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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 (Ω , F ,P ) for a random experiment whose outcome is a recording of the results of n independent rolls of a balanced sixsided dice(including their order). Compute the expectation of the random variable D ( ω ) which counts the number of different faces of the dice recorded in these n rolls. ANS: Ω = { 1 , 2 ,..., 6 } n , F = 2 Ω 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 ith face has been recorded and I i ( ω ) = I A i the corresponding indicator random variable. Since D ( ω ) = ∑ 6 i =1 I i ( ω ), 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 ω ∈ Ω 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 RV’s with X n ↑ X . Then X n I  X  >M is a sequence of simple RV’s 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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 Spring '09
 Probability theory, Xn

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