Mathematical Modeling in Finance with Stochastic Processes

110 CHAPTER 3. FIRST STEP ANALYSIS FOR STOCHASTIC PROCESSES

Proof that the duration is finite

The following discussion of finiteness of the duration of the game is adapted
from [49] by J. Michael Steele.
When we check the arguments for the probability of ruin or the duration
of the game, we find a logical gap. We have assumed that the durationDT 0 of
the game is finite. How do we know for sure that the gambler’s net winnings
will eventually reachaor 0? This important fact requires proof.
The proof uses a common argument in probability, an “extreme case argu-
ment”. We identify an “extreme” event with a small but positive probability
of occurring. We are interested in the complementary “good” event which
at least avoids the extreme event. Therefore the complementary event must
happen with probability not quite 1. The avoidance must happen infinitely
many independent times, but the probability of such a run of “good” events
must go to zero.
For the gambler’s ruin, we are interested in the event of the game contin-
uing forever. Consider the extreme event that the gambler winsatimes in
a row. If the gambler is not already ruined (at 0), then such an streak ofa
wins in a row is guaranteed to boost his fortune aboveaand end the game
in victory for the gambler. Such a run of luck is unlikely, but it has positive
probability, in fact, probabilityP =pa. We letEk denote the event that
the gambler wins on each turn in the time interval [ka,(k+ 1)a−1], so the
Ek are independent events. Hence the complementary eventsECk = Ω−Ek
are also independent. ThenD > naat least implies that all of theEk with
0 ≤k≤nfail to occur. Thus, we find

P[DT 0 > na]≤P

[n

⋂

k=0

EkC

]

= (1−P)n.

Note that

P[DT 0 =∞|T 0 =z]≤P[D > na|T 0 =z]

for alln. Hence,P[DT 0 =∞] = 0, justifying our earlier assumption.

Sources

This section is adapted from [49] with additional background information
from [15].