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HW7 - Mathematics Department UCLA T Richthammer spring 09...

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Mathematics Department, UCLA T. Richthammer spring 09, sheet 7 May 8, 2009 Homework assignments: Math 170B Probability, Sec. 1 65. Here we will show that E ( 1 X +1 ) = 1 - (1 - p ) n +1 ( n +1) p for a binomial RV X with parameters n, p : Let X 1 , . . . , X n +1 be a Bernoulli sequence with success rate p , X ± = X 1 + . . . + X n +1 and X = X 2 + . . . + X n +1 . Calculate E ( X 1 X ± 1 { X ± > 0 } ) in the following two ways: (a) by using symmetry and P ( X ± > 0) (b) by conditioning on the value of X 1 . Answer: (a) By symmetry we have E ( X 1 X ± 1 { X ± > 0 } ) = E ( X i X ± 1 { X ± > 0 } ), which implies n +1 i =1 E ( X i X ± 1 { X ± > 0 } ) = ( n + 1) E ( X 1 1 { X ± > 0 } ). On the other hand n +1 i =1 E ( X i X ± 1 { X ± > 0 } ) = E ( X 1 + ... + X n +1 X ± )1 { X ± > 0 } ) = E ( X ± X ± 1 { X ± > 0 } ) = P ( X ± > 0) = 1 - q n +1 , and thus E ( X 1 X ± 1 { X ± > 0 } ) = 1 - q n +1 n +1 . (b) X = X ± - X 1 is a binomial RV with parameters n, p that is independent from X 1 . We have E ( X 1 X ± 1 { X ± > 0 } | X 1 = 1) = E ( 1 1+ X ) and E ( X 1 X ± 1 { X ± > 0 } | X 1 = 0) = 0, so E ( X 1 1 { X ± > 0 } ) = p E ( 1 1+ X ). Combining (a) and (b) gives the desired result. 66.* In the lecture we considered N x := min { n : X 1 + . . . + X n x } , where X 1 , X 2 , . . . are independent RVs with uniform distribution on [0 , 1]. For m ( x ) := E ( N x ) we showed m ( x ) = e x for 0 x 1 by conditioning on the value of X 1 . (a) Calculate m ( x ) for 1 x 2 by conditioning, and describe a procedure how to obtain m ( x ) for larger x . (b) Calculate m ( x ) for 0 x 1 using indicators. Answer: (a) Let x 1. By conditioning on X 1 we get. E ( N x | X 1 = x 1 ) = E ( N x - x 1 ) + 1 = m ( x - x 1 ) + 1. So m ( x ) = E ( N x ) = R 1 0 dx 1 m ( x - x 1 ) + 1. This recursive equation can be used to calculate m ( x ) for x [ n, n + 1) if you already know m ( x ) for x [ n - 1 , n ). If m n ( x ) = m ( n + x ) for x [0 , 1] we get m n ( x ) = R x 0 dx 1 m n ( x - x 1 ) + R 1 x dx 1 m n - 1 (1 + x - x 1 ) + 1 = R x 0 m n ( u ) du + R 1 x m n - 1 ( u ) du + 1. So m ± n ( x ) = m n ( x ) - m n - 1 ( x ). This gives differential equations for m n ( x ) that can be solved recursively, using the initial conditions m n (0) = m n - 1 (1) and m 0 ( x ) = e x . For instance for m 1 we get m ± 1 ( x ) = m 1 ( x ) - e x and m 1 (0) = e . Solving this differential equation by means of variation of the constant we get m 1 ( x ) = ( e - x ) e x , which gives m ( x ) = ( e - x + 1) e x - 1 for 1 x 2. (b) We have
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HW7 - Mathematics Department UCLA T Richthammer spring 09...

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