PostLecture09EN(2).pdf - Lecture 9 Chapter 6.3 6.4 the...

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Lecture 9 Chapter 6.3, 6.4, the distribution of a sum of 2 independent variables Plan for today I The CDF method for 2 variables I The inverse function method I The Jacobian method I The method of moment generating functions for finding the distribution of a sum of random variables
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Example - CDF method for 2 variables I The arrival times of a bus and a passenger are mutually independent and uniformly distributed on ( 0 , 1 ) . Calculate the probability density function of X 2 - X 1 , with X 1 = the arrival time of the bus and X 2 = the arrival time of the passenger. I Let Y = X 2 - X 1 . I P ( Y y ) = P ( X 2 - X 1 y ) . I A = { ( x 1 , x 2 ) : 0 x 1 1 , 0 x 2 1 } . I D y = { ( x 1 , x 2 ) : x 2 - x 1 y } ∩ A . I P ( Y y ) = R R D y f ( x 1 , x 2 ) dA . I X 1 and X 2 are mutually independent, so f ( x 1 , x 2 ) = f 1 ( x 1 ) f 2 ( x 2 ) . I
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I For y [ 0 , 1 ] , integrate over the blue area I For y [ 0 , 1 ] ( y is a constant here ) P ( X 2 - X 1 y ) = 1 - Z 1 y Z x 2 - y 0 f ( x 1 , x 2 ) dx 1 dx 2 = 1 - Z 1 y ( x 2 - y ) dx 2 = 1 - ( 1 - y ) 2 2 I For y > 1, P ( X 2 - X 1 y ) = 1.
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I For y [ - 1 , 0 ] , integrate over I For - 1 < y 0, P ( Y y ) = Z 1 - y Z x 1 + y 0 dx 2 dx 1 (in this case -y > 0) = Z 1 - y ( x 1 + y ) dx 1 = ( 1 + y ) 2 2
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Example- conclusion I F Y ( y ) = P ( Y y ) = 1 for y > 1 1 - ( 1 - y ) 2 2 y [ 0 , 1 ] ( 1 + y ) 2 2 for y [ - 1 , 0 ] 0 otherwise I f Y ( y ) = dF Y ( y ) dy = 1 - y u ( 0 , 1 ) 1 + y for u ( - 1 , 0 ) The CDF method for 2 variables I Let Y = u ( X 1 , X 2 ) Question: Calculate f Y ( y ) . I Step 1: Draw the borders of the area A on which f ( x 1 , x 2 ) 6 = 0 . I Step 2: For different values of y , draw the graph of u ( x 1 , x 2 ) = y . I Step 3: Find the values of y for which the cross section D y = { ( x 1 , x 2 ) : u ( x 1 , x 2 ) y } ∩ A changes. I Step 4: For all values of y found in Step 3, calculate P ( Y y ) = Z Z D y f ( x 1 , x 2 ) dA . I Calculate f ( y ) = dF Y ( y ) dy . Homework: Study the examples 6.2.4 in the textbook.
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The inverse function method for discrete variables Example 6.3.1 in the textbook Let X be a geometrically distributed random variable with parameter p and Y = u ( X ) = X - 1. Find the probability mass function of Y . I P ( Y = y ) = P ( X - 1 = y ) = P ( X = y + 1 ) = ( 1 - p ) y p , for y = 0 , 1 , 2 ... Note that if y = x - 1, x = y + 1, so w ( y ) = y + 1 is the inverse of u . Theorem ( Th. 6.3.1 ) If X is a discrete random variable and u ( · ) a bijective function with inverse w, then the probability mass function of Y = u ( X ) is given by P ( Y = y ) = P ( u ( X ) = y ) = P ( X = w ( y )) .
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Example I Let Y be a random variable defined by Y = e X , with X N ( μ, σ 2 ) . Find the probability density function of Y. I F Y ( y ) = P ( Y y ) = P ( e X y ) I Define u ( x ) = e x . I For y 0, P ( Y y ) = 0. I For y > 0: The function u ( · ) is increasing and bijective I u - 1 ( y ) = ln ( y ) is also increasing, hence I P ( Y y ) = P ( u ( X ) y ) = P ( X u - 1 ( y )) = F X ( u - 1 ( y )) . I The pdf. of Y can be found by f Y ( y ) = dF Y ( y ) dy = dF X ( u - 1 ( y )) dy By using the chain rule, = du - 1 ( y ) dy f X ( u - 1 ( y )) .
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The inverse function method(method of transformations) Theorem (Th. 6.3.2 textbook) Let X be a random variable with probability density function f X and u an invertible, increasing or decreasing function on A = { x | f X ( x ) > 0 } , with inverse w.
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  • Fall '19
  • Probability theory, probability density function, CDF, dy, y P

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