Test2-Solutions-F09

Test2-Solutions-F09 - ACTSC 431/831 Loss Models 1 Fall 2009 TEST#2 1 12 marks The ground-up loss X has density f X x = p& 1& e& 1& x

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Unformatted text preview: ACTSC 431/831: Loss Models 1 Fall 2009 - TEST #2 1. ( 12 marks ) The ground-up loss X has density f X ( x ) = p & 1 & e & 1 & x ¡ + (1 & p ) ( ¢ 1 & £ 2 xe & 1 & x ) , x > , for < p < 1 . (a) (3 marks) Find E [ X ] and V ar ( X ) . Solution: First, we remark that X is a mixture of an exponential distribution and an Erlang-2 distrib- ution (with the same scale parameter). By de&nition, E [ X ] = p& + (1 & p ) (2 & ) = & (2 & p ) and E ¤ X 2 ¥ = p ¦ & 2 + & 2 § + (1 & p ) ¨ (2 & ) 2 + 2 & 2 © = p ¦ 2 & 2 § + (1 & p ) ¦ 6 & 2 § = & 2 (6 & 4 p ) . One concludes that V ar ( X ) = & 2 (6 & 4 p ) & ( & (2 & p )) 2 = & 2 ¦ 6 & 4 p & ¦ 4 & 4 p + p 2 §§ = & 2 ¦ 2 & p 2 § . (b) ( 4 marks ) Determine whether the random variable X is IFR, DFR or neither. Solution: From the handout, we have F X ( x ) = Z 1 x f X ( y ) dy = Z 1 x p & 1 & e & 1 & y ¡ + (1 & p ) ( ¢ 1 & £ 2 ye & 1 & y )! dy = p &Z 1 x 1 & e & 1 & y dy ¡ + (1 & p ) ( Z 1 x ¢ 1 & £ 2 ye & 1 & y dy ) = pe & 1 & x + (1 & p ) n¨ 1 + x & © e & x & o . 1 The hazard rate of X is given by h X ( x ) = f X ( x ) F X ( x ) = p n 1 & e & 1 & x o + (1 & p ) n & 1 & ¡ 2 xe & 1 & x o pe & 1 & x + (1 & p ) n & 1 + x & ¡ e & x & o = 1 & p + (1 & p ) x & p + (1 & p ) & 1 + x & ¡ = 1 & 1 & (1 & p ) p + (1 & p ) & 1 + x & ¡ ! . Di/erentiating h X ( x ) , one concludes that h X ( x ) > for x > . As a result, X is IFR....
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This note was uploaded on 11/23/2010 for the course ACTSC act431 taught by Professor Na during the Spring '09 term at Waterloo.

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Test2-Solutions-F09 - ACTSC 431/831 Loss Models 1 Fall 2009 TEST#2 1 12 marks The ground-up loss X has density f X x = p& 1& e& 1& x

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