 Marsden
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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). bearish
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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 $\lim_{t \rightarrow \infty} S_t = 0 \: a.s.$ Marsden
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

My abbreviated version of my "over time" and "into the tails" comment omits the "over time" part, in case that wasn't clear. Gamal
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Joined: February 26th, 2004, 8:41 am

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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 $\lim_{t \rightarrow \infty} S_t = 0 \: a.s.$
Replace α by -α in the exponent. bearish
Posts: 4875
Joined: February 3rd, 2011, 2:19 pm

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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 $\lim_{t \rightarrow \infty} S_t = 0 \: a.s.$
Replace α by -α in the exponent.
Que? bearish
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Joined: February 3rd, 2011, 2:19 pm

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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. Marsden
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

Bearish, I suspect that what you meant to write was something like
If 0.5σ2 > μ - r > 0 then St = eμt + σW(t) has the property that limt→∞E[S- ert] = ∞ but limt→∞(St - ert) = 0 almost surely.
Or were you getting at something else entirely? bearish
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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. Gamal
Posts: 954
Joined: February 26th, 2004, 8:41 am

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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 $\lim_{t \rightarrow \infty} S_t = 0 \: a.s.$
Replace α by -α in the exponent.
Que?
All right, nonsense is not forbidden, you may write whatever you want. Marsden
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

I should have written

If 0.5σ2 > r - μ > 0 then St = eμt + σW(t) has the property that limt→∞E[S- ert] = ∞ but limt→∞(St - ert) < 0 almost surely.
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. bearish
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Joined: February 3rd, 2011, 2:19 pm

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

I should have written

If 0.5σ2 > r - μ > 0 then St = eμt + σW(t) has the property that limt→∞E[S- ert] = ∞ but limt→∞(St - ert) < 0 almost surely.
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.

Apologies. The definition of the process should be $S_t = e^{( \alpha - 0.5 \sigma^2) t + \sigma W_t}$. Gamal
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

Apologies accepted. Or St = e -at + σW(t)  as I suggested. Forgot a bit Ito formula? bearish
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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. Gamal
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Joined: February 26th, 2004, 8:41 am

### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

Yes, it will as long 0.5 \sigma^2 > \alpha > 0. It's just a reparametrisation of your version. I know, university knowledge goes. Marsden
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### Re: "Option Pricing, Risk Premium, and Arbitrage: An Argument for Volatility-Modified Risk-Neutral Prices."

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.
I just subscribe to a different convention. 