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Unformatted text preview: 2) 1 X ≥ 2 ] = E [( X1) 1 X ≥ 1 ]P ( X > 1) = 2 3. 4 = 4 15 E [( X3) 1 X ≥ 3 ] = E [( X2) 1 X ≥ 2 ]P ( X > 2) = 4 15. 4 2 = 8 75 Alternatively, let x3 = u so ∞ X x =3 ( x3)(0 . 6)(0 . 4) x1 = 0 . 4 3 ∞ X u =0 u (0 . 6)(0 . 4) u1 = 0 . 4 3 E [ X ] UNSW Week 12 ACTL1001 Tutorial Examples Q2  Deductible and Reinsurance The insurer’s loss is given by if Y ≤ 50; Y50 if 50 < Y ≤ 300; 30050 = 250 if 300 < Y ≤ 500 . and the density function of Y is f ( y ) = d dy P ( Y ≤ y ) = 1 500 . Hence, the insurer’s expected loss is Z 50 (0) 1 500 dy + Z 300 50 ( y50) 1 500 dy + Z 500 300 250 1 500 dy = 0 + ± ( y50) 2 1000 ² 300 50 + 100 = 250 2 1000 + 100 = 325 2 = 162 . 5 UNSW Week 12 ACTL1001 Tutorial...
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