851_lectures17_24

# 851_lectures17_24 - Statistics 851(Fall 2013 Prof Michael...

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Statistics 851 (Fall 2013) October 16, 2013 Prof. Michael Kozdron Lecture #17: Expectation of a Simple Random Variable Recall that a simple random variable is one that takes on finitely many values. Definition. Let ( , F , P ) be a probability space. A random variable X : R is called simple if it can be written as X = n i =1 a i 1 A i where a i R , A i F for i = 1 , 2 , . . . , n . We define the expectation of X to be E ( X ) = n i =1 a i P { A i } . Example 17.1. Consider the probability space ( , B 1 , P ) where = [0 , 1], B 1 denotes the Borel sets of [0 , 1], and P is the uniform probability on . Suppose that the random variable X : R is defined by X ( ω ) = 4 i =1 a i 1 A i ( ω ) where a 1 = 4, a 2 = 2, a 3 = 1, a 4 = 1, and A 1 = [0 , 1 2 ) , A 2 = [ 1 4 , 3 4 ) , A 3 = ( 1 2 , 7 8 ] , A 4 = [ 7 8 , 1] . Show that there exist finitely many real constants c 1 , . . . , c n and disjoint sets C 1 , . . . , C n B 1 such that X = n i =1 c i 1 C i . Solution. We find X ( ω ) = 4 , if 0 ω < 1 / 4 , 6 , if 1 / 4 ω < 1 / 2 , 2 , if ω = 1 / 2 , 3 , if 1 / 2 < ω < 3 / 4 , 1 , if 3 / 4 ω < 7 / 8 , 0 , if ω = 7 / 8 , 1 , if 7 / 8 < ω 1 , so that X = 7 i =1 c i 1 C i where c 1 = 4, c 2 = 6, c 3 = 2, c 4 = 3, c 5 = 1, c 6 = 0, c 7 = 1 and C 1 = [0 , 1 4 ) , C 2 = [ 1 4 , 1 2 ) , C 3 = { 1 2 } , C 4 = ( 1 2 , 3 4 ) , C 5 = [ 3 4 , 7 8 ) , C 6 = { 7 8 } , C 7 = ( 7 8 , 1] . 17–1

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Proposition 17.2. If X and Y are simple random variables, then E ( α X + β Y ) = α E ( X ) + β E ( Y ) for every α , β R . Proof. Suppose that X and Y are simple random variables with X = n i =1 a i 1 A i and Y = m j =1 b j 1 B j where A 1 , . . . , A n F and B 1 , . . . , B m F each partition . Since α X = α n i =1 a i 1 A i = n i =1 ( α a i )1 A i we conclude by definition that E ( α X ) = n i =1 ( α a i ) P { A i } = α n i =1 a i P { A i } = α E ( X ) . The proof of the theorem will be completed by showing E ( X + Y ) = E ( X ) + E ( Y ). Notice that { A i B j : 1 i n, 1 j m } consists of pairwise disjoint events whose union is and X + Y = n i =1 m j =1 ( a i + b j )1 A i B j . Therefore, by definition, E ( X + Y ) = n i =1 m j =1 ( a i + b j ) P { A i B j } = n i =1 m j =1 a i P { A i B j } + n i =1 m j =1 b j P { A i B j } = n i =1 a i P { A i } + m j =1 b j P { B j } and the proof is complete. Fact. If X and Y are simple random variables with X Y , then E ( X ) E ( Y ) . Exercise 17.3. Prove the previous fact. 17–2
Having already defined E ( X ) for simple random variables, our goal now is to construct E ( X ) in general. To that end, suppose that X is a positive random variable . That is, X ( ω ) 0 for all ω . (We will need to allow X ( ω ) [0 , + ] for some consistency.) Definition. If X is a positive random variable, define the expectation of X to be E ( X ) = sup { E ( Y ) : Y is simple and 0 Y X } . That is, we approximate positive random variables by simple random variables. Of course, this leads to the question of whether or not this is possible. Fact. For every random variable X 0, there exists a sequence ( X n ) of positive, simple random variables with X n X (that is, X n increases to X ).

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