HW2_F09 - Stat 219 - Stochastic Processes Homework Set 2,...

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Unformatted text preview: Stat 219 - Stochastic Processes Homework Set 2, Fall 2009, Due: October 7 1. Exercise 1.2.22. Write ( , F ,P ) for a random experiment whose outcome is a recording of the results of n independent rolls of a balanced six-sided dice(including their order). Compute the expectation of the random variable D ( ) which counts the number of different faces of the dice recorded in these n rolls. ANS: = { 1 , 2 ,..., 6 } n , F = 2 and P ( A ) = | A | 6- n for A F , where | A | denotes the number of elements in the set A . Now, let A i denote the event that the i-th face has been recorded and I i ( ) = I A i the corresponding indicator random variable. Since D ( ) = 6 i =1 I i ( ), we have by linearity of the expectation that: E ( D ) = 6 X i =1 E I i = 6 X i =1 P ( A i ) = 6 X i =1 | A | 6- n . The complement of A 1 consists of all sequences whose components are from { 2 , 3 , 4 , 5 , 6 } , so it has 5 n elements. Consequently, | A 1 | = 6 n- 5 n . The same size calculation applies for each A i , leading to E ( D ) = 6(1- (5 / 6) n ). 2. Exercise 1.2.27. Show that a R.V. X is integrable if and only if E | X | I | X | >M 0 as M . ANS: ( ) Since lim M E | X | I | X | >M = 0, M and > 0 s.t. E | X | = Z | X | M | X | dP + Z | X | >M | X | dP M P ( {| X | M } ) + < since P is a probability measure. ANS: ( ) The easiest way to show this is to use monotone convergence or dominated convergence. (Monotone convergence) As M , | X | I | X | M | X | and so by the monotone convergence theorem E I | X | M E | X | . Also, since E | X | = E I | X | M + E I | X | >M taking limits we have E | X | = E | X | + lim M E I | X | >M . But since E | X | < we must have lim M E I | X | >M = 0. (Dominated convergence) | X | I | X | >M 0 a.s. as M . Then since | X | I | X | >M | X | and E | X | < we have lim M E | X | I | X | >M = 0 by the dominated convergence theorem. (Argument based on definition) Since we are dealing only with | X | it suffices to consider nonnegative X . Clearly lim M E | X | I | X | >M = 0 if X is a simple RV since it has finite range. Now let X be a nonnegative RV and X n a sequence of simple RVs with X n X . Then X n I | X | >M is a sequence of simple RVs with limit XI | X | >M . By definition E XI | X | >M = lim n E X n I | X | >M . Thus, lim M E | X | I | X | >M = lim M lim n E | X n | I | X | >M = lim n lim M E | X n | I | X | >M = 0 1 The interchange of limits above is justified by the following: for fixed M , E | X n | I | X |...
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HW2_F09 - Stat 219 - Stochastic Processes Homework Set 2,...

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