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 8 ± 63 9 µ X 9 σ X ² = 1 8( 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 ) = ³ 1 / 64 x 10 0 otherwise (2) (b) From Theorem 7.1, we know that [ M 16 ( X ) [ X ] 16 = 3 16 (3) 272
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(c) P [ X 1 9] = Z 9 f X 1 ( x ) dx = Z 10 9 ( 1 / 6 ) = 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 8 ± 144 16 µ X 16 σ X ² = 1 8( 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 = p [ M n ( Y ) ]= ³ [ Y ] n . (1) Since X is a uniform ( 0 , 1 ) random variable, E [ Y ] = E ´ X 2 = Z 1 0 x 2 = 1 / 3 , (2) E ´ Y 2 = E ´ X 4 = Z 1 0 x 4 = 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 = ³ 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 [ Y n [ X 2 n 1 ]+ [− X 2 n 2Var [ 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 ,...
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This note was uploaded on 12/12/2011 for the course EE EE980 taught by Professor Ji during the Spring '11 term at Akademia Rolnicza w Wrocławiu.

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확률_및_랜덤변수_솔&e

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