Stat219sample_midterm_f08_sol

# Stat219sample_midterm_f08_sol - Math136/Stat219 Fall 2008...

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Math136/Stat219 Fall 2008 Sample Midterm 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 3. Homework problems and solutions posted in Coursework 4. Your own graded homework papers 5. Your own notes taken during lecture 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 in a calculator or computer, consultation with anyone during the exam (except for the Instructor). 1. (4x2) Let Ω = IR with event space B and the uniform probability measure U on (0 , 1). a) Find the distribution functions F X ( x ), F Y ( y ) and F V ( v ) that correspond to the random variables X ( ω ) = ω , Y ( ω ) = 0 . 5 I A ( ω ) and V ( ω ) = I B ( ω ), where A = (0 , 1 / 3) and B = [1 / 4 , 1 / 3). ANS: Recall that U (( −∞ , x ]) = x when 0 < x < 1, with U (( −∞ , x ]) = 1 when x 1 and U (( −∞ , x ]) = 0 when x 0. Consequently, F X ( x ) = U (( −∞ , x ]) = xI [0 , 1) ( x ) + I [1 , ) ( x ). We also know that P ( Y = 0 . 5) = U ( A ) = 1 / 3 and P ( Y = 0) = U ( A c ) = 2 / 3. Hence, F Y ( y ) = (2 / 3) I [0 , 0 . 5) ( y ) + I [0 . 5 , ) ( y ). Similarly, V is an indicator RV with P ( V = 1) = U ( B ) = 1 / 3 1 / 4 = 1 / 12, hence F V ( v ) = (11 / 12) I [0 , 1) ( v ) + I [1 , ) ( v ). 1

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b) Compute the characteristic function Φ Z ( θ 1 , θ 2 ) of the random vector Z = ( X, Y ). ANS: By Definition 3.2.1, since Y = 0 . 5 I A ( X ) we have that Φ Z ( θ 1 , θ 2 ) = E [exp( 1 X + 0 . 5 2 I A ( X ))] = E [ I A c ( X ) e 1 X ] + e 2 / 2 E [ I A ( X ) e 1 X ] . Recall that X has the density f X ( x ) = I (0 , 1) ( x ), hence E [ I A c ( X ) e 1 X ] = integraldisplay 1 1 / 3 e 1 x dx = ( e 1 e 1 / 3 ) / ( 1 ) (see text following Example 3.2.3 for similar calculations). Likewise, E [ I A ( X ) e 1 X ] = ( e 1 / 3 1) / ( 1 ), so that Φ Z ( θ 1 , θ 2 ) = ( e 1 e 1 /
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