Using this with p zn 1 ezn gives the desired result

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Unformatted text preview: Having found the martingale it is easy now to conclude: Theorem 5.16. Consider a left continuous random walk with positive mean. Let a < x and Va = min{n : Sn = a}. Px (Va < 1) = e↵(x a) Proof. Again if one argues casually e↵x = Ex (exp(↵Va )) = e↵a Px (Va < 1) but we have to prove that there is no contribution from {Va = 1}. To do this note that Theorems 5.13 and 5.11 give e↵x = E0 exp(↵SVa ^n ) = e↵a P0 (Va n) + E0 (exp(↵Sn ); Va > n) exp(↵Sn ) e↵a on Va > n but since P (Va = 1) > 0 this is not enough to make the last term vanish. The strong law of large numbers implies that on Va = 1, Sn /n ! µ > 0, so the second term ! 0 as n ! 1 and it follows that e↵x = e↵a P0 (Va < 1). 171 5.4. APPLICATIONS When the random walk is not left continuous we cannot get exact results on hitting probabilities but we can still get a bound. Example 5.13. Cram´r’s estimate of ruin. Let Sn be the total assets of e an insurance company at the end of year n. D...
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