If μ > r, then for any {ε > 0, ζ > 0} there exists a put option P

_{t}(S,X,T) such that Pr{P

_{T}(S,X,T) > 0} < ε and E{P

_{T}(S,X,T)} < ζ P

_{0}(S,X,T).

Tightened up:

If μ > r, then for any {ε > 0, ζ > 0} there exists a put option P_{t}(S,X,T) such that Pr{P_{T}(S,X,T) > 0} < ε and E{P_{T}(S,X,T)} < ζ P_{0}(S,X,T).

If μ > r, then for any {ε > 0, ζ > 0} there exists a put option P

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 [$] \lim_{t \rightarrow \infty} S_t = 0 \: a.s. [$]

My abbreviated version of my "over time" and "into the tails" comment omits the "over time" part, in case that wasn't clear.

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 [$] \lim_{t \rightarrow \infty} S_t = 0 \: a.s. [$]

I mean, you could do that, and the result would be very different. Or you could raise the whole thing to the power of 17, and it would also be different. And, likewise, probably not in an interesting way.

Bearish, I suspect that what you meant to write was something like

Or were you getting at something else entirely?If0.5σthen^{2}> μ - r > 0Shas the property that_{t}= e^{μt + σW(t)}lim[_{t→∞}ES_{t }- e^{rt}]= ∞butlim_{t→∞}(S_{t}- e^{rt})= 0almost surely.

I’m pretty sure I wrote what I meant to write. Your statements are true, too, but I saw no particular reason to bring the money market account into play. We can set [$] r = 0 [$] and we’ll all be happy.

All right, nonsense is not forbidden, you may write whatever you want.Que?Replace α by -α in the exponent.

I should have written

*lim*_{t→∞}*S*_{t} *= 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.

Bearish, your "If0.5σthen^{2}> r - μ > 0Shas the property that_{t}= e^{μt + σW(t)}lim[_{t→∞}ES_{t }- e^{rt}]= ∞butlim_{t→∞}(S_{t}- e^{rt})< 0almost surely.

I should have written

Bearish, your "If0.5σthen^{2}> r - μ > 0Shas the property that_{t}= e^{μt + σW(t)}lim[_{t→∞}ES_{t }- e^{rt}]= ∞butlim_{t→∞}(S_{t}- e^{rt})< 0almost surely.lim_{t→∞}S_{t}= 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.

Apologies. The definition of the process should be [$] S_t = e^{( \alpha - 0.5 \sigma^2) t + \sigma W_t} [$].

Apologies accepted. Or *S*_{t} = e^{ -at} ^{+ σW(t)} as I suggested. Forgot a bit Ito formula?

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.

Yes, it will as long 0.5 \sigma^2 > \alpha > 0. It's just a reparametrisation of your version. I know, university knowledge goes.

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.