Ch5 Transformation of Random Variables

Ch5 Transformation of Random Variables - Stat1801...

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Stat1801 Probability and Statistics: Foundations of Actuarial Science Fall 2010-2011 Chapter V Transformation of Random Variables § 5.1 Distribution of Functions of A Single Random Variable In general, functions of random variables are random variables. Example 5.1 1. If , then ( 1 , 0 ~ N Z ) ( ) 2 1 , 2 1 ~ 2 Γ = Z Y , ( ) 2 , ~ σμ μσ N Z X + = . 2. If () λ α , ~ Γ X , then Γ = a aX Y , ~ . 3. If , then ( 1 , 0 ~ U X ) Exp X Y ~ ln 1 = . Theorem Let X be a continuous random variable on distributed on a space S with pdf ( ) x f X . Let where g is a function such that exists. Then the pdf of Y can be ( X g Y = ) 1 g obtained by () () 1 1 y g dy d y g f y f X Y = , ( ) S g y . Proof Let be the distribution function of X . The distribution function of Y is x F X () ( ) ( ) ( ) y X g P y Y P y F Y = = . If g is a strictly increasing function, then () ( ) ( ) ( ) ( ) y g F y g X P y F X Y 1 1 = = ( ) y g F dy d y f X Y 1 = y g dy d y g f X 1 1 = . p. 91
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Stat1801 Probability and Statistics: Foundations of Actuarial Science Fall 2010-2011 On the other hand, if g is a strictly decreasing function, then () ( ) ( ) ( ) ( ) y g F y g X P y F X Y 1 1 1 = = () y g F dy d y f X Y 1 = y g dy d y g f X 1 1 = . Since y g dy d 1 is positive when g is increasing, and is negative when g is decreasing, the pdf of Y can be expressed as () () 1 1 y g dy d y g f y f X Y = , ( ) S g y . Example 5.2 Let and ( 1 , 0 ~ U X ) X X g Y log 1 λ = = . 1 = x f X , 1 0 < < x 1 , 0 = S , = , 0 S g Since y e x x y x g = = log 1 , the inverse function exists. The pdf of Y is y e y g = 1 y y X Y e dy d e f y f = y y e e = × = 1 , . 0 > y Therefore Exp X Y ~ log 1 = . p. 92
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Stat1801 Probability and Statistics: Foundations of Actuarial Science Fall 2010-2011 Example 5.3 Let and . ( 1 , 0 ~ N X ) () X Y Φ = = , S , () ( ) 1 , 0 = Φ S Since Φ is a one-one function, 1 Φ exists. The pdf of Y is given by () () 1 1 y dy d y f y f X Y Φ Φ = y y 1 1 ' 1 Φ Φ Φ = φ , 1 = ( ) 1 , 0 y (as x x = Φ ') Hence . 1 , 0 ~ U Y Example 5.4 ( Log-normal Distribution ) Let ( ) 2 , ~ σ μ N X and . X e X g Y = = = 2 2 2 2 exp 2 1 πσ x x f X , < < x = , S , = , 0 S g Since , the inverse x y y e y x g x log = = = ( ) y y g log 1 = exists. The pdf of Y is given by log log y dy d y f y f X Y = ( ) = 2 2 2 2 log exp 2 1 y y , . 0 > y It is called the log-normal distribution which is commonly used to model multiplicative product of random quantities such as return rate on stock investment. p. 93
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Stat1801 Probability and Statistics: Foundations of Actuarial Science Fall 2010-2011 Example 5.5 Let and . ( 1 , 0 ~ N X ) 2 X Y = () () 2 2 2 1 x X e x x f = = π φ , < < x () = , S , () ( ) = , 0 S g Since y x y x y x g ± = = = 2 does not have a unique solution, does not exists. We cannot use the formula. y g 1 We may start from the distribution function. Consider () ( ) ( ) y X P y Y P y F Y = = 2 ( ) y X y P = ( ) 1 2 Φ = y ' 2 ' y dy d y y F y f Y Y Φ = = 2 1 2 1 exp 2 1 2 2 y y = 2 2 1 2 1 2 1 y e y = 2 1 2 1 2 1 2 1 2 1 y e y Γ = , . 0 > y Hence 2 1 , 2 1 ~ 2 Γ = X Y .
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Ch5 Transformation of Random Variables - Stat1801...

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