I wolud like to know if there is an analytic formula to to valuate a up-knock-in forward, it means \begin{equation*} (S_{H_B}-S_T)1_{[H_B\leq T]} \end{equation*} where $H_B=\inf[t\geq0 | S_t=B]$ for some barrier $B>0$. Is possible under the Black-Scholes model compute in analytic way this derivative?. I have been read the book of S. Shreve of stochastic calculus for finance II, but there is only formulas to knock-out options, if anyone knows where to read about theses subjets, or has some answer I really appreciate it. Thank you.


1 Answer 1


We note that $\{H_B \le T \} = \{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}$ and $S_{H_B} = B$, then, it suffices to compute $$V:=\mathbb{E}((B-S_T)\cdot \mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}}) =B\cdot\mathbb{P}(\underset{0 \leq t \leq T}{\max}S_t \ge B) -\mathbb{E}(S_T\cdot \mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}}) \tag{1}$$ Here, for the sake of simplicity, we ignore the discount factor in $(1)$ by supposing that $r=0$.

The dynamic of $S_t$ is described by $$\frac{dS_t}{S_t} = \sigma dW_t \iff S_t =S_0\cdot e^{\sigma W_t -\frac{1}{2}\sigma^2t}$$

Using the Proprosition10.4 in Chapter 10, Maximum of Brownian Motion, Privault, the first term of $(1)$ can be computed analytically, by denoting $\mu = -\frac{1}{2}\sigma$. Indeed, we have $$\mathbb{P}(\underset{0 \leq t \leq T}{\max}S_t \ge B) = \mathbb{P}\left(\underset{0 \leq t \leq T}{\max}(W_t-\frac{1}{2}\sigma t) \ge \frac{1}{2}\ln\left(\frac{B}{S_0}\right)\right)$$ and it suffices to apply the formula $(10.13)$

For the second term of $(1)$, we will use the Proprosition 10.3 for example, with $\tilde{W}_T := W_T +\mu T = W_T-\frac{1}{2}\sigma T $ we have: $$\begin{align} \mathbb{E}(S_T\cdot \mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}})=\mathbb{E}\left(S_0\cdot \exp\left(\sigma \tilde{W}_T \right)\cdot \mathbf{1}_{\left\{ \underset{0 \leq t \leq T}{\max}\tilde{W}_t \ge \frac{1}{2}\ln\left(\frac{B}{S_0}\right) \right\}}\right) \tag{2} \end{align}$$ we apply the formula $(10.11)$ and compute numerically $(2)$ with a double integral.

Remark: I'm pretty sure that $(2)$ can be computed analytically with a more elegant method as follows:

  • First, make a change or measure by using the $S_t$-neutral measure, you can eliminate the term $S_T$ in $\mathbb{E}(S_T\cdot \mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}})$ $$\mathbb{E}(S_T\cdot \mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}}) = \mathbb{E}^{\mathbb{Q}_{S}}(\mathbf{1}_{\{ \underset{0 \leq t \leq T}{\max}S_t \ge B \}}) = \mathbb{P}^{\mathbb{Q}_{S}}\left(\underset{0 \leq t \leq T}{\max}\bar{W}_t \ge \frac{1}{2}\ln\left(\frac{B}{S_0}\right) \right)$$
    where $\bar{W}_t: = W_t + \left(\alpha - \frac{1}{2}\sigma \right)t$. I let you find the right $\alpha $ in the new measure $\mathbb{Q}_S$ (I think $\alpha = 1$ but not sure)
  • Second, applying the Proprosition10.4 in Chapter 10, Maximum of Brownian Motion, Privault, and use the same technique for the first term of $(1)$, you deduce directly the closed-form formula of the second term of $(1)$
  • $\begingroup$ Thank you very much! $\endgroup$
    – Don P.
    Sep 9, 2023 at 17:44
  • $\begingroup$ @DonP. you’re welcome! $\endgroup$
    – NN2
    Sep 9, 2023 at 18:30

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