Without having to use Black-Scholes, how do I price this option using a basic no-arbitrage argument?


Assume zero interest rate and a stock with current price at \$$1$ that pays no dividend. When the price hits level \$$H$ ($H>1$) for the first time you can exercise the option and receive \$$1$. What is the fair price $C$ of the option today?

My thoughts so far

According to my book, the answer is $\frac{1}{H}$. I'm stuck on the reasoning.

Clearly I'm not going to pay more than \$$\frac{1}{H}$ for this option. If $C > \frac{1}{H}$ then I would simply sell an option and buy $C$ shares with $0$ initial investment. Then:

  • If the stock reaches $H$ I pay off the option which costs \$$1$ but I have $\$CH > 1$ worth of shares.
  • If the stock does not reach $H$ I don't owe the option owner anything but I still have $CH>0$ shares.

What if $C<\frac{1}{H}$? Then $CH<1$ and I could buy $1$ option at \$$C$ by borrowing $C$ shares at \$$1$ each. Then:

  • If the stock reaches $H$ then I receive $1-CH > 0$ once I pay back the $C$ shares at $\$H$ each.
  • But if the stock does not reach $H$, then I do not get to exercise my option and I still owe $C S_t $ where $S_t$ is the current price of the stock. This is where I am stuck.
  • 1
    $\begingroup$ Isn't the point exactly the assumption that it will hit $H$ at some point? $\endgroup$
    – torbonde
    Commented Mar 24, 2015 at 13:19
  • $\begingroup$ @torbonde If that is true, would that imply the market is not efficient? $\endgroup$ Commented Mar 24, 2015 at 13:22
  • 1
    $\begingroup$ Fushialatitude : what additional detail are you looking for? $\endgroup$
    – q.t.f.
    Commented May 18, 2015 at 20:18

7 Answers 7


This option is a perpetual one touch option. Its price depends on the model used; additional assumptions are required to get a model-independent price.

Let us first consider 3 important example models for stock price $S$.

Constant: $S(t) \equiv 1.$ There is $0$ probability that the perpetual one touch pays off, so its price is $0.$

Black-Scholes: $S$ follows geometric Brownian motion with volatility $\sigma > 0.$ Option price $C(S,t)$ satisfies a PDE $C_t + 1/2 \sigma^2 S^2 C_{ss} = 0.$ Since it is perpetual, $C(S,t)$ cannot depend on $t$ and so $C_t = 0.$ Then the PDE reduces to an ODE $C_{ss}=0.$ With boundary values $C(0)=0$ and $C(H)=1$ the solution is $C(S)=S/H.$ With $S(0)=1$ option value is $1/H.$

Bachelier: $S$ follows arithmetic Brownian motion with volatility $\sigma > 0$ and no drift. Since Brownian motion is recurrent, with probability one $S$ will reach the level $H$. Thus the perpetual one touch has value $1.$

Note: Geometric Brownian motion is not guaranteed to reach the level $H.$ When we take log of GBM, it is an arithmetic Brownian motion with drift $-1/2 \sigma^2 dt.$ This negative drift is enough to allow some paths of log-spot to stay below the barrier level at $log(H).$ The probability of hitting the barrier is the option price $C(S,t)$ we calculated by PDE above.

Now lets return to the original question about making a model-independent no arbitrage price. Clearly from the examples it is impossible; different models give different prices.

We can get a little farther by assuming that $S(t) \ge 0.$ In this case the original poster correctly argues the fair value has $C \le 1/H.$ But we still get a range of prices. The Black-Scholes model with zero rates and positive volatility gives $C = 1/H.$ But for the constant model the fair value is 0. Any value $0 \le C \le 1/H$ is possible: consider the model where at time $0$ with risk-neutral probability $HC$ the stock follows a GBM with volatility $\sigma > 0$ and with probability $1-HC$ it remains fixed at 1 forever. The expected value under the risk neutral measure is $HC \cdot 1/H + (1-HC) \cdot 0 = C.$

There is no obvious choice for a hidden assumption to rule out these other models. So there is not a model-free fair value of this option.

