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EXERCISE R.7 R.7*Find E(X 2 ) for X defined in Exercise R.2. [R.2*A random variable X is defined to be the larger of the numbers when two dice are thrown, or the number if they are the same. Find the probability distribution for X.] 1

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Definition of E[g(X)], the expected value of a function of X: To find the expected value of a function of a random variable, you calculate all the possible values of the function, weight them by the corresponding probabilities, and sum the results. 2 EXERCISE R.7

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Definition of E[g(X)], the expected value of a function of X: Example: For example, the expected value of X 2 is found by calculating all its possible values, multiplying them by the corresponding probabilities, and summing. 3 EXERCISE R.7

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x i p i g(x i ) g(x i ) p i x 1 p 1 g(x 1 )g(x 1 ) p 1 x 2 p 2 g(x 2 ) g(x 2 ) p 2 x 3 p 3 g(x 3 ) g(x 3 ) p 3 ………...……... x n p n g(x n ) g(x n ) p n g(x i ) p i EXERCISE R.7 4 The calculation of the expected value of a function g(x) is shown in abstract in the table. The expected value is the sum of the terms g(x i )p i.

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x i p i g(x i ) g(x i ) p i x i p i x 1 p 1 g(x 1 )g(x 1 ) p 1 11/36 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/36 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/36 ………...……... 47/36 ………...……... 59/36 x n p n g(x n ) g(x n ) p n 611/36 g(x i ) p i EXERCISE R.7 In this exercise, X is the random variable defined in Exercise R.2. The 6 possible values of X and the corresponding probabilities are listed. 5

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x i p i g(x i ) g(x i ) p i x i p i x i 2 x 1 p 1 g(x 1 )g(x 1 ) p 1 11/361 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/364 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/369 ………...……... 47/3616 ………...……... 59/3625 x n p n g(x n ) g(x n ) p n 611/3636 g(x i ) p i EXERCISE R.7 First you calculate the possible values of X 2. 6

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x i p i g(x i ) g(x i ) p i x i p i x i 2 x i 2 p i x 1 p 1 g(x 1 )g(x 1 ) p 1 11/3611/36 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/364 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/369 ………...……... 47/3616 ………...……... 59/3625 x n p n g(x n ) g(x n ) p n 611/3636 g(x i ) p i EXERCISE R.7 The first value is 1 since x is equal to 1. The probability of X being equal to 1 is 1/36. Hence the weighted square is 1/36. 7

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x i p i g(x i ) g(x i ) p i x i p i x i 2 x i 2 p i x 1 p 1 g(x 1 )g(x 1 ) p 1 11/3611/36 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/36412/36 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/36945/36 ………...……... 47/3616112/36 ………...……... 59/3625225/36 x n p n g(x n ) g(x n ) p n 611/3636396/36 g(x i ) p i EXERCISE R.7 Similarly for all the other possible values of X. 8

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x i p i g(x i ) g(x i ) p i x i p i x i 2 x i 2 p i x 1 p 1 g(x 1 )g(x 1 ) p 1 11/3611/36 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/36412/36 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/36945/36 ………...……... 47/3616112/36 ………...……... 59/3625225/36 x n p n g(x n ) g(x n ) p n 611/3636396/36 g(x i ) p i 791/36= 21.97 EXERCISE R.7 The expected value of X 2 is the sum of its weighted values in the final column. It is equal to 21.97. It is the average value of the figures in the previous column, taking the differing probabilities into account. 9

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x i p i g(x i ) g(x i ) p i x i p i x i 2 x i 2 p i x 1 p 1 g(x 1 )g(x 1 ) p 1 11/3611/36 x 2 p 2 g(x 2 ) g(x 2 ) p 2 23/36412/36 x 3 p 3 g(x 3 ) g(x 3 ) p 3 35/36945/36 ………...……... 47/3616112/36 ………...……... 59/3625225/36 x n p n g(x n ) g(x n ) p n 611/3636396/36 g(x i ) p i 791/36= 21.97 Note that E(X 2 ) is not the same thing as E(X), squared. In the previous sequence we saw that E(X) for this example was 4.47. Its square is 19.98. EXERCISE R.7 10 E(X 2 ) = 21.97E(X) = 4.47 [E(X)] 2 = 19.98

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Copyright Christopher Dougherty 1999–2006. This slideshow may be freely copied for personal use. 26.08.06

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