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Using Images to Price Double-Barrier Knock-Out Options

Article Quant Q&A · Author: joesyc
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Summary

The document asks whether a double-barrier knock-out call can be assembled from an up-and-out option and a down-and-out option. The accepted response says this decomposition does not work: satisfying one barrier condition does not by itself enforce both absorbing boundaries. It points instead to the method of images as an approach for valuing double-barrier options, including cases with exponential boundaries and arbitrary payoffs.

Key ideas

  • A double-barrier knock-out option cannot generally be priced by combining single-barrier knock-out options.
  • The price must satisfy zero-value conditions at both barrier levels.
  • The method of images is identified as a framework for double-barrier valuation.
  • The cited references cover exponential boundaries, with straight barriers described as a special case.
  • The excerpt gives references rather than a full derivation or worked pricing formula.

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OptionsDerivatives pricing
Full text
# Derivation using Ito's Lemma of price process


# Derivation using Ito's Lemma of price process












Define $q(t)$ as the log price minus a linear trend

$$ q(t) = \ln P(t) - \mu t $$

Assume the log price process = Equation 1: $$ dq(t) = - \Theta q(t) dt + \sigma dW(t) $$

Can you show that the solution to Equation 1 is: $$ \ln P(t+h) - \ln P(t) = \mu h + (\exp(-h \Theta) - 1) \ln P(t) + \sigma \int_t^{t+h} \exp(-\Theta(t-u))dW_u $$

## Answer by user16891 (score 1)

https://quant.stackexchange.com/a/19134

by application of Ito's lemma , we have $$d\left(q(t)e^{\Theta\,t}\right)=\Theta \,q(t)e^{\Theta\,t}dt+e^{\Theta\,t}dq(t)+0$$ then $$d\left(q(t)e^{\Theta\,t}\right)=\sigma e^{\Theta\,t}dW_t$$ in other words $$q(t+h)e^{\Theta\,(t+h)}-q(t)e^{\Theta\,t}=\sigma\int_{t}^{t+h}e^{\Theta\,u}dW_u\Rightarrow$$ $$q(t+h)-q(t)=\left(e^{-h\Theta}-1\right)q(t)+\sigma\int_{t}^{t+h}e^{-\Theta\,(t+h-u)}dW_u$$ By substituting $\ln P(t)-\mu\,t$ to last equation , we have $$ \ln P(t+h) -\ln P(t) =\mu h+\left(e^{-h\Theta}-1\right)(\ln P(t)-\mu t) + \sigma\int_{t}^{t+h}e^{-\Theta\,(t+h-u)}dW_u $$

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