s2 - Stat 5101 Lecture Slides Deck 2 Charles J. Geyer...

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Unformatted text preview: Stat 5101 Lecture Slides Deck 2 Charles J. Geyer School of Statistics University of Minnesota 1 Axioms An expectation operator is a mapping X 7 E ( X ) of random variables to real numbers that satisfies the following axioms: E ( X + Y ) = E ( X ) + E ( Y ) for any random variables X and Y , E ( X ) for any nonnegative random variable X (one such that X ( s ) for all s in the sample space), E ( aX ) = aE ( X ) for any random variable X and any constant a , and E ( Y ) = 1 when Y is the constant random variable s 7 1. 2 Axioms (cont.) The fourth axiom is usually written, a bit sloppily, as E (1) = 1 The reason this is sloppy is that on the left-hand side 1 must indicate a random variable, because the argument of an expecta- tion operator is always a random variable, and on the right-hand side 1 must indicate a real number, because the value of an expectation operator is always a real number. When we have a constant as an argument of an expectation op- erator, we always take this to mean a constant random variable. 3 Axiom Summary E ( X + Y ) = E ( X ) + E ( Y ) (1) E ( X ) , when X (2) E ( aX ) = aE ( X ) (3) E (1) = 1 (4) (3) and (4) together imply E ( a ) = a, for any constant a It can be shown (but we wont here) that when the sample space is finite these axioms hold if and only if the expectation operator is defined in terms of a PMF as we did before. 4 Axioms (cont.) E ( X + Y ) = E ( X ) + E ( Y ) says an addition operation can be pulled outside an expectation. X implies E ( X ) says nonnegativity can be pulled outside an expectation. E ( aX ) = aE ( X ) says a constant can be pulled outside an expectation. 5 Axioms (cont.) Many students are tempted to overgeneralize, and think anything can be pulled outside an expectation. Wrong! In general E ( XY ) 6 = E ( X ) E ( Y ) E ( X/Y ) 6 = E ( X ) /E ( Y ) E { g ( X ) } 6 = g E { X } although we may have equality for certain special cases. 6 Axioms (cont.) We do have E ( X- Y ) = E ( X )- E ( Y ) because E ( X- Y ) = E { X + (- 1) Y } = E ( X ) + (- 1) E ( Y ) by axioms (1) and (3). 7 Axioms (cont.) We do have E ( a + bX ) = a + bE ( X ) because E ( a + bX ) = E ( a ) + E ( bX ) = a + bE ( X ) by axioms (1), (3), and (4). 8 Axiom Summary (cont.) E ( X Y ) = E ( X ) E ( Y ) addition and subtraction come out X implies E ( X ) nonnegative comes out E ( aX ) = aE ( X ) constants come out E ( a + bX ) = a + bE ( X ) linear functions come out. But thats all! 9 Linearity of Expectation By mathematical induction the addition comes out axiom ex- tends to any finite number of random variables. For any random variables X 1 , ... , X n E ( X 1 + + X n ) = E ( X 1 ) + + E ( X n ) More generally, for any random variables X 1 , ... , X n and any constants a 1 , ... , a n E ( a 1 X 1 + + a n X n ) = a 1 E ( X 1 ) + + a n E ( X n ) This very useful property is called linearity of expectation ....
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s2 - Stat 5101 Lecture Slides Deck 2 Charles J. Geyer...

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