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Statistics 510: Notes 20 Reading: Sections 7.1-7.3 In Chapter 7, we study more properties of expected values. I. Expectations of Sums of Random Variables (Section 7.2) Recall Proposition 4.1 of Chapter 4: E[ g ( X )] g ( x) P( X x) possible values of x A two-dimensional analog of this proposition is Proposition 2.1 of Chapter 7: If X and Y have a joint probability mass function p( x, y ) , then E[ g ( X , Y )] g ( x, y ) p( x, y ) y x Similarly, for continuous random variables, E[ g ( X , Y )] g ( x, y) f ( x, y)dxdy An important application of Proposition 2.1 is to sums of random variables. Suppose E ( X ), E (Y ) are both finite and let g ( X , Y ) X Y . Then, in the continuous case, 1 E[ X Y ] ( x y) f ( x, y)dxdy xf ( x, y )dxdy yf ( x, y )dxdy xf ( x, y )dydx yf ( x, y )dxdy xf ( x) dx yf ( y )dy X Y E[ X ] E[Y ] The same result holds for discrete random variables that E[ X Y ] E[ X ] E[Y ] . By induction, if E[ X i ] is finite for all i 1, , n, then E[ X1 X n ] E[ X1 ] E[ X n ] (1.1) Example 1: Let X be a binomial random variable with parameters n and p , i.e., X is the number of successes in n independent trials when each trial has probability p of being a success. Find E ( X ) . 2 Example 2: The Sample Mean. Let X 1 , , X n be independent and identically distributed random variables having cdf F and expected value . Such a sequence of random variables is said to constitute a random sample from F. The quantity X , defined by n X X i , i 1 n is called the sample mean. Compute E ( X ) . Example 3: A group of N people throw their jackets into the center of a room. The jackets are mixed up and each person randomly selects one. Find the expected number of people that select their own jacket. 3 Example 4: Shuffle an ordinary deck of 52 playing cards. Then turn up cards from the top until the first ace appears. What is the expected value of the number of cards required to produce the first ace? 4 II. Moments of the Number of Events That Occur (Chapter 7.3) Example 1 continued: Consider n independent trials, with each trial being a success with probability p. Let X be the number of successes in the n trials; X is a binomial ( n, p ) random variable. In Section 7.2, we used the following strategy to find E ( X ) . We noted that X is the number of the events A1 , , An that occur where Ai is the event that the ith trial is a success. We then defined indicator variables 1, if Ai occurs Ii 0, otherwise We then computed E ( X ) via the formula E ( X ) E ( I1 I n ) E ( I1 ) E ( I n ) . 5 The use of indicator random variables for events to compute the expected value can be extended to compute higher order moments (e.g., variances) in the following way. Consider the number of pairs of events A1 , , An that occur. Because I i I j will equal 1 if both Ai and A j occur, and will equal 0 otherwise, it follows that the number of pairs is equal to I i I j . But as X is the number of events that i j occur, it also follows that the number of pairs of events that X occur is 2 . Consequently, X Ii I j 2 i j Taking expectations yields, X E E[ I i I j ] P( Ai Aj ) i j 2 i j (1.2) or X ( X 1) E P( Ai Aj ) 2 i j giving that E X 2 E[ X ] 2 P( Ai Aj ) i j (1.3) 2 2 2 which yields E[ X ] and thus Var ( X ) E ( X ) ( E[ X ]) . 6 Moreover, by considering the number of distinct subsets of k events that all occur, we see that X I i1 I i 2 I ik . k i1 i2 ik Taking expectations gives the identity X E E[ I i1I i 2 I ik ] P( Ai1 Ai 2 Aik ) i1 i2 ik k i1 i2 ik Example 1 continued: Ai is the event that the ith trial is a 2 success. When i j, P( Ai Aj ) p . Consequently, by (1.2) X n 2 2 E p p 2 2 i j or equivalently by (1.4) E[ X 2 ] E[ X ] n(n 1) p 2 . n Using that E[ X ] P( Ai ) np , this yields that i 1 Var[ X ] E[ X 2 ] ( E[ X ])2 n(n 1) p 2 np (np) 2 np(1 p) Example 3 continued: The Matching Problem. A group of N people throw their jackets into the center of a room. The jackets are mixed up and each person randomly selects one. We showed that the expected number of people that select 7 their own jacket is 1. Find the variance of the number of people that select their own jacket. 8