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확률_및_랜덤변수_솔&e

확률_및_랜덤변수_솔&e

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Problem Solutions – Chapter 7 Problem 7.1.1 Solution Recall that X 1 , X 2 . . . X n are independent exponential random variables with mean value µ X = 5 so that for x 0, F X ( x ) = 1 e x / 5 . (a) Using Theorem 7.1, σ 2 M n ( x ) = σ 2 X / n . Realizing that σ 2 X = 25, we obtain Var [ M 9 ( X ) ] = σ 2 X 9 = 25 9 (1) (b) P [ X 1 7] = 1 P [ X 1 7] (2) = 1 F X ( 7 ) = 1 ( 1 e 7 / 5 ) = e 7 / 5 0 . 247 (3) (c) First we express P [ M 9 ( X ) > 7 ] in terms of X 1 , . . . , X 9 . P [ M 9 ( X ) > 7] = 1 P [ M 9 ( X ) 7] = 1 P [ ( X 1 + . . . + X 9 ) 63] (4) Now the probability that M 9 ( X ) > 7 can be approximated using the Central Limit Theorem (CLT). P [ M 9 ( X ) > 7] = 1 P [ ( X 1 + . . . + X 9 ) 63] 1 63 9 µ X 9 σ X = 1 ( 6 / 5 ) (5) Consulting with Table 3.1 yields P [ M 9 ( X ) > 7 ] ≈ 0 . 1151. Problem 7.1.2 Solution X 1 , X 2 . . . X n are independent uniform random variables with mean value µ X = 7 and σ 2 X = 3 (a) Since X 1 is a uniform random variable, it must have a uniform PDF over an interval [ a , b ] . From Appendix A, we can look up that µ X = ( a + b )/ 2 and that Var [ X ] = ( b a ) 2 / 12. Hence, given the mean and variance, we obtain the following equations for a and b . ( b a ) 2 / 12 = 3 ( a + b )/ 2 = 7 (1) Solving these equations yields a = 4 and b = 10 from which we can state the distribution of X . f X ( x ) = 1 / 6 4 x 10 0 otherwise (2) (b) From Theorem 7.1, we know that Var [ M 16 ( X ) ] = Var [ X ] 16 = 3 16 (3) 272
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(c) P [ X 1 9] = 9 f X 1 ( x ) dx = 10 9 ( 1 / 6 ) dx = 1 / 6 (4) (d) The variance of M 16 ( X ) is much less than Var [ X 1 ] . Hence, the PDF of M 16 ( X ) should be much more concentrated about E [ X ] than the PDF of X 1 . Thus we should expect P [ M 16 ( X ) > 9 ] to be much less than P [ X 1 > 9 ] . P [ M 16 ( X ) > 9] = 1 P [ M 16 ( X ) 9] = 1 P [ ( X 1 + · · · + X 16 ) 144] (5) By a Central Limit Theorem approximation, P [ M 16 ( X ) > 9] 1 144 16 µ X 16 σ X = 1 ( 2 . 66 ) = 0 . 0039 (6) As we predicted, P [ M 16 ( X ) > 9 ] P [ X 1 > 9 ] . Problem 7.1.3 Solution This problem is in the wrong section since the standard error isn’t defined until Section 7.3. How- ever is we peek ahead to this section, the problem isn’t very hard. Given the sample mean estimate M n ( X ) , the standard error is defined as the standard deviation e n = Var [ M n ( X ) ] . In our problem, we use samples X i to generate Y i = X 2 i . For the sample mean M n ( Y ) , we need to find the standard error e n = Var [ M n ( Y ) ] = Var [ Y ] n . (1) Since X is a uniform ( 0 , 1 ) random variable, E [ Y ] = E X 2 = 1 0 x 2 dx = 1 / 3 , (2) E Y 2 = E X 4 = 1 0 x 4 dx = 1 / 5 . (3) Thus Var [ Y ] = 1 / 5 ( 1 / 3 ) 2 = 4 / 45 and the sample mean M n ( Y ) has standard error e n = 4 45 n . (4) Problem 7.1.4 Solution (a) Since Y n = X 2 n 1 + ( X 2 n ) , Theorem 6.1 says that the expected value of the difference is E [ Y ] = E [ X 2 n 1 ] + E [ X 2 n ] = E [ X ] E [ X ] = 0 (1) (b) By Theorem 6.2, the variance of the difference between X 2 n 1 and X 2 n is Var [ Y n ] = Var [ X 2 n 1 ] + Var [− X 2 n ] = 2 Var [ X ] (2) 273
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(c) Each Y n is the difference of two samples of X that are independent of the samples used by any other Y m . Thus Y 1 , Y 2 , . . . is an iid random sequence. By Theorem 7.1, the mean and variance of M n ( Y ) are E [ M n ( Y ) ] = E [ Y n ] = 0 (3) Var [ M n ( Y ) ] = Var [ Y n ] n = 2 Var [ X ] n (4) Problem 7.2.1 Solution If the average weight of a Maine black bear is 500 pounds with standard deviation equal to 100
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