A06ans - EXERCISES IN STATISTICS Series A No 6 1 Let x1 and...

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Series A, No. 6 1. Let x 1 and x 2 have the joint p.d.f f ( x 1 ,x 2 )= x 1 + x 2 with 0 x 1 ,x 2 1. Find the conditional mean of x 2 given x 1 . Answer. First let us show that Z x 2 Z x 1 f ( x 1 ,x 2 ) dx 2 dx 1 =1: We have Z x 2 Z x 1 ( x 1 + x 2 ) dx 1 dx 2 = Z x 2 x 2 1 2 + x 2 x 1 1 0 dx 2 = Z x 2 ± 1 2 + x 2 dx 2 = x 2 2 + x 2 2 2 1 0 =1 . From this we can see that f ( x 2 )=[ 1 2 + x 2 ]. Likewise, by an argument of symmetry, we have f ( x 1 )=[ 1 2 + x 1 ]. Next consider E ( x 2 | x 1 )= Z x 2 x 2 f ( x 1 ,x 2 ) f ( x 1 ) dx 2 = 1 ( 1 2 + x 1 ) Z x 2 x 2 ( x 1 + x 2 ) dx 2 = 1 ( 1 2 + x 1 ) x 1 x 2 2 2 + x 3 2 3 1 x 2 =0 = 1 ( x 1 + 1 2 ) (3 x 1 +2) 6 = 3 x 1 +2 6 x 1 +3 . 2. Find the P ( x<y | x< 2 y ) when f ( x, y )= e - ( x + y ) . Draw a diagram to represent the event. Answer. P ( x<ay )= Z y =0 Z ay x =0 f ( x, y ) dxdy = Z y =0 e - y ‰Z ay x =0 e - x ² dy = Z y =0 e - y £ - e - x / ay 0 dy = Z
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A06ans - EXERCISES IN STATISTICS Series A No 6 1 Let x1 and...

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