Suppose an option is defined as follows. There is an upper barrier at $H$ and a lower barrier at $0$. If the stock price touches the upper barrier you get a payoff of $1$ and the trade terminates immediately. Likewise if the stock price touches the lower barrier, you get a payoff of $0$ and the trade terminates. What is the price of the option? (Assume zero interest rates and the usual Geometric Brownian motion for the underlying)

I am not sure how this option should be dealt with, any solution to this problem is greatly appreciated.

  • 2
    $\begingroup$ Is this correctly specified? The probability of a geometric brownian hitting zero is zero, so the price of this contract is trivially 1. $\endgroup$
    – dm63
    Commented Mar 5, 2017 at 16:10
  • 1
    $\begingroup$ Is it perpetual? I don't see any reference to expiration $\endgroup$ Commented Mar 5, 2017 at 16:34
  • $\begingroup$ @dm63 yes, it is correctly specified. I don't understand your conclusion, could you please elaborate it in an answer? Thanks a lot $\endgroup$
    – RandomGuy
    Commented Mar 5, 2017 at 17:38
  • $\begingroup$ @NivelEgres That's right, in another source they called it "infinite expiry". Can you figure out a pricing formula? Thanks. $\endgroup$
    – RandomGuy
    Commented Mar 5, 2017 at 17:39
  • 3
    $\begingroup$ ok well suppose the price is K<1. Borrow K at zero interest rate, buy the contract for K. Wait until the upper barrier is hit, receive 1 and pay back the loan of K for a 1-K>0 profit. suppose the price is K>1. Sell the contract for K, invest at zero in the bank, wait for the upper barrier to be hit, pay out the 1 and be left with K-1 >0. This all supposes that the lower barrier can never be hit (geometric brownian always >0) and that the upper barrier will eventually be hit (this happens with probability 1, I believe, although I admit i don't have a proof) $\endgroup$
    – dm63
    Commented Mar 5, 2017 at 17:52

1 Answer 1


As specified I will assume your option is perpetual; I will also assume that it is written on an asset whose price $(S_t)_{t \geq 0}$ follows a Geometric Brownian Motion (GBM) with drift coefficient $rS_t$ and diffusion coefficient $\sigma S_t$ under the risk-neutral measure $\mathbb{Q}$ $-$ we assume a constant risk-free rate:

$$ dS_t = rS_tdt + \sigma S_t dW_t$$

where $W_t = W_t^{\mathbb{Q}}$. Let us now define the two hitting times of interest here $-$ where time $0$ is today:

  • The time at which the stock price breaches the barrier $H$: $\tau_H=\min\{t \geq 0:S_t=H\}$;
  • The time at which the stock price reaches $0$: $\tau_0=\min\{t \geq 0:S_t=0\}$.

The 2$^{\text{nd}}$ is irrelevant: the probability of $S_t$ hitting $0$ is $0$. Thus we are left with a payoff of $1$ whenever the price reaches the barrier $H$, after which the trade expires. Letting then:

$$ \tau = \tau_H$$

By risk-neutral valuation theory the price today $C_0$ of this claim is given by its discounted payoff under the risk-neutral measure$^{(1)}$:

$$ C_0 = \mathbb{E}^{\mathbb{Q}}[e^{-r\tau}|\mathcal{F}_0] = \mathbb{E}^{\mathbb{Q}}[e^{-r\tau}] $$


  • If the interest rate is $0$, the problem is trivial and the claim's price is $1$, as explained by @dm63 in his comment.
  • On the other hand, if $r >0$ then we have to compute the Laplace transform of the hitting time of a GBM, which is equivalent to computing the Laplace transform of an arithmetic Brownian motion.


$$ \tau = \min\{t \geq 0 : W_t+\theta t = x\} $$

be the hitting time of the arithmetic Brownian motion (ABM) with volatility $1$ and $0<\alpha<1$. The solution, which I prove in the appendix below, is:

$$ \mathbb{E}^{\mathbb{Q}}\left[e^{-\alpha\tau}\right] = e^{(\theta-\sqrt{\theta^2+2\alpha})x}$$

But to be applied to our problem, it requires the following strong assumption$^{(1)}$:

$$ \frac{2r-\sigma^2}{2\sigma} \geq 0 \quad \Leftrightarrow \quad r \geq \frac{\sigma^2}{2}$$

