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Stat219midterm_f08_sol

# Stat219midterm_f08_sol - Math136/Stat219 Fall 2008 Midterm...

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Math136/Stat219 Fall 2008 Midterm Examination Friday, October 24, 2008, 11:00am - 12:30pm Write your name and sign the Honor code in the blue books provided. You have 90 minutes to solve all questions, each worth points as marked (maximum of 50). Complete reasoning is required for full credit. You may cite lecture notes and homework sets, as needed, stating precisely the result you use, why and how it applies. You may consult the following materials while taking the exam: 1. Stat219/Math136 Lecture notes, Fall 2008 version (the required text) 2. Kevin Ross’s Lecture slides posted in Coursework (Fall 2008 only) 3. Homework problems, sample exams, and solutions posted in Coursework (Fall 2008 only) 4. Your own graded homework papers 5. Your own notes, handwritten or typed Use of any other material is prohibited and constitutes a violation of the Honor Code. This includes, but is not limited to: other texts (including optional and recommended texts), photocopying of texts or notes, materials from previous sections of Stat219/Math136, the internet, programming formulas or other results in a calculator or computer, consultation with anyone during the exam (except for the Teaching Assistants or the Instructor). 1. (3 Points each) On a probability space (Ω , F , P ), let Y be a random variable with E ( Y 2 ) < and G ⊆ F be a σ -field. Define Var( Y |G ) = E ( Y 2 |G ) ( E ( Y |G ) 2 ) . Show the following. (Note: you must give a proof. Merely citing Exercise 2.3.7 will receive no credit.) a) Show that if Y is G -measurable then Var( Y |G ) = 0 almost surely. 1

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ANS. Since Y is G -measurable, Var( Y |G ) = E ( Y 2 |G ) ( E ( Y |G ) 2 ) = Y 2 ( Y ) 2 = 0 . b) Show that Var( Y ) = E (Var( Y |G )) + Var( E ( Y |G )) . ANS. Using the tower property and linearity of CE gives E (Var( Y |G )) = E [ E ( Y 2 |G ) ( E ( Y |G ) 2 )] = E ( Y 2 ) E [ E ( Y |G ) 2 ] . The definition of (unconditional) variance and the tower property implies Var( E ( Y |G )) = E [( E ( Y |G )) 2 ] ( E [ E ( Y |G )]) 2 = E [( E ( Y |G )) 2 ] ( E [ Y ]) 2 . Adding the above two equations yields E (Var( Y |G )) + Var( E ( Y |G )) = E ( Y 2 ) ( E ( Y )) 2 = Var( Y ) . c) Suppose that Y is G -measurable and X is a random variable on (Ω , F , P ) with E ( X 2 ) < . Show that Var( XY |G ) = Y 2 Var( X |G ) . ANS. Since Y is G -measurable, taking out what is known yields Var( XY |G ) = E (( XY ) 2 |G ) ( E ( XY |G ) 2 ) = Y 2 E ( X 2 |G ) ( Y E ( X |G )) 2 = Y 2 Var( X |G ) . 2. (3 Points each) Consider the probability space (Ω , F , P ), where Ω = (0 , 1); F is the Borel σ -field on (0 , 1), that is, F = σ ( { ( a, b ) : 0 < a < b < 1 } ); and P is the uniform probability measure. For n = 1 , 2 , . . . define X n ( ω ) = 2 n I A n ( ω ), ω Ω, where A n = parenleftbigg 1 2 1 2 n , 1 2 + 1 2 n parenrightbigg .
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