Ch10: More Expectations and Variances Part I: Expected Values of - - PowerPoint PPT Presentation

Ch10: More Expectations and Variances

Part I: Expected Values of Sums of Random Variables

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10.1 Expected values of sums of random variables

Theorem 10.1 For random variables X1,X2, . . . ,Xn defined on the same sample space, Corollary Let X1,X2, . . . ,Xn be random variables on the same sample space. Then E(X1 + X2 +· · ·+Xn) = This is the generalization of the corollaries of Thm 8.1 and Thm 8.2.







) (

n 1 i i i

X E 

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Example 10.1

A die is rolled 15 times. What is the expected value of the sum of the

• utcomes?
• 1. Let X be the sum of the outcomes, and for

i=1,2,…,15. Let Xi be the outcome of the ith roll. Then X = X1 + X2 +· · ·+X15. Thus 2. 5 . 52 ) 2 / 7 ( 15 ) ( Hence, ) ( ) (     X E X E X E

i

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Example 10.2

A well-shuffled ordinary deck of 52 cards is divided randomly into four piles of 13 each. Counting jack, queen, and king as 11, 12, and 13, respectively, we say that a match

• ccurs in a pile if the j th card is j .

What is the expected value of the total number of matches in all four piles?

Example 10.2

• 1. Let Xi, i=1,2,3,4 be the number of matches in the ith
• pile. X=X1+X2+X3+X4 is the total number of matches.

2.

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4 1 1 1 1 ) ( ) ( ) ( ) ( ) ( 1 13 1 ) ( ) E(X Hence, ) ( ) P(A Now . X that have we defining by Then ). 13 1 , 4 i (1 j is pile ith in the card jth the event that the be A Let , 4 1 ), ( calculate To ) (

4 3 2 1 13 1 13 1 i ij 13 1 i ij

                       

  

  

X E X E X E X E X E X E X E X X j i X E X E

j j ij ij ij j ij i

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Example 10.3

Exactly n married couples are living in a small town. What is the expected number of intact couples after m deaths occur among the couples? Assume that the deaths occur at random, there are no divorces, and there are no new marriages.

Example 10.3

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) 1 2 ( 2 ) 1 2 )( 2 ( ) 1 ( ) ( ) 1 ( ) E(X where ) ( ... ) ( ) ( ) ( Hence,

• therwise

intact left is couple ith the if 1 define n 1,2,..., i for and death, m after couples intact

• f

number the be X Let . 1

i 2 1

                    n m n m n X P n X E X P X E X E X E X E X

i i n i

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Example 10.4

• Dr. Windler’s secretary accidentally threw a

patient’s file into the wastebasket. A few minutes later, the janitor cleaned the entire clinic, dumped the wastebasket containing the patient’s file randomly into one

• f the seven garbage cans outside the clinic,

and left. Determine the expected number of cans that

• Dr. Windler should empty to find the file.

Example 10.4

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           ) ( therefore, 7 ... 2 1 Then,

• therwise.

, X and empty, ill Windler w Dr. that can garbage ith in the is file s patient' the if 1 X let 1,2,...,7, i For 2. file. s patient' the find to empty should Windler Dr. that cans garbage

• f

number the be X Let . 1

7 2 1 i i

X E X X X X

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Example 10.5

A box contains nine light bulbs, of which two are defective. What is the expected value of the number of light bulbs that one will have to test (at random and without replacement) to find both defective bulbs?

Example 10.5

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67 . 6 ) ( Therefore,

• therwise.

X and defective, are examined be to bulbs light jth and ith the if j X let i, j and 8 ,..., 2 , 1 i For . 1

ij ij

      X E

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Example 10.6

Let X be a binomial random variable with parameters (n, p). Recall that X is the number of successes in n independent Bernoulli trials. Thus, for i = 1, 2, . . . , n, letting we get X = X1 + X2 +· · ·+Xn, (10.1) where Xi is a Bernoulli random variable for i = 1, 2, . . . , n.

   

• therwise

success a is ith trial the if 1

i

X

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Example 10.6

Now, since ∀i, 1 ≤ i ≤ n, E(Xi) = (10.1) implies that E(X) = E(X1) + E(X2)+· · ·+E(Xn) = np

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Example 10.7

Let X be a negative binomial random variable with parameters (r, p). Then in a sequence of independent Bernoulli trials each with success probability p, X is the number of trials until the r th success. Let X1 be the number of trials until the first success, X2 be the number of additional trials to get the second success, X3 the number of additional ones to obtain the third success, and so on.

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Example 10.7

Then clearly X = X1 + X2 +· · ·+Xr , where for i = 1, 2, . . . , n, the random variable Xi is geometric with parameter p. This is because P(Xi = n) = (1−p)n−1p by the independence of the trials. Since E(Xi) = 1/p (i = 1, 2, . . . , r), E(X) =

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Example 10.8

Let X be a hypergeometric random variable with probability mass function Then X is the number of defective items among n items drawn at random and without replacement from an urn containing D defective and N-D nodefective items.

n D N D n n N x n D N x D x X P x p ,..., 2 , 1 , x ), , min( , ) ( ) (                                

Example 10.8

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N nD X E n X E X E X E X E n

i n

            E(X) Therefore, ) ( , ,..., 2 , 1 i for where ) ( ... ) ( ) ( ) ( Hence,

• therwise
• ccurs

A if 1 X let , ,..., 2 , 1 i for Also, defective. is drawn item ith the event that the be A Let E(X), calculate To

2 1 i i i