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FRE6083_HW2_solutions

# FRE6083_HW2_solutions - Solutions Question 1 Similar to the...

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Unformatted text preview: Solutions Question 1 Similar to the example in the notes, differing only in the form of the starting pdf. u1D439 u1D44C ( u1D466 ) = u1D443 ( u1D44C ≤ u1D466 ) = u1D443 ( u1D44B 2 ≤ u1D466 ) = u1D443 ( − √ u1D466 ≤ u1D44B ≤ √ u1D466 ) = u1D439 u1D44B ( √ u1D466 ) − u1D439 u1D44B ( − √ u1D466 ) The pdf is obtained from the distribution function by differentiation, u1D453 u1D44C ( u1D466 ) = u1D451 u1D451u1D466 u1D443 ( u1D44C ≤ u1D466 ) = u1D453 u1D44B ( √ u1D466 ) 1 2 uni221A.alt01 ( u1D466 ) − u1D453 u1D44B ( − √ u1D466 ) − 1 2 uni221A.alt01 ( u1D466 ) = 1 / √ u1D466 u1D70B (1 + u1D466 ) for u1D466 ≥ Question 2 Solving for u1D44B 1 and u1D44B 2 : u1D44B 1 = ( u1D44C 1 u1D44C 2 ) 1 / 2 and u1D44B 2 = uni0028.alt03 u1D44C 2 u1D44C 1 uni0029.alt03 1 / 2 . Domains: 0 < u1D44B 1 < 1 and 0 < u1D44B 2 < 1 are equivalent to the conditions u1D44C 1 > 0, u1D44C 2 > 0, u1D44C 1 u1D44C 2 < 1, and ( u1D44C 2 /u1D44C 1 ) < 1. That is, for values ( u1D465 1 , u1D465 2 ) of u1D44B 1 and u1D44B 2 varying over the the set u1D446 = points ( u1D465 1 , u1D465 2 ) such that 0 < u1D465 1 < 1 and 0 < u1D465 2 < 1, u1D44C 1 and u1D44C 2 will vary over the set u1D447 containing all points ( u1D466 1 , u1D466 2 ) such that u1D466 1 > 0, u1D466 2 > 0, u1D466 1 u1D466 2 < 1, and ( u1D466 2 /u1D466 1 ) < 1 You might find it helpful to draw the domains of ( u1D44B 1 , u1D44B 2 ) and ( u1D44C 1 , u1D44C 2 ). The transformation from ( u1D44B 1 , u1D44B 2 ) to ( u1D44C 1 , u1D44C 2 ) is one-to-one, so for points ( u1D466 1 , u1D466 2 ) in u1D447 the relationship is u1D465 1 = u1D460 1 ( u1D466 1 , u1D466 2 ) = ( u1D466 1 u1D466 2 ) 1 / 2 , and u1D465 2 = u1D460 2 ( u1D466 1...
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FRE6083_HW2_solutions - Solutions Question 1 Similar to the...

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