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I am trying to understand the no arbitrage argument for to determine the price of an option that pays $1$ when the stock hits $\$H$ for the first time. The current price of the stock is $\$1$. The argument goes as follows. I can buy $1/H$ of the stock now, which will give me $\$1$ when the stock becomes $\$H$. Thus the option can not be more than $1/H$. On the other hand, if the option price is $C$ less than $\$1/H$ then I can buy one option by borrowing $C$ shares of the stock. Once it hits $\$H$, exercise the option to make profit $1 - CH > 0$. So the price of the option can not be less than $1/H$.

What I don't understand about this argument is that how come one doesn't take into account the possibility that the stock never gets to $H$. In both cases, the argument assumes that the stock gets hits $H$ and deduces the conclusion. Any clarification would be appreciated. Thank you

Johnny T.
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  • How do you buy an option (at whichever price) by "borrowing $C$ shares of the stock" ? – Kurt G. Oct 28 '22 at 17:26
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    I remember a result that the probability that stock never gets to $H$ over a period $(0,T)$ tends to $0$ when $T \to +\infty$. In other words, almost surely that the stock hits $H$. – NN2 Oct 28 '22 at 17:36
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    @KurtG. apparently it means to borrow, then sell, and use that money to buy the option, exercise the option, buy back the stock and return – Johnny T. Oct 28 '22 at 17:46
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    I can understand it better when we change the face value of the option payout to be $H$ when the stock hits $H$. The price for this option today is clearly the stock price today. Let's call this $S_0$. If the option price were less than $S_0$ you could buy the option for less than $S_0$ borrow a stock, sell it at at $S_0$ and pocket a profit. When the stock hits $H$ you get $H$ from the option which you use to buy back the stock and return it to the lender (zero sum game at the end). Profit was pocketed at the beginning. ... – Kurt G. Oct 28 '22 at 18:37
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    ...The reason that this works without volatility is because we silently assumed that the option has a perpetual maturity. In reality such options expire after a while which changes the picture completely. – Kurt G. Oct 28 '22 at 18:37
  • A more appropriate site for this would be https://quant.stackexchange.com/ – whoisit Oct 29 '22 at 05:13

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For simplicity let's change the face value of the option payout to be $H$ when the stock hits the upper level $H\,.$ The price for this option today is clearly the stock price today. Let's call this $S_0\,.$

Proof (I guess that's what you had in mind). If the option price were less than $S_0$ you could buy the option for less than $S_0$ and borrow a stock, sell it at at $S_0$ and pocket a profit. When the stock hits $H$ you get $H$ from the option which you use to buy back the stock and return it to the lender (zero sum game at the end). Profit was pocketed at the beginning. $\quad\quad\quad\quad\quad\Box$

Clearly, when the face value is one the option price will be $S_0/H\,.$

The reason that this works without volatility is because we silently assumed that the option has a perpetual maturity. In reality such options expire after a while which changes the picture completely:

Using the well-known Reiner & Rubinstein (1991b) formula for this option with maturity $T$ which we can find in [1] we get the following price-maturity relationship that depends on volatility $\sigma$ but -as expected- approaches the stock price of $S_0=100$ for large $T$:

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The Reiner & Rubinstein (1991b) option pricing formula is $$ H\Big(\frac{H}{S_0}\Big)^{\mu+\lambda}\Phi(-z)+H\Big(\frac{H}{S_0}\Big)^{\mu-\lambda}\Phi(-z+2\lambda\sigma\sqrt{T}) $$ where $\mu=\frac{r-q+\sigma^2/2}{\sigma}$ and $\lambda=\sqrt{\mu^2+\frac{2r}{\sigma^2}}$ and $z=\frac{\log(H/S_0)}{\sigma\sqrt{T}}+\lambda\sigma\sqrt{T}\,.$

Here $q$ is the continuous dividend yield of the stock. In our case it must be zero. If not, the lender will typically demand a fee which we have ignored in the above proof. A non zero dividend yield will change the option price even for very large maturities.

[1] E.G. Haug, The Complete Guide to Option Pricing Formulas.

Kurt G.
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  • I'm not convinced that this assuages OP's doubt. Even with an option with perpetual maturity, the stock price could never hit the upper level $H$. – Raskolnikov Oct 29 '22 at 07:11
  • You mean when $\sigma=0$ ? This case is totally trivial imho. – Kurt G. Oct 29 '22 at 07:13
  • Why should $\sigma=0$? – Raskolnikov Oct 29 '22 at 07:15
  • In that case the stock never gets to $H$. In the case $\sigma>0$ it does. – Kurt G. Oct 29 '22 at 07:16
  • Why? It seems to me you're implicitly assuming a Wiener process driving the stock price. The original argument proposed by OP tries to avoid that. Maybe it's unwarranted to make such a general argument, but that is exactly what I interpret OP's question to be about. – Raskolnikov Oct 29 '22 at 07:18
  • Yes. Wiener. It is OP's call now to confirm this. We can surely dream up lots of models where the stock never hits $H$. However I am wondering if they all can satisfy the requirement that the discounted stock price must be a martingale. Wiener will lurk behind those scenes. – Kurt G. Oct 29 '22 at 07:20
  • Thank you for the comments and answer. In the question I was working on, I now came to the conclusion that it must be assuming, though not stated explicitly, that maturity is infinity and that at some point it will be $H$. Then the solution in the book makes sense. – Johnny T. Oct 31 '22 at 15:51