Pricing exotic option whose payout depends on the stopping time

I am struggling with this question:

Let $B$ be a standard Brownian motion. In a Black-Scholes model, at time $t$, the stock price is given by $$S_t = \exp \{ \sigma B_t + ( r- \frac{1}{2} \sigma^2 ) t \}.$$ where $\sigma >0$ and $r$ are constants. Let $a>0$. We want to calculate the time-0-price of an exotic option which will pay $1$ at the time $\tau = \inf \{ t \in [0,T]: S_t > e^{\sigma a} \}$ if the time happens before the expiry $T>0$, otherwise it pays nothing.

The following fact is given: For any $c \in \mathbb{R}$, $a>0$, $$\mathbb{P} \, ( \inf \{ t \in [0,T]: B_t + ct =a \} \leq T ) = 1- \Phi \bigg( \frac{a-cT}{\sqrt{T}} \bigg) + e^{2ac} \Phi \bigg( \frac{-a-cT}{\sqrt{T}} \bigg),$$ where $\Phi$ denotes the cumulative distribution function of the $N(0,1)$ distribution.

I was only taught about the pricing of European options. What do we need to do in this case?

Here is to continue the above answer of Emcor to make it more explicit. Note that the fact given in the question should instead be \begin{align*} P(\inf \big\{t \in [0, T], B_t +ct = a \big\} \geq T) = 1- \Phi\Big(\frac{a-cT}{\sqrt{T}}\Big) + e^{2ac}\Phi\Big(\frac{-a-cT}{\sqrt{T}}\Big). \end{align*} Then, for $0<t_0\leq T$, \begin{align*} P(\tau \leq t_0) &= P\Big(\inf \big\{t \in [0, T], S_t >e^{a\sigma} \big\} \leq t_0 \Big)\\ &= P\Big(\inf \big\{t \in [0, T], B_t +ct>a \big\}\leq t_0 \Big)\\ &= P\Big(\inf \big\{t \in [0, t_0], B_t +ct>a \big\} \leq t_0 \Big)\\ &= P\Big(\inf \big\{t \in [0, t_0], B_t +ct = a \big\} \leq t_0 \Big)\\ &= \Phi\Big(\frac{a-ct_0}{\sqrt{t_0}}\Big) - e^{2ac}\Phi\Big(\frac{-a-ct_0}{\sqrt{t_0}}\Big). \end{align*} Let $\phi$ denote the density function of a standard normal random variable. Then the density of $\tau$ over the interval $[0, T]$ is given by (by differentiating the above function with respect to $t_0$) \begin{align*} \phi_{\tau}(t_0) &= \frac{a}{\sqrt{t_0^3}}\phi\Big(\frac{a-ct_0}{\sqrt{t_0}}\Big) \\ &=\frac{a}{\sqrt{2\pi t_0^3}}e^{-\frac{1}{2}\big(\frac{a^2}{t_0} - 2ac + c^2 t_0 \big)}. \end{align*} The option value is then \begin{align*} &\int_0^T e^{-r t_0} \frac{a}{\sqrt{2\pi t_0^3}}e^{-\frac{1}{2}\big(\frac{a^2}{t_0} - 2ac + c^2 t_0 \big)} dt_0\\ =&e^{a\big(c-\sqrt{c^2 + 2 r}\big)}\int_0^T \frac{a}{\sqrt{2\pi t_0^3}}e^{-\frac{1}{2}\big(\frac{a^2}{t_0} - 2a\sqrt{c^2 + 2 r} + (\sqrt{c^2 + 2 r})^2 t_0 \big)} dt_0\\ =& e^{a\big(c-\sqrt{c^2 + 2 r}\big)}\bigg[\Phi\bigg(\frac{a-\sqrt{c^2 + 2 r}\,t_0}{\sqrt{t_0}}\bigg) -e^{2a\sqrt{c^2 + 2 r}}\Phi\bigg(\frac{-a-\sqrt{c^2 + 2 r}\,t_0}{\sqrt{t_0}}\bigg)\bigg]_0^T\\ =& e^{a\big(c-\sqrt{c^2 + 2 r}\big)}\Phi\bigg(\frac{a-\sqrt{c^2 + 2 r}\,T}{\sqrt{T}}\bigg) + e^{a\big(c+\sqrt{c^2 + 2 r}\big)}\Phi\bigg(\frac{-a-\sqrt{c^2 + 2 r}\,T}{\sqrt{T}}\bigg). \end{align*}

$S_t$ is already under $Q$ (riskfree drift), so you not need to change the measure here.

Note that $c:=\left(\frac{r}{\sigma}-\frac{1}{2}\sigma\right)$ and $E\left(1_A\right)=P(A)$.

So one computes the European option price as the discounted payoff expectation: $$C=e^{-rT}E\left(1_{\tau\leq T}\right)=e^{-rT}P(\tau\leq T).$$

The option price equals the discounted probability of the hitting time.

If the option is of American type, the discount factor becomes stochastic:

$$C=E\left(e^{-r\tau}1_{\tau\leq T}\right)=\int_0^T e^{-r\tau}\,f(\tau)\,d\tau$$

The expression can also be calculated by Laplace transform.

• But to be very precise, the payoff of the option at time $T$ is not $\mathbf{1}_{\tau \leq T}$. (This can only be interpreted as the "accumulated" payoff over the period [0,T]. Right?) Consider the situation when the first hitting of $\{ S_t > e^{ \sigma a} \}$ occurs at time $T/2$. Then an amount of 1 is paid at time $T/2$ but 0 at time T. – ashburn Dec 6 '14 at 1:12
• Another question: The stopping time $\tau$ in the question is equivalent to $\inf \{ t \in [0,T] : B_t + (\frac{r}{\sigma} - \frac{\sigma}{2})t >a \}$. The fact given in the last line can only be used if $''>a''$ is replaced by $''=a''$. Or is it true that $\inf \{ t \in [0,T] : B_t + (\frac{r}{\sigma} - \frac{\sigma}{2})t >a \} = \inf \{ t \in [0,T] : B_t + (\frac{r}{\sigma} - \frac{\sigma}{2})t =a \}$? – ashburn Dec 6 '14 at 1:21
• @ashburn The Infimum coincides with the equal sign, as the function is continuous. I edited the answer to include the American payoff case. – emcor Dec 6 '14 at 10:05
• What is the function $f$? Also, do you know any source that derives the formula of the pricing of American options in Black Scholes by Laplace transforms (I haven't seen it before)? – ashburn Dec 6 '14 at 12:28
• $f$ is the density of $P$ for the expectation (it is just the derivative of $P$). You may search for hitting times laplace transform, it is usually related to barrier options (math.stackexchange.com/questions/26258/…). – emcor Dec 6 '14 at 12:47

If you assume the payoff is paid at time T, you just have to compute P(tau < T). In this case, you have everything you need to do it. If the payoff is paid at time tau, you need to compute the density of the stopping time.

• $P(\tau<T)$ must be discounted. – emcor Dec 7 '14 at 14:35
• If the option is european the discount factor is deterministic : it is exp(-rT). – Onyxx Dec 7 '14 at 17:03
• Did you read my answer? – emcor Dec 7 '14 at 17:40