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Unformatted text preview: Math 136  Stochastic Processes Homework Set 4, Winter 2008, Due: February 6 1. 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 ) . Since X is bounded, we know that U = E ( TX  X ) = XW by Proposition 2.3.15. (b) Find the value of E (( T W )sin( e X )). ANS: Since sin( e X ) L 2 ( , ( X ) , P ) = H X and W = E ( T  X ) for T squareintegrable, this is zero by Proposition 2.1.2. (c) Without any computation decide whether E W 2 E T 2 is negative, zero, or positive. Explain your answer. ANS: Recall Proposition 2.1.2 that E (( T W ) W ) = 0 (since W H X ). Hence, with T = W + ( T W ) we have that E T 2 = E W 2 + 2 E ( T W ) W + E ( T W ) 2 = E W 2 + E ( T W ) 2 . By part (a) we already know that P ( T negationslash = W ) > 0, hence E ( T W ) 2 > 0, implying that E W 2 E T 2 is negative. We end by remarking that a perhaps shorter and more geometric argument can be 1 made by recalling that the conditional expectation...
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 Fall '08
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