This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Stat219 / Math 136  Stochastic Processes Homework Set 1, Fall 2008. Due: Wednesday, October 1 For questions on grading, see TBA 1. Exercise 1.1.3 . Let ( , F , IP) be a probability space and A,B,A i events in F . Prove the following properties of IP. (a) Monotonicity. If A B then IP( A ) IP( B ). ANS: A B implies that B = A ( B \ A ). Hence, IP( B ) = IP( A ) + IP( B \ A ). Thus since IP( B \ A ) 0, we get IP( A ) IP( B ). (b) Subadditivity. If A i A i then IP( A ) i IP( A i ). ANS: For each i set B i = A i \ uniontext i 1 j =1 A j . Then the B i are disjoint and we let C = uniontext i =1 A i = uniontext i =1 B i . Since A C , from part (a), IP( A ) IP( C ) . Also, IP( C ) = i =1 IP( B i ) and B i A i therefore IP( B i ) IP( A i ) so IP( C ) i =1 IP( A i ) and hence IP( A ) i =1 IP( A i ). (c) Continuity from below: If A i A , that is, A 1 A 2 ... and i A i = A , then IP( A i ) IP( A ). ANS: Construct the disjoint sets B 1 = A 1 and B i = A i \ A i 1 for i 2, noting that A i = j i B j and A = j B j . Therefore, IP( A i ) = i j =1 IP( B j ) j =1 IP( B j ) = IP( j B j ) = IP( A ). (d) Continuity from above: If A i A , that is, A 1 A 2 ... and i A i = A , then IP( A i ) IP( A ). ANS: Apply part (c) to the sets A c i A c to have that 1 IP( A i ) = IP( A c i ) IP( A c ) = 1 IP( A )....
View
Full
Document
This note was uploaded on 11/08/2009 for the course STAT 219 taught by Professor 2 during the Fall '08 term at Stanford.
 Fall '08
 2
 Probability

Click to edit the document details