deciliter (mg/dl) and the cholesterol level, Y is usually between 120 and 240 mg/dl. Assume that for a healthy individual in this age group, X is uniformly distributed in the range (8.5, 10.5), and Y is uniformly distributed on (120, 240). Thus, the density function is f X,Y (x, y) = 1 240, 8.5 x 10.5 and 120 y 240; 0, o.w. 1. Verify that f X,Y is a proper probability density function. Solution: (1) For any x R and y R, f (x, y) is non-negative. (2) f (x, y)dydx = dydx = 1. 9 / 33

10 2. Compute the probability that a healthy person in this age group will have calcium level between 9 and 10 mg/dl and cholesterol level between 125 and 140 mg/dl. Solution: This is to compute P(9 X 10, 125 Y 140). P(9 X 10, 125 Y 140) = = = dydx ( )dx 10 / 33

11 Definition: Let (X, Y ) be a two-dimensional continuous random variable with joint density f X,Y. The marginal density for X, denoted by f X, is given by f X (x) = f X,Y (x, y)dy. The marginal density for Y, denoted by f Y, is given by f Y (y) = f X,Y (x, y)dx. 11 / 33

12 Example: Continued with the last example. 1. Compute the marginal distributions for X and Y. Solution: f X (x) = When x < 8.5 or x > 10.5, f X (x) = 0. f Y (y) = When y < 120 or y > 240, f Y (y) = dy = 1, 8.5 x dx = 1, 120 y / 33

13 2. Compute the probability that a healthy individual has a cholesterol level between 150 and 200. Solution: This is to compute P(150 Y 200). P(150 Y 200) = dydx = We can also use marginal density to compute this as follows. P(150 Y 200) = dy = / 33

14 Example: Let 1, 0 x 1, 0 y 1. f (x, y) = 0, o.w. Compute the probability that 1 2 X + Y / 33

15 Example: Let e (x+y), x, y 0 f (x, y) = 0, o.w. Compute the probability that Y X / 33

16 Example: In studying the behavior of air support roofs, the random variables X, the inside barometric pressure (in inches of mercury), and Y, the outside pressure, are consdered. Assume that the joint density for (X, Y ) is given by f X,Y (x, y) = c x when 27 y x 33, and 1 c = ln Compute the marginal density functions of X and Y. 16 / 33

17 2. Compute the probability that X 30 and Y / 33

18 Independence Definition: Let X and Y be random variables with joint density f XY and marginal densities f X and f Y, respectively. X and Y are independent if and only if for all x and y. f XY (x, y) = f X (x)f Y (y) 18 / 33

19 Example: Continue with Example on slide no. 4. x/y f X (x) f Y (y) X and Y are not independent, because f XY (x = 0, y = 0) = 0.840, and f X (x = 0)f Y (y = 0) = = f XY (x = 0, y = 0). 19 / 33

20 Example: Continue with Example on slide no. 9 and no. 12. We have already derived the joint pdf and marginal pdf s as follows , 8.5 x 10.5 and 120 y 240; f X,Y (x, y) = 0, o.w. f X (x) = f Y (y) = 1 2, 8.5 x 10.5; 0, o.w , 120 y 240; 0, o.w. Thus, for any pair of (x, y) R 2, f X,Y (x, y) = f X (x)f Y (y). X and Y are independent. 20 / 33

21 Expectation For (X, Y ) discrete, E(X ) = xf X (x) = xf XY (x, y), and x R x R y R E(Y ) = yf Y (y) = yf XY (x, y). y R x R y R For (X, Y ) continuous, E(X ) = xf X (x)dx = xf XY (x, y)dxdy, and x R E(Y ) = y R yf Y (y)dy = x R y R y R x R yf XY (x, y)dxdy. 21 / 33

