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# hw11 - EY = E E Y | X we get EY = E E Y | X = E X 2 = 1 4...

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MATH 471 HW 11 Solutions Pengsheng Ji November 28, 2006 4.4.3 Let X be the highest number that was rolled. We find P ( X = 1) = 1 / 36, P ( X = 2) = 3 / 36, . . . , P ( X = k ) = (2 k - 1) / 36. Thus, EX = 6 X k =1 k · 2 k - 1 36 = 4 . 472 , and EX 2 = 6 X k =1 k 2 · 2 k - 1 36 = 21 . 972 . Therefore, the variance of X is given by V ar ( X ) = EX 2 - ( EX ) 2 = 1 . 97 . 4.4.8 Set f ( c ) = E ( X - c ) 2 = EX 2 - 2 cEX + c 2 . Differentiating it, we get df dc = - 2 EX + 2 c. Setting the derivative to zero, we get c=EX, which corresponds to the minimum since the second derivative f 00 = 2 > 0. 4.4.17 Let T n be the number of cards we need to buy to see n cards. Note T n - T n - 1 , T n - 1 - T n - 2 , · · · , T 2 - T 1 , T 1 are mutually independent and T i - T i - 1 has a geometric ( n - i +1 n ) distribution. Since the variance of T i - T i - 1 is given by var ( T i - T i - 1 ) = 1 - n - i +1 n ( n - i +1 n ) 2 = n ( i - 1) ( n - i + 1) 2 . Therefore, var ( T n ) = var [( T n - T n - 1 ) + ( T n - 1 - T n - 2 ) + · · · + ( T 2 - T 1 ) + T 1 ] = n X i =1 n ( i - 1) ( n - i + 1) 2 .

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4.6.6 Since the expectation of Y conditional on X is X/2, using the fact
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Unformatted text preview: EY = E ( E ( Y | X )), we get EY = E ( E ( Y | X )) = E ( X/ 2) = 1 / 4 . Also, we can ﬁst get the marginal density of Y by its conditional density and the marginal density of X, and then compute the expectation by deﬁnition. 4.6.7 Since Y denote the number of 6’s that show up, we have n-y rolls that are not 6’s. 3/5 those should be odd(1, 3, and 5). Thus, the distribution of X conditional on Y is binomial (3 / 5) and the conditional expectation is E ( X | Y ) = 3 5 ( n-Y ) . Similarly, the distribution of Y given X is binomial (1 / 3) and the conditional expectation is E ( Y | X ) = 1 3 ( n-X ) ....
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hw11 - EY = E E Y | X we get EY = E E Y | X = E X 2 = 1 4...

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