Valuing an Up-and-In Forward Option with a Barrier Event
Summary
The question asks whether Black–Scholes provides an analytic value for a payoff that activates when the underlying first reaches an upper barrier and then depends on the terminal asset price. The answer observes that the barrier event is equivalent to the running maximum reaching the barrier, and that the asset value at the first hit is the barrier itself. With the interest rate set to zero for simplicity, it decomposes expected payoff into the barrier level multiplied by the probability of a hit, minus the expected terminal asset value conditional on a hit.
The response relates the hit probability to the maximum of a Brownian motion with drift, citing formulas for that maximum. It expresses the second expectation through a shifted Brownian motion and suggests numerical evaluation by a double integral. It then proposes a change of measure that may turn the expectation involving the terminal asset value into a barrier hitting probability under a stock numeraire, but leaves a parameter unresolved. Thus, the answer outlines a route to valuation rather than presenting a verified closed-form result; discounting and general interest rates are not worked through.
Key ideas
- The payoff activates when the asset's running maximum reaches the upper barrier before maturity.
- At the first barrier hit, the asset price equals the barrier, simplifying the payoff decomposition.
- The expected payoff can be expressed using a barrier hitting probability and a terminal-price expectation on the hit event.
- A change of measure may simplify the terminal-price expectation, but the response leaves the derivation incomplete.
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Full text
# Hedge up-knock-in forward option
# 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.
## Answer by NN2 (score 1, accepted)
https://quant.stackexchange.com/a/76541
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 $(1)$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.