examsoln1 - | n flips of 2 nd coin heads]= ( 2 3 )( 3 4 )...

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EECS 501 SOLUTIONS TO EXAM #1 Fall 2001 1a. 1 = c R 1 0 dX X R X 0 dY Y = c R 1 0 dX X X 2 2 = c X 4 8 | 1 0 c = 8. 1b. x and y are NOT independent, since pdf has nonsquare support. HARD WAY: Compute marginal pdfs f x ( X ) = R f x,y ( X,Y ) dY and f y ( Y ) = R f x,y ( X,Y ) dX and show that f x,y ( X,Y ) 6 = f x ( X ) f y ( Y ). 1c. f x ( X ) = R X 0 dY 8 XY = 4 XY 2 | X 0 = 4 X 3 , 0 < X < 1; 0, otherwise. 1d. f y | x ( Y | X ) = 8 XY 4 X 3 = 2 Y/X 2 , 0 < Y < X < 1 = 8 Y, 0 < Y < 1 / 2. 1e. F z ( Z ) = Pr [ z Z ] = Pr [( y x ) Z ] = Pr [ y xZ ] = 8 R 1 0 dX X R XZ 0 dY Y = 4 R 1 0 dX X ( XZ ) 2 = Z 2 , 0 < Z < 1 f z ( Z ) = dF z dZ = 2 Z, 0 < Z < 1. Note if Z > 1 then upper integral limit changes to X and F z ( Z ) = 1. This makes sense: y x z = y/x 1 Pr [ z Z > 1] = 1. 1f. Pr [( x + y ) < 1] = 8 R 1 / 2 0 dY Y R 1 - Y Y dX X = 4 R 1 / 2 0 dY Y ((1 - Y ) 2 - Y 2 ) = 4( Y 2 2 - 2 Y 3 3 ) | 1 / 2 0 = 1 / 6. 2a. Pr[second coin heads]=( 2 3 )( 3 4 ) + ( 1 3 )( 4 5 ) = 23 / 30. 2b. Pr[A heads | second coin heads]= Pr [ coinAheads coinB heads ] Pr [2 nd coinheads ] = (2 / 3)(3 / 4) 23 / 30 = 15 / 23. 2c. Pr[ n flips of second coin heads]=( 2 3 )( 3 4 ) n + ( 1 3 )( 4 5 ) n . 2d. Pr[A heads
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Unformatted text preview: | n flips of 2 nd coin heads]= ( 2 3 )( 3 4 ) n / [( 2 3 )( 3 4 ) n + ( 1 3 )( 4 5 ) n ]. 2e. LIM n →∞ [answer to (d)]=0 since ( 4 5 ) n dominates ( 3 4 ) n . 2f. Coin C is more likely than coin B to land heads indefinitely. 2g. Pr [ E ] = Pr [ E ∩ F ] + Pr [ E ∩ F ] → Pr [ E ∩ F ] = Pr [ E ]-Pr [ E ∩ F ] = Pr [ E ]-Pr [ E ] Pr [ F ] = Pr [ E ](1-Pr [ F ]) = Pr [ E ] Pr [ F ] → E,F are independent. Q.E.D. All these from Problem Set #1. 3a. C,Y,0. 3b. U,Y,1. 3c. C,Y,0. 3d. U,Y,0....
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This note was uploaded on 07/22/2011 for the course EECS 370 taught by Professor Lehman during the Spring '10 term at University of Florida.

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