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Unformatted text preview: Math 136  Stochastic Processes Homework Set 4, Fall 2009 1. Exercise 2.3.8. The left hand side is the smallest distance from G 2 , while by embeddedness, right hand side is a distance from G 2 , and hence the inequality holds.(no need for algebra) 2. Exercise 2.3.16. Let Z = ( X,Y ) be a uniformly chosen point on (0 , 1) 2 . That is, X and Y are independent random variables, each having the U (0 , 1) measure of Example 1.1.11. Set T = I A ( Z ) + 5 I B ( Z ) where A = { < x < 1 / 4 , 3 / 4 < y < 1 } and B = { 3 / 4 < x < 1 , < y < 1 / 2 } . (a) Find an explicit formula for the conditional expectation W = E ( T  X ) and use it to determine the conditional expectation U = E ( TX  X ). ANS: Note A = A 1 × A 2 for A 1 = { x ∈ (0 , 1 / 4) } ,A 2 = { y ∈ (3 / 4 , 1) } hence I A ( x,y ) = I A 1 ( x ) I A 2 ( y ). Similarly I B ( x,y ) = I B 1 ( x ) I B 2 ( y ) for B 1 = { x ∈ (3 / 4 , 1) } ,B 2 = { y ∈ (0 , 1 / 2) } . Consequently, T = I A 1 ( X ) I A 2 ( Y )+5 I B 1 ( X ) I B 2 ( Y ). Thus, by the linearity of the C.E. and “taking out what is known” (Proposition 2.3.15) we have that W = E ( T  X ) = I A 1 ( X ) E ( I A 2 ( Y )  X ) + 5 I B 1 ( X ) E ( I B 2 ( Y )  X ) . Further, since X and Y are independent, I A 2 ( Y ) and I B 2 ( Y ) are independent of X . Thus, we have that E ( I A 2 ( Y )  X ) = E I A 2 ( Y ) = P ( Y ∈ A 2 ) = 1 4 , with the rightmost identity due to Y being uniformly chosen on (0 , 1) with A 2 an interval of length 1 / 4. Similarly, E ( I B 2 ( Y )  X ) = 1 / 2, so we have that W = 1 4 I A 1 ( X ) + 5 2 I B 1 ( X ) ....
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This document was uploaded on 04/12/2010.
 Spring '09
 Algebra

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