HW2_F09

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 six-sided 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 i-th 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 |...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online