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midterm_sol - Stat 310A/Math 230A Theory of Probability...

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Stat 310A/Math 230A Theory of Probability Midterm Solutions Andrea Montanari November 1, 2010 The midterm was long! This will be taken into account in the grading. We will assign points proportion- ally to the number of questions answered (e.g. Problem 1 counts for 4 questions) and then rescale upwards the grades distribution. This is also a good time to discuss any difficulty you encountered with the instructors. Problem 1 Let λ 2 be the Lebesgue measure on ( R 2 , B R 2 ). We know already that it is invariant under translation i.e. that λ 2 ( B + x ) = λ 2 ( B ) for any Borel set B and x R 2 (whereby B + x = { y R 2 : y - x B } ). (a) Show that it is invariant under rotations as well, i.e. that for any α [0 , 2 π ], and any Borel set B R 2 , λ 2 ( R ( α ) B ) = λ 2 ( B ) (whereby R ( α ) denotes a rotation by an angle α and R ( α ) B = { x R 2 : R ( - α ) x B } ). Solution : Throughout the solution we will use the fact that λ 2 = λ 1 × λ 1 whence we obtain the action of λ 2 on rectangles: λ 2 ( A 1 × A 2 ) = λ 1 ( A 1 ) λ 1 ( A 2 ). Also, for J 1 , J 2 R two intervals, let T J 1 ,J 2 be any triangle with two sides equal to J 1 (parallel to the first axis) and J 2 (equal to the second axis). from the addittivity of λ 2 it follows immediately that λ 2 ( T J 1 ,J 2 ) = | J 1 | · | J 2 | / 2. (We use here the fact that for a segment S = { x 0 + x 1 λ : λ [ a, b ] } , x 0 , x 1 R 2 , λ 2 ( S ) = 0, which can be proved by covering S with squares.) α β Consider next a rectangle A = [0 , a ) × [0 , b ), and let A 0 = R ( α ) A . Using again addittivity (see figure above) it follows that, for β = π/ 2 - α : λ 2 ( A 0 ) = ( a cos α + b cos β )( a sin α + b sin β ) - a 2 sin α cos α - b 2 sin β cos β = ab (cos α sin β + cos β sin α ) = ab sin( α + β ) = ab . Hence λ 2 ( A ) = λ 2 ( R ( α ) A ) and by translation invariance this holds for any A = [ a 1 , a 2 ) × [ b 1 , b 2 ) (not necessarily with a corner at the origin). Since the π -system P = { A = [ a 1 , a 2 ) × [ b 1 , b 2 ) : a 1 < a 2 , b 1 < b 2 } generates the Borel σ algebra, and recalling that λ 2 is σ -finite, this proves the claim by Caratheodory uniqueness theorem. (b) For s R + , and B R 2 Borel, let sB ≡ { x R 2 : s - 1 x B } . Prove that λ 2 ( sB ) = s 2 λ 2 ( B ). 1
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Solution : The proof is analogous to the previous one. Let μ be the measure defined by μ ( B ) s - 2 λ 2 ( sB ). For A = [ a 1 , a 2 ) × [ b 1 , b 2 ), a 1 < a 2 , b 1 < b 2 , we have sA = [ sa 1 , sa 2 ) × [ sb 1 , sb 2 ), whence μ ( A ) = 1 s 2 λ 2 ( sA ) = 1 s 2 ( sa 2 - sa 1 )( sb 2 - sb 1 ) = ( a 2 - a 1 )( b 2 - b 1 ) = λ 2 ( A ) .
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