px2 x1 x2 1 004060115 03606035 018060310 002060130 b

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Unformatted text preview: 63. 42. Refer to problem #41. a. (Recall that px2 |x1 (x2 |1) = x2 0 1 2 3 = 0, 1, 2, 3.) px2 |x1 (x2 |1) 0.04/0.60=1/15 0.36/0.60=3/5 0.18/0.60=3/10 0.02/0.60=1/30 b. Similarly, px1 |x2 (x1 |2) = x1 0 1 2 3 p(1,x2 ) p1 (1) , x2 px1 |x2 (x1 |2) 0.03/0.27=1/9 0.18/0.27=2/3 0.04/0.27=4/27 0.02/0.27=2/27 p(x1 ,2) p2 (2) , x1 = 0, 1, 2, 3 c. Based on a. and b. E(X2 |X1 = 1) = 3 ￿ x2 =0 x2 · px2 |x1 (x2 |1) 1 3 3 1 ) + 1( ) + 2( ) + 3( ) 15 5 10 30 0 + 18 + 18 + 3 39 = = 1.3 30 30 3 ￿ x1 · px1 |x2 (x1 |2) = 0( = E(X1 |X2 =...
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This document was uploaded on 02/15/2014 for the course MATH 231 at Lehigh University .

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