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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 > 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 +...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online