# 16 that p1 t0 1 e where 0 is the solution of e expxi

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Unformatted text preview: uring year n, premiums totaling c dollars are received, while claims totaling Yn dollars are paid, so Sn = Sn 1 +c Yn Let Xn = c Yn and suppose that X1 , X2 , . . . are independent random variables that are normal with mean µ &gt; 0 and variance 2 . That is the density function of Xi is (2⇡ 2 ) 1/2 exp( (x µ)2 /2 2 ) Let B for bankrupt be the event that the wealth of the insurance company is negative at some time n. We will show P (B ) exp( 2µS0 / 2 ) (5.14) In words, in order to be successful with high probability, µS0 / 2 must be large, but the failure probability decreases exponentially fast as this quantity increases. Proof. We begin by computing (✓) = E exp(✓Xi ). To do this we need a little algebra (x µ)2 2 2 + ✓(x and a little calculus Z (✓) = e✓x (2⇡ µ 2 2 ✓)2 2 + exp( (x µ)2 /2 2 ) dx ✓ Z (x 22 2 1 /2 = exp( ✓ /2 + ✓µ) (2⇡ ) exp 2 ) (x µ) + ✓µ = 22 ✓ + ✓µ 2 1 /2 µ 2 2 2 Since the integrand is the density of a normal with mean µ + 2 it follows that (✓) = exp( 2 ✓2 /2 +...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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