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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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 Spring '10
 DURRETT
 The Land

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