# Hedge up-knock-in forward option

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.

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 $$(1)$$
• Thank you very much! Sep 9 at 17:44
• @DonP. you’re welcome!
– NN2
Sep 9 at 18:30