Tightened up:
If μ > r, then for any {ε > 0, ζ > 0} there exists a put option Pt(S,X,T) such that Pr{PT(S,X,T) > 0} < ε and E{PT(S,X,T)} < ζ P0(S,X,T).
Replace α by -α in the exponent.The standard martingale theory approach to model replication, pricing, etc. always comes with a bounded time horizon, to avoid problems of this sort. The law of large numbers prevents non-trivially different measures from being equivalent in the limit as time goes to infinity. Another version of your problem is to note that if [$]0.5 \sigma^2 > \alpha > 0 [$], then [$] S_t = e^{\alpha t + \sigma W_t} [$] has the properties that [$] \lim_{t \rightarrow \infty} E[S_t] = \infty [$] but [$] \lim_{t \rightarrow \infty} S_t = 0 \: a.s. [$]
Que?Replace α by -α in the exponent.The standard martingale theory approach to model replication, pricing, etc. always comes with a bounded time horizon, to avoid problems of this sort. The law of large numbers prevents non-trivially different measures from being equivalent in the limit as time goes to infinity. Another version of your problem is to note that if [$]0.5 \sigma^2 > \alpha > 0 [$], then [$] S_t = e^{\alpha t + \sigma W_t} [$] has the properties that [$] \lim_{t \rightarrow \infty} E[S_t] = \infty [$] but [$] \lim_{t \rightarrow \infty} S_t = 0 \: a.s. [$]
Or were you getting at something else entirely?If 0.5σ2 > μ - r > 0 then St = eμt + σW(t) has the property that limt→∞E[St - ert] = ∞ but limt→∞(St - ert) = 0 almost surely.
All right, nonsense is not forbidden, you may write whatever you want.Que?Replace α by -α in the exponent.The standard martingale theory approach to model replication, pricing, etc. always comes with a bounded time horizon, to avoid problems of this sort. The law of large numbers prevents non-trivially different measures from being equivalent in the limit as time goes to infinity. Another version of your problem is to note that if [$]0.5 \sigma^2 > \alpha > 0 [$], then [$] S_t = e^{\alpha t + \sigma W_t} [$] has the properties that [$] \lim_{t \rightarrow \infty} E[S_t] = \infty [$] but [$] \lim_{t \rightarrow \infty} S_t = 0 \: a.s. [$]
Bearish, your "limt→∞St = 0" is wrong, which is probably what is causing the confusion. I don't know if you were intending to get at my original interpretation or maybe something about the certainty of ruin, but the certainty of ruin is not reflected in the model you described.If 0.5σ2 > r - μ > 0 then St = eμt + σW(t) has the property that limt→∞E[St - ert] = ∞ but limt→∞(St - ert) < 0 almost surely.
I should have written
Bearish, your "limt→∞St = 0" is wrong, which is probably what is causing the confusion. I don't know if you were intending to get at my original interpretation or maybe something about the certainty of ruin, but the certainty of ruin is not reflected in the model you described.If 0.5σ2 > r - μ > 0 then St = eμt + σW(t) has the property that limt→∞E[St - ert] = ∞ but limt→∞(St - ert) < 0 almost surely.
I just subscribe to a different convention.I’d like to blame Marsden for his unconventional use of [$] \mu [$], but that’s probably not fair. With a negative drift, my first property will not hold, though, since then the process will tend to go zero in all relevant senses.