  • $\begingroup$ I don't think your answer $C \le 1/H$ is correct. As many pointed out, the stock will hit the level $H$ when given enough time. $\endgroup$
    – wsw
    Commented Mar 25, 2015 at 3:16
  • 1
    $\begingroup$ The result that $C \le 1/H$ does depend on the assumption that stock price $S \ge 0$ but is otherwise correct as demonstrated by the original poster. The point is that $1/H$ units of stock has value $1$ when $S=H$, and value $>=0$ otherwise, so has at least as much value aa the option. $\endgroup$
    – q.t.f.
    Commented Mar 25, 2015 at 3:44
  • $\begingroup$ It is NOT true that $S $ is certain to reach $H.$ If it were true, then one could borrow $1$ dollar, buy a share of stock, wait until $S$ reaches $H$, sell the stock for $H>1$ dollars, repay the original $1$ dollar loan, and keep $H-1$ dollars as arbitrage profit. $\endgroup$
    – q.t.f.
    Commented Mar 25, 2015 at 4:01
  • $\begingroup$ qtf: let me think about your no-arb argument. I don't get why you wrote that the "Black-Scholes model with zero rates and positive volatility gives C=1/H". This is an American binary call option, not a European one. $\endgroup$
    – wsw
    Commented Mar 25, 2015 at 5:29
  • $\begingroup$ Expanded answer. Hopefully it is complete enough now to convince you. $\endgroup$
    – q.t.f.
    Commented Mar 25, 2015 at 10:38

Let $T= \inf\{t>0: S_t = H\}$. Then the option payoff is given by $\mathbb{1}_{\{T < \infty\}}$, and the value of the option is given by $\mathbb{P}(T< \infty)$. We assume that the stock price process is a geometric Brownian motion, that is, for $t>0$ $$ S_t = \exp\big(-\frac{1}{2}\sigma^2 t + \sigma W_t\big),$$ where $\{W_t, t \geq 0\}$ is a standard Brownian motion, and $\sigma$ is the volatility. Then, \begin{align*} S_t = H \Leftrightarrow -\frac{1}{2}\sigma t + W_t = \frac{1}{\sigma}\ln H. \end{align*} Let $\nu= -\frac{1}{2}\sigma$ and $y= \frac{1}{\sigma}\ln H$. It is well known that the density of $T$ is given by \begin{align*} f(t) = \frac{y}{\sqrt{2\pi t^3}}\exp\big(-\frac{1}{2t}(y-\nu t)^2\big)\mathbb{1}_{\{t \geq 0\}}; \end{align*} see, for example, "Mathematical Methods for Financial Markets" by Jeanblanc et. al. Then, \begin{align*} \mathbb{P}(T< \infty) &= e^{2 \nu y}\\ &= \frac{1}{H}. \end{align*} That is, $\frac{1}{H}$ is indeed the option price under the geometric Brownian motion stock price assumption.


Consider a portfolio where I sell $\frac{1}{H}$ in stock and use that to buy an option. This is a 0 cost portfolio. When I hit the barrier the price of this portfolio is also 0. Law of one price would suggest that this portfolio should be zero cost at all times. So the price of the option at any time must be $$ C_t = \frac{1}{H}*S_t $$

Also, the option should eventually get exercised.


Important assumptions:
- we have zero interest rate,
- option is perpetual,


  • with probability 1, share price will hit the barrier $H$ (in fact this is a hidden assumption that price changes continuously or we can at least trade at the very moment when $S_t = H$).