Now, observe that by letting $S_0$ be the current stock price, we have:

$$ S_t = S_0e^{(r-\frac{\sigma^2}{2})t+\sigma W_t} = H \quad \Leftrightarrow \quad W_t + \frac{2r-\sigma^2}{2\sigma}t = \frac{1}{\sigma}\log \frac{H}{S_0}$$


$$ \begin{align} & \alpha := r \\[6pt] & \theta := \frac{2r-\sigma^2}{2\sigma} \\[6pt] & x := \frac{1}{\sigma}\log \frac{H}{S_0} \end{align} $$

We have:

$$ \begin{align} \left(\theta - \sqrt{\theta^2+2\alpha}\right)x & = \left(\frac{2r-\sigma^2}{2\sigma}-\sqrt{\frac{(2r-\sigma^2)^2}{4\sigma^2}+2r}\right)\frac{1}{\sigma}\log \frac{H}{S_0} \\[6pt] & = \left(\frac{2r-\sigma^2}{2\sigma}-\sqrt{\frac{4r^2-4r\sigma^2+\sigma^4+8r\sigma^2}{4\sigma^2}}\right)\frac{1}{\sigma}\log \frac{H}{S_0} \\[6pt] & = \left(\frac{2r-\sigma^2}{2\sigma}-\sqrt{\frac{(2r+\sigma^2)^2}{4\sigma^2}}\right)\frac{1}{\sigma}\log \frac{H}{S_0} \\[6pt] & = \log \frac{S_0}{H} \end{align} $$


$$ \boxed{C_0 = \mathbb{E}^{\mathbb{Q}}\left[e^{-r\tau}\right] = \frac{S_0}{H}}$$

Note that for the claim to make sense, we need $H > S_0$, so we have $0 \leq C_0 < 1$. The price has a striking feature: it does not depend on the volatility of the stock price. This can be explained by the fact that, on a perpetual time scale and given that we are almost sure the stock price will hit the barrier (see appendix), the volatility becomes irrelevant $-$ although one might think that a higher volatility should increase the price of this claim.

Appendix: Laplace transform of the hitting time of an ABM

We are looking for a closed-form formula for:

$$\mathcal{L_{\alpha}}(\tau) = \mathbb{E}^{\mathbb{Q}}\left[e^{-\alpha\tau}\right] $$


$$ \tau = \min\{t \geq 0 : W_t+\theta t = x\} $$

We will assume that:

$$ \begin{align} x > 0 & \quad (\text{A}) \\[6pt] 1 > \lambda > 0 & \quad (\text{B}) \\[6pt] \theta \geq 0 & \quad (\text{C}) \end{align} $$


$$ \hat{W}_t = W_t+\theta t $$

We first define the auxiliary process:

$$ X_t = e^{\lambda \hat{W}_t - \lambda\theta t - \frac{\lambda^2}{2}t} = e^{\lambda W_t - \frac{\lambda^2}{2}t}$$

It can be easily shown that $X_t$ is a martingale$^{(2)}$. Therefore the stopped process $X_{\min(t,\tau)}$ is also a martingale. By the martingale property and given $W_0=0$:

$$ \mathbb{E}^{\mathbb{Q}}[X_{\min(t,\tau)}] = \mathbb{E}^{\mathbb{Q}}[X_{\min(0,\tau)}] = \mathbb{E}^{\mathbb{Q}}[X_0] = 1 $$

Let us now assess the behaviour of the stopped process as $t \rightarrow \infty$ depending on whether $\tau$ is finite or not:

  • If the hitting time is finite, then it comes:

$$ \lim_{t \, \rightarrow \, \infty} X_{\min(t,\tau)} = e^{\lambda \hat{W}_{\tau} - \lambda\theta\tau - \frac{\lambda^2}{2}\tau} $$

  • If the hitting time is infinite, i.e. the arithmetic Brownian motion never hits $x$, then we know that the exponential of $\lambda \hat{W}_t$ remains bounded between $0$ and $e^{\lambda x}$. Under assumptions $(\text{B})$ and $(\text{C})$ the rest of the exponential converges to $0$ as $t \rightarrow \infty$, thus:

$$ \lim_{t \, \rightarrow \, \infty} X_{\min(t,\tau)} = 0 $$

Combining these two results, we get:

$$ \lim_{t \, \rightarrow \, \infty} X_{\min(t,\tau)} = X_{\infty} $$


$$ X_{\infty} = \mathbf{1}_{\{\tau \, < \, \infty\}}\left(e^{\lambda \hat{W}_{\tau} - \lambda\theta\tau - \frac{\lambda^2}{2}\tau}\right) $$

