Chapter6

# Chapter6 - Chapter 6 lecture notes Math 431, Spring 2011...

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Unformatted text preview: Chapter 6 lecture notes Math 431, Spring 2011 Instructor: David F. Anderson Chapter 6: Jointly Distributed Random Variables Really need to be reading book now. Formally discuss the possibility of more than one random variable at a time. Case 1: Discrete Random Variables. Suppose that X and Y are discrete random variables built on same sample space (experiment). Thus, X,Y : S R . We define the joint probability mass function for X and Y by p ( x,y ) = P { X = x,Y = y } . Note that p X ( x ) = P { X = x } = P { X = x,Y R } = X y : p ( x,y ) > p ( x,y ) , and p Y ( y ) = X x : p ( x,y ) > p ( x,y ) Example 1. Flip a fair coin twice. Let X = 1 if head on first flip, 0 if tail. Let Y be number of heads. Find p ( x,y ) and p X , p Y (GIVE CHART: CALLED MARGINALS BECAUSE....) Solution: The ranges for X and Y are { , 1 } , { , 1 , 2 } , respectively. We have p (0 , 0) = P ( X = 0 ,Y = 0) = P ( X = 0) P ( Y = 0 | X = 0) = (1 / 2)(1 / 2) = 1 / 4 p (0 , 1) = P ( X = 0 ,Y = 1) = P ( X = 0) P ( Y = 1 | X = 0) = (1 / 2)(1 / 2) = 1 / 4 p (0 , 2) = P ( X = 0 ,Y = 2) = 0 p (1 , 0) = P ( X = 1 ,Y = 0) = 0 p (1 , 1) = P ( X = 1 ,Y = 1) = P ( X = 1) P ( Y = 1 | X = 1) = (1 / 2)(1 / 2) = 1 / 4 p (1 , 2) = P ( X = 1 ,Y = 2) = P ( X = 1) P ( Y = 2 | X = 1) = (1 / 2)(1 / 2) = 1 / 4 . 1 x . . . y 1 2 Row sum = P { X = x } 1 / 4 1 / 4 1 / 2 1 1 / 4 1 / 4 1 / 2 Column Sum = P { Y = y } 1 / 4 1 / 2 1 / 4 Further, p X (0) = 2 X y =0 p (0 ,y ) = p (0 , 0) + p (0 , 1) + p (0 , 2) = 1 / 2 p X (1) = 2 X y =0 p (1 ,y ) = p (1 , 0) + p (1 , 1) + p (1 , 2) = 1 / 2 p Y (0) = 2 X x =0 p ( x, 0) = p (0 , 0) + p (1 , 0) = 1 / 4 p Y (1) = 2 X x =0 p ( x, 1) = p (0 , 1) + p (1 , 1) = 1 / 2 p Y (2) = 2 X x =0 p ( x, 2) = p (0 , 2) + p (1 , 2) = 1 / 4 Example 2. A college has 90 male and 30 female profs. A committee of 5 is selected at random. Let X and Y be the # of men and women selected respectively. Find joint p.m.f. p ( x,y ) and marginal p.m.f. p X ( x ) and p Y ( y ). Solution: The range for both RVs is { , 1 , 2 , 3 , 4 , 5 } . The joint p.m.f. is p ( x,y ) = ( 90 x )( 30 y ) / ( 120 5 ) x,y { , 1 , 2 , 3 , 4 , 5 } and x + y = 5 else . To find p X ( x ) we note that for x in range of X , p X ( x ) = 5 y =0 p ( x,y ), which is only nonzero when x + y = 5. Thus, p X ( x ) = p ( x, 5- x ) = 90 x 30 5- x / 120 5 , and for y in the range of Y p Y ( y ) = p (5- y,y ) = 90 5- y 30 y / 120 5 . Both are hypergeometric. 2 Case 1: Continuous Random Variables. We say that X and Y are jointly continuous if there exists a function f ( x,y ), defined for all x,y , such that for any C R 2 P { ( X,Y ) C } = ZZ ( x,y ) C f ( x,y ) dxdy. f ( x,y ) is called the joint probability density function of X and Y . Thus, note P { X A,Y B } = Z B Z A f ( x,y ) dxdy....
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## This note was uploaded on 04/17/2011 for the course MATH 431 taught by Professor Balazs during the Spring '05 term at Wisconsin.

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Chapter6 - Chapter 6 lecture notes Math 431, Spring 2011...

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