No, we can't assume that, because , as @q.t.f noted, it would imply arbitrage. In fact with no arbitrage and zero interest rate we have that share price process $S_t$ is a martingale with respect to some "arbitrage" probability. Thus $\mathbb{E}(S_t) = S_0$ for all $t\geq 0$, and assuming $S_t$ is positive, by Doob's inequality $\mathbb{P}(\sup_t S_t > H) \leq \sup_t \mathbb{E}(S_t)/H = S_0/H$.

At time 0 you create zero-cost portfolio:
- you buy 1 share at current price $S_0 = 1$,
- you sell $1/C_0$ options each worth $C_0$.

At the moment of exercise:
- have to pay 1 for every $1/C_0$ option, which costs you $1/C_0$,
- you can sell your share for $H$.
This gives you $H - 1/C_0$, and be no arbitrage, this can't be be positive. Thus we have $H - 1/C_0 \leq 0 \iff H \leq 1/C_0 \iff C_0 \leq 1/H$.

And in fact it is only a different presentation of arguments given in the question post.

Now IMO this (and bound $C_0 \geq 0$, since option is no obligation) is all we can get by pure arbitrage arguments without any further assumptions. And again I agree with @q.t.f: price is model dependent and any price in the range $[0, 1/H]$ is possible.

Let us consider the following model: at time $t=1$ price either jumps up to $S^u = 3$ or falls down to $S^d = 1/2$. Let $H=2$. I claim that $C_0 = 1/5$, since this is the cost of replicating portfolio constructed with $2/5$ long shares and borrowed $1/5$ of cash.

This price is calculated as $C_0 = \frac{S_0 - S^d}{S^u - S^d}$, so letting $S^u \to \infty$, we can get as close to zero as we want.

  • $\begingroup$ The reasoning for the other side is not symmetric -- it is is the problem that the original poster encountered. Can you please provide more details? $\endgroup$
    – Gordon
    Commented May 19, 2015 at 17:12
  • $\begingroup$ @Gordon : you are right, I edited my answer. $\endgroup$
    – RKucharski
    Commented May 22, 2015 at 13:46

Is the option perpetual? If so, the $C=1/H$ answer looks suspicious and $C=1$ is more plausible for the reasons detailed below.

If $C<1$, you borrow \$$C$, buy the option, wait until the underlying hits the barrier, receive \$1 payout, repay the \$$C$ debt (we have assumed 0 interest) and pocket the difference. Similarly, if $C>1$ then one can arbitrage it by selling the option.

In other words, the fair value of the option is the risk-neutral expected value of the discounted payout. If we model the underlying as a geometric Brownian motion, then it will hit $H$ almost surely, so the payout will be $1 with probability 1. We need to discount this amount to today, and we don't know when the payout will happen (the hitting time would have a Lévy distribution), but it's not an obstacle, because with zero interest rate, discount factor will always equal to 1 no matter the time. So again we arrive at fair value of \$1.

  • $\begingroup$ This is wrong, at least for the case stock price $S \ge 0.$ It is right for $S$ a standard brownian motion (Bachelier model), but wrong for geometric brownian motion (Black-Scholes model). With GBM, it is NOT true that $S$ is guaranteed to reach $H.$ $\endgroup$
    – q.t.f.
    Commented Mar 25, 2015 at 4:24
  • $\begingroup$ qtf: the OP wanted to find out the price of the option while using model-free arguments. $\endgroup$
    – wsw
    Commented Mar 25, 2015 at 4:28
  • $\begingroup$ I think the key point here is "perpetual". $\endgroup$
    – Matt Wolf
    Commented Mar 25, 2015 at 7:41
  • $\begingroup$ I expanded my answer to address the misconception everyone is having that GBM is guaranteed reach a level. See below. $\endgroup$
    – q.t.f.
    Commented Mar 25, 2015 at 10:37

Unfortunately I cannot upvote user2142's answer because I lack the reputation, but his reasoning makes sense to me: the price is $\$1/H$ because as the seller of the option you buy $1/H$ shares for the premium. You sell them when the $S_t$ hits $H$ to obtain the $\$1$ you have to pay to the option buyer.

