Skip to content
All library documents

Replicating a Path-Dependent Payoff in Black–Scholes

Article Quant Q&A · Author: user47393

Summary

The document poses a derivative-replication exercise in a Black–Scholes market. Its payoff combines a time integral of the log price relative to its initial value with terms depending on the terminal log price, including a squared term. The question asks for a replicating strategy and notes that the payoff is not solely a function of the terminal asset price, unlike many standard examples.

The prompt suggests checking square integrability under the risk-neutral measure as a route to establishing replicability, but it does not provide a proof, a strategy, or supporting calculations. The key learning challenge is therefore how to handle the path-dependent integral and derive a hedge, rather than apply a terminal-payoff formula directly. The document is an exercise statement; it does not establish the proposed integrability condition or explain the assumptions needed for replication.

Key ideas

  • The payoff depends on the price path through an integral as well as on the terminal price.
  • A payoff with path dependence cannot be handled directly as a function of the terminal asset price alone.
  • The exercise asks whether risk-neutral square integrability supports replication and how to derive the hedge.
  • No proof, hedge strategy, or verification of the stated condition is included.

Tags

Full text
# Black-Scholes market and payoff with integrals


# Black-Scholes market and payoff with integrals












I am struggling with the following exercise:

> Prove that on Black-Scholes market, with some parameters $r, \mu, \sigma >0$, a payoff $$X=\int_{0}^{T}\ln \frac{S_t}{S_0}\mathrm{d}t+\frac{1}{\sigma}\Big(\ln^2\frac{S_T}{S_0} + (\sigma^2-2\mu)T\ln\frac{S_T}{S_0}\Big),$$ is replicable. Find the appropriate strategy.

I know that if $ \ \mathbb{E}_{Q}X^2<\infty, \ $ then $X$ is replicable. I guess that this can be easily checked here. But I do not know how to look for replication strategy. This $X$ is not in the form of $X=f(S_T)$, which I have seen in some materials. I will be grateful for any help.

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.