Now, under assumptions $(\text{A})$ and $(\text{B})$, we have:

$$ 0 \leq X_{\min(t,\tau)} \leq e^{\lambda \hat{W}_{\min(t,\tau)}} \leq e^{\lambda \hat{W}_{\tau}} \leq e^{\hat{W}_{\tau}} = e^{x} $$

Therefore by the dominated convergence theorem we can state that:

$$ \lim_{t \rightarrow \infty} \mathbb{E}^{\mathbb{Q}}[X_{\min(t,\tau)}] = \mathbb{E}^{\mathbb{Q}}\left[\lim_{t \rightarrow \infty}X_{\min(t,\tau)}\right] = \mathbb{E}^{\mathbb{Q}}[X_{\infty}]$$

That is:

$$ \mathbb{E}^{\mathbb{Q}}\left[\mathbf{1}_{\{\tau \, < \, \infty\}}\left(e^{\lambda \hat{W}_{\tau} - \lambda\theta\tau - \frac{\lambda^2}{2}\tau}\right)\right] = 1 $$

Now making $X_{\infty}$ dependent on $\lambda$, we see it is bounded from above by $e^{x}>1$ and it converges $\text{a.s.}$ to $\mathbf{1}_{\{\tau \, < \, \infty\}}$ when $\lambda \rightarrow 0$. Applying again the dominated convergence theorem, we get:

$$ \lim_{\lambda \rightarrow 0}\mathbb{E}^{\mathbb{Q}}[X_{\infty}(\lambda)] = \mathbb{E}^{\mathbb{Q}}\left[\lim_{\lambda \rightarrow 0}X_{\infty}(\lambda)\right] = \mathbb{E}^{\mathbb{Q}}\left[\mathbf{1}_{\{\tau \, < \, \infty\}}\right] = \mathbb{Q}\left(\tau \, < \, \infty\right)=1$$

The hitting time is $\text{a.s}$ finite, hence we can get rid of the indicator function and write:

$$ \mathbb{E}^{\mathbb{Q}}\left[e^{\lambda \hat{W}_{\tau} - \lambda\theta\tau - \frac{\lambda^2}{2}\tau}\right] = 1 \quad \Leftrightarrow \quad \mathbb{E}^{\mathbb{Q}}\left[e^{- (\lambda\theta + \frac{\lambda^2}{2})\tau}\right] = e^{-\lambda x} $$

Let us define:

$$ \alpha = \lambda\left(\theta+\frac{\lambda}{2}\right) $$

This identity defines a quadratic equation in $\lambda$ with solutions:

$$ \begin{align} \lambda_1 = -\theta + \sqrt{\theta^2+2\alpha} \\[6pt] \lambda_2 = -\theta - \sqrt{\theta^2+2\alpha} \end{align} $$

Given we have assumed $\lambda>0$ and $(\text{C})$, the solution must be $\lambda_1$, thus:

$$ \mathbb{E}^{\mathbb{Q}}\left[e^{- \alpha\tau}\right] = e^{(\theta - \sqrt{\theta^2+2\alpha}) x} $$

Note $(1)$: alternatively you could also interpret the payoff such that:

$$ C_0 = \mathbb{E}^{\mathbb{Q}}[\mathbf{1}_{\{\tau \, < \, \infty\}}e^{-r\tau}] $$

In this case you are explicitly requiring the hitting time to be finite.

There is one strong advantage from this representation, as we do not require any longer the strong assumption we made at the beginning, which purpose is to allow us to get rid of the indicator function in $X_{\infty}$:

$$ r \geq \frac{\sigma^2}{2} $$

Note $(2)$: let us pick two times $s<t$:

$$ \begin{align} \mathbb{E}^{\mathbb{Q}}[X_t|\mathcal{F}_s] & = \mathbb{E}^{\mathbb{Q}}[e^{\lambda W_t-\frac{\lambda^2}{2}t}|\mathcal{F}_s] \\[3pt] & = \mathbb{E}^{\mathbb{Q}}[e^{\lambda W_t-\frac{\lambda^2}{2}t+\lambda(W_s-W_s)}|\mathcal{F}_s] \\[3pt] & = e^{\lambda W_s-\frac{\lambda^2}{2}t}\mathbb{E}^{\mathbb{Q}}[e^{\lambda(W_t-W_s)}|\mathcal{F}_s] \\[3pt] & = e^{\lambda W_s-\frac{\lambda^2}{2}t}\mathbb{E}^{\mathbb{Q}}[e^{\lambda(W_t-W_s)}] \\[3pt] & = e^{\lambda W_s-\frac{\lambda^2}{2}t}e^{\frac{\lambda^2}{2}(t-s)} \\[3pt] &=X_s \end{align} $$


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