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Solving a Mean-Reverting Log-Price Process with Ito's Lemma

Article Quant Q&A · Author: A.Oreo

Summary

The document derives the finite-horizon change in log price when log price, after subtracting a linear time trend, follows an Ornstein–Uhlenbeck process. The drift pulls the detrended log price toward zero at a rate set by the mean-reversion parameter, while Brownian noise enters with constant volatility. The derivation applies an integrating factor, multiplying the process by an exponential in time, then integrates the resulting stochastic differential equation.

Key ideas

  • Subtracting a linear trend from log price defines the state variable that follows a mean-reverting process.
  • Multiplication by an exponential integrating factor removes the mean-reverting drift.
  • Integrating the transformed equation gives the state at a future time as a decayed initial value plus a stochastic integral.
  • Substituting the detrended log price back yields the log-price change and its trend contribution.
  • The result depends on the stated mean-reversion model and does not establish that market prices follow it.

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Full text
# How to use reflection principle to solve the analytic solution of double barrier-out-call


# How to use reflection principle to solve the analytic solution of double barrier-out-call












We consider `up/down-out-call` whose payment $$V(T,S_T) = \Psi(S_T)\mathbb{II}(S_T),\ V(t,B) = 0.$$ Here the range constraint function is `indictor function` such that $\mathbb{II}(S_T)$ = \begin{cases} \mathbb{II}_{\{S_T < B\}} & \textrm{up-out-call}\\ \mathbb{II}_{\{S_T > B\}} & \textrm{down-out-call}\\ \end{cases} We see that the only difference between the `barrier option` and `vanilla option` is the barrier condition: $V(t,B) = 0,$ so we can suppose $$V(t,S) = \widehat V(t,S) - \widetilde V(t,S)$$ s.t $$\widehat V(T,S) = \Psi(S_T),\quad \widetilde V(T,S) = 0$$ $$\widehat V(t,B) = \widetilde V(t,B)$$ And use the `reflection principle` $$\widetilde V(t,S) = \left(\dfrac{S}{B}\right)^{2\alpha}\widehat V(t,\dfrac{B^2}{S})$$ Since if $\mathbb{II}(S)$ is not zero, $\mathbb{II}(\dfrac{B^2}{S})$ must be zero, then $\widetilde V(T,S) = 0.$

But how to use this method to deal with the `double barrier-out-call` i.e $$V(t,B_1) = V(t,B_2) = 0,\quad B_1 < B_2$$ or is it possible to use `down-out-call` with barrier $B_1$ and `up-out-call` with barrier $B_2$ to construct double barrier?

## Answer by LocalVolatility (score 3, accepted)

https://quant.stackexchange.com/a/34548

No, the pricing of a double barrier knock-out option cannot be decomposed into single barrier options.

Here are a few references that apply the method of images to the valuation of double barrier options:

- A very clear and easy to follow exposition can be found in Chapter 3.5 of the Ph.D. thesis by Konstandatos (2003).

- If you don't have access to that, then I believe it is also reproduced in the author's book Konstandatos (2008).

- Another reference is the paper Buchen and Konstandatos (2008). Here, they consider exponential barriers. You seem to be interested in the special case when the exponential "bending" is zero.

References

Buchen, Peter W. and Otto Konstandatos (2009) "A New Approach to Pricing Double-Barrier Options with Arbitrary Payoffs and Exponential Boundaries," Applied Mathematical Finance, Vol. 16, No. 6, pp. 245-259

Konstandatos, Otto (2003) "A New Framework for Pricing Barrier and Lookback Options," Ph.D. Thesis, University of Sydney

Konstandatos, Otto (2008) Pricing Path Dependent Exotic Options: VDM Verlag Dr. Müller

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.