STAT 430/510 Probability Lecture 14: Joint Probability Distribution, Continuous Case

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1 STAT 430/510 Probability Lecture 14: Joint Probability Distribution, Continuous Case Pengyuan (Penelope) Wang June 20, 2011

2 Joint density function of continuous Random Variable When X and Y are two continuous random variables, the joint density function f (x, y) is a function defined for each pair of numbers (x,y) by f (x, y) Since the total probability of the all possible pairs of (x,y) is 1, the joint density function must satisfy f (x, y)dxdy = 1 Also x,y f (x, y) 0

pair of numbers (x,y) by f (x, y) Since the total probability of the all possible pairs

3 Marginal pdf of Continuous Random Variable The marginal pdf s of X and Y, denoted by f X (x) and f Y (y), respectively, are given by f X (x) = f (x, y)dy y f Y (y) = f (x, y)dx x

4 Example The joint density { function of X and Y is given by 2e f (x, y) = x e 2y, 0 < x <, 0 < y < 0, otherwise Check that f (x, y) is a joint density function. Check 1: Check 2: x,y f (x, y) 0 f (x, y)dxdy = 1

5 Example-continue The joint density { function of X and Y is given by 2e f (x, y) = x e 2y, 0 < x <, 0 < y < 0, otherwise Compute the marginal density of X and Y.

6 Example-continue The joint density { function of X and Y is given by 2e f (x, y) = x e 2y, 0 < x <, 0 < y < 0, otherwise Compute the marginal density of X and Y. f X (x) = 0 2e x e 2y dy = e x f Y (y) = 0 2e x e 2y dx = 2e 2y

7 Usage 1: compute probability X and Y are two continuous r.v. s. P[(X, Y ) A] = (x,y) A f (x, y)dxdy In the last example, compute (a) P(X > 1, Y < 1) (b) P(X < Y )

8 Usage 1: compute probability X and Y are two continuous r.v. s. P[(X, Y ) A] = (x,y) A f (x, y)dxdy In the last example, compute (a) P(X > 1, Y < 1) (b) P(X < Y ) 1 P(X > 1, Y < 1) = 1 0 2e x e 2y dxdy = e 1 e 3 P(X < Y ) = y 0 0 2e x e 2y dxdy = 1/3

9 Usage 2: compute marginal Expected Value X and Y are two continuous r.v. s and they have marginal distribution f X (x) and f Y (y). E[X] = xf X (x)dx E[Y ] = x y yf Y (y)dy In the last example, what is the expected value of X? EX = 0 xe x dx = 1.

10 Usage 3: compute Expected Value of a function of X and Y If X and Y have a joint probability mass function p(x, y), then E[g(X, Y )] = g(x, y)p(x, y) y x X and Y are two continuous r.v. s. E[g(X, Y )] = g(x, Y )f (x, y)dxdy x,y In the last example, what is the expected value of e 1 2 X+Y? E[e 1 2 X+Y ] = y 0 0 e 1 2 X+Y 2e x e 2y dxdy = y 0 2e 1 2 X Y dxdy = 4. 0

11 Usage 4: compute Conditional probability X and Y are two continuous r.v. s and they have marginal distribution f X (x) and f Y (y). f X Y (x y) = f (x, y) f Y (y) f Y X (y x) = f (x, y) f X (x) What is f X Y (x y)? Given that y = 2, what is the distribution of x? f X Y (x y) = e x, for any y.

f X Y (x y) = f (x, y) f Y (y) f Y X (y x) = f (x, y) f X (x) What is f X

12 Conditional Expectation E[X Y = y] = xf X Y (x y)dx. What is E[X Y = 2]? E[X Y = y] = xe x dx = 1.

13 Comments Again, conditional expectation satisfies all of the properties of ordinary expectation, for example E[ n i=1 X i Y = y] = n i=1 E[X i Y = y]

14 Usage 5: Check independence When X and Y are continuous, X and Y are independent if and only if f (x, y) = f X (x)f Y (y), for all x, y Are X and Y independent? They are, since f (x, y) = 2e x e 2y = f X (x)f Y (y)

15 Example The joint density of X and Y is given by f (x, y) = { 12 5 x(2 x y), 0 < x < 1, 0 < y < 1 0, otherwise Compute the conditional density of X given that Y = y, where 0 < y < 1.

16 Example The joint density of X and Y is given by f (x, y) = { 12 5 x(2 x y), 0 < x < 1, 0 < y < 1 0, otherwise Compute the conditional density of X given that Y = y, where 0 < y < 1. f X Y (x y) = = = f (x, y) f Y (y) x(2 x y) x(2 x y)dx 6x(2 x y) 4 3y

17 Example 1: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. Find out f X (x). Given X = 0.5, find f Y X (y x = 0.5) and E[Y X = 0.5].

18 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function.

19 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. 1 x=0 1 x y=0 f (x, y)dydx = 1.

20 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. 1 x=0 1 x y=0 Find out f X (x). f (x, y)dydx = 1.

21 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. 1 x=0 1 x y=0 f (x, y)dydx = 1. Find out f X (x). f X (x) = 1 x y=0 f (x, y)dy = 12x(1 x)2.

22 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. 1 x=0 1 x y=0 f (x, y)dydx = 1. Find out f X (x). f X (x) = 1 x y=0 f (x, y)dy = 12x(1 x)2. Given X = 0.5, find f Y X (y x = 0.5) and E[Y X = 0.5].

23 Example: f (x, y) = 24xy, where 0 < x < 1, 0 < y < 1, 0 < x + y < 1, and it equals to 0 otherwise. Show that f (x, y) is a joint probability density function. 1 x=0 1 x y=0 f (x, y)dydx = 1. Find out f X (x). f X (x) = 1 x y=0 f (x, y)dy = 12x(1 x)2. Given X = 0.5, find f Y X (y x = 0.5) and E[Y X = 0.5]. f Y X (y x = 0.5) = f (X = 0.5, y)/f X (0.5) = 8y, for y (0, 1 0.5). E[Y X = 0.5] = y=0 yf Y X (y x = 0.5)dy = y=0 8y 2 dy = 1/ y (1 0.5) 2 =

24 Example 2 A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P.M. Then f (x, y) = (1/60) 2, 0 < x < 60, 0 < y < 60, where x represents the number of minutes after 12 noon when the man arrives, and y is for the woman. (why?) find the probability that the first to arrive has to wait longer than 10 minutes.

25 Example: Solution X and Y denote, respectively, the time past 12 that the man and the woman arrive. X and Y are independent uniform random variables over (0,60). = = P(X + 10 < Y ) + P(Y + 10 < X) f (x, y)dxdy + {x+10<y} 60 y = 25/36 (1/60) 2 dxdy {y+10<x} 60 x f (x, y)dxdy (1/60) 2 dydx

26 Example 3 An accident occurs at a point X that is uniformly distributed on a road of length L. At the time of the accident, an ambulance is at a location Y that is also uniformly distributed on the road. Assuming that X and Y are independent, find the expected distance between the ambulance and the point of the accident.

27 Example: Continued Need to compute E X Y The joint density function of X and Y is f (x, y) = 1 L 2, 0 < x < L, 0 < y < L L L 1 E[ X Y ] = x y dydx 0 0 L2 = 1 L x L 2 ( (x y)dy + 0 = 1 L 2 L = L ( L2 2 + x 2 x)dx L x (y x)dy)dx

Statistics 100A Homework 7 Solutions

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