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Test1-Solutions-S09-simplified

# Test1-Solutions-S09-simplified - ACTSC 431 Loss Models 1...

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ACTSC 431 - Loss Models 1 TEST #1 1. (20 marks) Suppose that the ground-up loss random variable X has density function f X ( x ) = °e ° °x ° a + 2 (1 ° a ) e ° °x ± , x ± 0 , for a 2 (0 ; 1) and ° > 0 . (a) (1 mark) Determine the survival distribution of the random variable X . Solution : By integrating the density from x to 1 , one °nds that F X ( x ) = Z 1 x f X ( y ) dy = Z 1 x °e ° °y ° a + 2 (1 ° a ) e ° °y ± dy = ae ° °x + (1 ° a ) e ° 2 °x , for x ± 0 . (b) (3 marks) Using (a), °nd the hazard rate of the random variable X . Is the random variable X DFR, IFR or neither? Solution : By de°nition, the hazard rate of the random variable X is given by h X ( x ) = f X ( x ) F X ( x ) = °e ° °x ° a + 2 (1 ° a ) e ° °x ± ae ° °x + (1 ° a ) e ° 2 °x = ° ° a + 2 (1 ° a ) e ° °x ± a + (1 ° a ) e ° °x = ° ° a + (1 ° a ) e ° °x + (1 ° a ) e ° °x ± a + (1 ° a ) e ° °x = ° + ° (1 ° a ) e ° °x a + (1 ° a ) e ° °x = ° + ° (1 ° a ) ae °x + (1 ° a ) . (1) Given that a 2 (0 ; 1) , the hazard rate is a decreasing function in x . To convince yourself, one can di/erentiate (1) with respect to x to show that d dx h X ( x ) = ° ° (1 ° a ) a°e °x ( ae °x + (1 ° a )) 2 < 0 . One concludes that X is DFR . (c) (4 marks) Show that the Value-at-Risk at level 100 p % , namely V ar p ( X ) , is given by V aR p ( X ) = ° 1 ° ln p a 2 + 4 (1 ° a ) (1 ° p ) ° a 2 (1 ° a ) !

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Test1-Solutions-S09-simplified - ACTSC 431 Loss Models 1...

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