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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 \begin{equation} S_t = \exp \{ \sigma B_t + ( r- \frac{1}{2} \sigma^2 ) t \}. \end{equation} 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 $c \in \mathbb{R}$, $a>0$, \begin{equation} \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), \end{equation} 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?

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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*}

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$S_t$ 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.

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  • $\begingroup$ 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. $\endgroup$ – ashburn Dec 6 '14 at 1:12
  • $\begingroup$ 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 $\inf \{ t \in [0,T] : B_t + (\frac{r}{\sigma} - \frac{\sigma}{2})t >a \} = \inf \{ t \in [0,T] : B_t + (\frac{r}{\sigma} - \frac{\sigma}{2})t =a \}$? $\endgroup$ – ashburn Dec 6 '14 at 1:21
  • $\begingroup$ @ashburn The Infimum coincides with the equal sign, as the function is continuous. I edited the answer to include the American payoff case. $\endgroup$ – emcor Dec 6 '14 at 10:05
  • $\begingroup$ 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)? $\endgroup$ – ashburn Dec 6 '14 at 12:28
  • $\begingroup$ $f$ 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/…). $\endgroup$ – emcor Dec 6 '14 at 12:47
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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.

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

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