bst631-fall-2008-lecture-05

bst631-fall-2008-lecture-05 - Lecture 5 on BST 631:...

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Lecture 5 on BST 631: Statistical Theory I – Kui Zhang, 09/02/2008 1 Review for the previous lecture Definition: cdf, pmf, pdf, identically distributed Theorem: How to determine the cdf, pmf, or pdf. Example : how to calculate cdf, pmf, or pdf. Chapter 2 – Transformations and Expectations Chapter 2.1 – Distributions of Functions of a Random Variable Problem: Let X be a random variable with cdf () X Fx . If we define any function of X , say Yg X = , then X = is also a random variable whose distribution depends on X F and the function g . Specifically, for any set A , ( ( ) ) . PY A Pg X A =∈ Formally, if X X and Y Y , then ( ) gX is a mapping from the sample space of X , X , to the sample space of Y , Y , i.e., ( ): gx XY . To go from Y back to X , we define the inverse function of g , denoted by 1 g , as 1 () { :() } gA x g x A = ∈∈ X . If A is a set that only contains a single point, say { } A y = , then 1 : () } gy x g x y = ∈= X .
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Lecture 5 on BST 631: Statistical Theory I – Kui Zhang, 09/02/2008 2 Therefore, for the random variable ( ) Yg X = , 1 () ( ( ) ) ( {: ( ) } ) ( ( ) ) P Y AP g X x g xA P X gA ∈= = X . However, we have mentioned that the distribution of Y can be determined by its cdf and pdf (or pmf). Actually, we only need to calculate the following: 1 ( ) ( ( ) ) ({ : ( ) }) ( (( , ])) P Y yP g X x g xy P X g y ≤= = X . Discrete Case: Let X be a discrete random variable with pmf ( ) X f x. In this case X is a countable set. Define X = for some function g so that { : ( ), } Yy y g x x = =∈ X is also a countable set. Then the pmf of Y is given by. 11 ( ) ( ) Y X xg y f Y y P X x f x −− ∈∈ == = = = . Example 2.1.1 (Binomial transformation): Let be a discrete random variable with pmf (known as the Binomial pmf): ( ) ( 1 ) , 0 , 1 , , xn x X n f xP X x p p x n x ⎛⎞ = = ⎜⎟ ⎝⎠ " , where n is a positive integer and 0 1 p . In this case, we call and p as parameters of a Binomial distribution. Consider ( ) X nX so that {0,1, , } n = " Y has the same sample space with X. Then ( ) ( ) ( 1 ), 0 , 1 ny y Y n f Y n X y p py n =− = = = " . Y also has a Binomial distribution but with parameters n and 1 p .
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Lecture 5 on BST 631: Statistical Theory I – Kui Zhang, 09/02/2008 3 Example: Suppose that X has the following pmf: exp( 1), 0; exp( 1)/(2* !), 1,2, ; () exp( 1)/(2*( )!), 1, 2, ; 0,otherwise. X x xx fx −= = −− = " " Find the pmf of exp(| |) YX = . Solution: The support set if Y is {1,exp(1),exp(2), } A = " . For any yA . We have: exp( 1), 1; ( ) ( ) (exp(| |) ) ( log( )) exp( 1)/(log( ))!, exp(1),exp(2), ; 0,otherwise.
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This note was uploaded on 03/17/2010 for the course STATISTIC 472 taught by Professor Amjad during the Spring '08 term at Yarmouk University.

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bst631-fall-2008-lecture-05 - Lecture 5 on BST 631:...

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