22 Extension: Let (X, Y ) be a two-dimensional random variable with joint density f XY. Let h(x, Y ) be a random variable. The expected value of h(x, Y ), denoted by E(h(X, Y )), is given by E(h(X, Y )) = h(x, Y )f XY (x, y), x R y R when (X, Y ) is discrete and E(h(X, Y )) exists; or E(h(X, Y )) = h(x, Y )f XY (x, y)dxdy, x R y R when (X, Y ) is continuous and E(h(X, Y )) exists. 22 / 33

23 Example: The joint pdf of X and Y are in the following table. Compute E(X ) and E(X + Y ). x/y f X (x) f Y (y) / 33

24 Example: The joint density for the random variable (X, Y ), where X denotes the calcium level and Y denotes the cholesterol level in the blood of a healthy individual, is given by f X,Y (x, y) = Compute E(X ) and E(XY ) , 8.5 x 10.5 and 120 y 240; 0, o.w. 24 / 33

25 Covariance Definition: Let X and Y be random variables with means µ X and µ Y, respectively. The covariance between X and Y, denoted by Cov(X, Y ) or σ XY is given by Cov(X, Y ) = E((X µ X )(Y µ Y )). Using definition to compute covariance is often inconvenient. Instead we use the following formula to compute covariance. Cov(X, Y ) = E((X µ X )(Y µ Y )) = E(XY ) E(X )E(Y ). 25 / 33

26 Theorem: Let (X, Y ) be a two-dimensional random variable with joint density f XY. If X and Y are independent then E(XY ) = E(X )E(Y ). Thus, If X and Y are independent, then Cov(X, Y ) = E(XY ) E(X )E(Y ) = 0. That is, independence covariance is 0. However, when covariance is 0 for two random variables, it is not always true that they are independent. 26 / 33

27 Example (book page 169): The joint density for X and Y is in the following table. x/y f X (x) 1 0 1/4 1/4 0 1/2 4 1/ /4 1/2 f Y (y) 1/4 1/4 1/4 1/4 1 From the table, we have E(X ) = 5/2, E(Y ) = 0, and E(XY ) = 0, yielding a covariance of 0. However, X and Y are not independent. 27 / 33

28 Correlation Definition: Let X and Y be random variables with means µ X and µ Y and variances σx 2 and σ2 Y, respectively. The correlation, ρ XY, between X and Y is given by ρ XY = Cov(X, Y ) σ 2 X σ 2 Y = Cov(X, Y ) σ X σ Y. This is also called the Pearson coefficient of correlation. 28 / 33

29 Property: The Pearson correlation coefficient ρ XY ranges from -1 to 1, and it measures linearity of X and Y. Specifically, When ρ XY =1, Y can be written as Y = β 0 + β 1 X and β 1 is positive. When ρ XY =-1, Y can be written as Y = β 0 + β 1 X and β 1 is negative. When ρ XY = 0, there is no linear relationship between X and Y, but they could have other relationships. 29 / 33

30 Example: The joint pdf of X and Y are in the following table. Compute ρ XY. x/y f X (x) f Y (y) / 33

31 Conditional Density Definition: Let (X, Y ) be a two-dimensional random variable with joint density f XY and marginal densities f X and Y. Then The conditional density for X given Y = y is given by f X Y =y (x) = f XY (x, y), when f Y (y) > 0. f Y (y) The conditional density for Y given X = x is given by f Y X =x (y) = f XY (x, y), when f X (x) > 0. f X (x) Note: When two r.v. s are independent, the marginal density is the same with the conditional density, i.e. f X Y (x) = f (x). and same for Y. 31 / 33

32 Example: The joint pdf of X and Y are in the following table. Compute P(X 1 y = 0). x/y f X (x) f Y (y) / 33

33 Chapter Summary For both the discrete and continuous cases, know the definition of joint pdf, and use the joint pdf to compute marginal pdf s. Given the joint pdf, be able to compute probability of X and Y satisfying specific requirements, such as P(X + Y 1), P(X < Y < 0.5), etc. Expectations computed from marginal density or joint density. Covariance and independence. Correlation. Relationship of marginal density, conditional density and joint density. 33 / 33