I think the price is model free for any model with continuous $S_t$. If you assume that paths of $S_t$ "typically" contain upward jumps it's less clear to me and I would expect the price to become model dependent.

  • $\begingroup$ Can you show why $C<1/H$ is not true? $\endgroup$
    – Gordon
    Commented May 19, 2015 at 17:14
  • $\begingroup$ I can try with a no arbitrage argument. Let's assume that $S_t$ is continuous. If $C<1/H$ then I buy the option for $C$ and short $1/H$ shares. This leaves me with $\epsilon = 1/H - C$ in cash, the option, and short $1/H$ shares. When the stock hits $H$, I get 1 from the option and buy $1/H$ shares and have $\epsilon > 0$ in cash. This happens almost surely in finite time - maybe that's another ingredient for the model free result!. This would be an almost certain profit with no risk against the law of one price. $\endgroup$ Commented May 19, 2015 at 22:12
  • $\begingroup$ PS: With the GBM in your derivation you hit $H$ almost surely in finite time. Does my derivation - although not very rigorous - make sense? $\endgroup$ Commented May 19, 2015 at 22:14
  • $\begingroup$ PPS: And actually I just realized that you don't need the hitting in finite time. You are left with $\epsilon > 0$ in cash no matter if $S_t$ ever hits $H$ or not. $\endgroup$ Commented May 19, 2015 at 22:21
  • 1
    $\begingroup$ The cash amount $C-1/H<1$ above should be read as "the cash amount $1/H-C<1$. $\endgroup$
    – Gordon
    Commented May 20, 2015 at 0:36

Maybe it is better to use martingale theory to characterise whether it is an equality or not.

Let $S_t$ be a (right)-continuous positive martingale with $S_0 < H$.

Let $\tau = \inf \{ t > 0| S_t = H \}$.

The option pays 1 unit of cash at $\tau$, and there is no maturity (perpetual option). What is the price of the option? I.e. compute $$P_0 = \mathbb{E} \left(I_{\{ \tau < \infty \}}\right) = \mathbb{P}\left ( \tau < \infty\right).$$

By the optional sampling theorem, the stopped process $S_{t \wedge \tau}$ is also a martingale. Therefore, we have $$S_0 = \mathbb{E}\left(S_{t \wedge \tau}\right) = \mathbb{E}\left(S_t I_{\{ t \leq \tau\}}\right) + \mathbb{E}\left(S_{\tau} I_{\{ \tau < t\}}\right).$$

Taking the limit $t \rightarrow \infty$ on both side and using the fact that $S_{\tau} = H$, we have $$S_0 = \lim_{t \rightarrow \infty} \mathbb{E}\left(S_t I_{\{ t \leq \tau\}}\right) + H \mathbb{P}\left ( \tau < \infty\right).$$

By the dominated convergence theorem and the fact that any (right)-continuous positive martingale converges almost surely, we have $$\lim_{t \rightarrow \infty} \mathbb{E}\left(S_t I_{\{ t \leq \tau\}}\right) = \mathbb{E}\left(S_{\infty} I_{\{ \tau = \infty\}}\right).$$

Hence, $P_0 = S_0/H$ if only if $\mathbb{E}\left(S_{\infty} I_{\{ \tau = \infty\}}\right) = 0$. In particular, if the martingale converges to zero this holds true. For example we can think of the GBM $S_t = S_0 \exp\left(\sigma B_t - 0.5 \sigma^2 t\right)$. It is a martingale converging to zero a.s. However, as a counter-example, we can think of $S_t + x$ with a positive constant $x > 0$. This process is still a martingale but it converges to $x > 0$ and $$P_0 = \mathbb{P}\left ( \tau < \infty\right) < S_0/H.$$

Finally, without any model assumption, I think that we can only claim that $P_0 \leq S_0/H$.


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