1
$\begingroup$

I would like to write down the PDE for the price of an up-and-in call option under the Black-Scholes model as follows. The payoff of the option at expiry $T$ is

$$C_T := \max(S_T-K,0)1_{M_T \geq L}$$

where $M_t = \sup_{u\leq t}S_u$ and $L > K > 0$. The price of the option at time $t < T$ is given by

$$C_t = e^{-r(T-t)}E[\max(S_T-K,0)1_{M_T \geq L}\mid \mathcal{F}_t]$$ where $\mathcal{F}_t = \sigma(S_u: u\leq t)$. It is a well-known fact that the vector-valued process consisting of Brownian motion and its running maximum is Markov. I am assuming that this applies to geometric Brownian motion and its running maximum as well (I haven't checked this though). If that is the case, then $$C_t = f(t,S_t,M_t)$$ for some measurable function $f$. I don't know whether this function is smooth enough to apply Ito's lemma but again assuming that this is the case one obtains $$dC_t = \left(f_t(t,S_t,M_t) + f_S(t,S_t,M_t)rS_t + \frac{1}{2}f_{SS}(t,S_t,M_t)\sigma^2S_t^2\right)dt + f_M(t,S_t,M_t)dM_t + f_S(t,S_t,M_t)\sigma S_tdW_t^Q$$ The terms with $f_{MM}$ and $f_{MS}$ do not appear because $M$ is a continuous non-decreasing process. So its quadration variation as well as its covariation with $S$ are zero.

If $C$ is a self-financing traded asset, then $e^{-rt}C_t$ must be a martingale. This translates to $$f_t(t,S,M) + f_S(t,S,M)rS + \frac{1}{2}f_{SS}(t,S,M)\sigma^2S^2 = rf(t,S,M)$$ and $$f_M(t,S,M) = 0$$ The latter condition kind of makes sense. If $M \geq L$, then the option has become an ordinary vanilla call so at least on $M \geq L$ the pricing function is constant in $M$. If $M < L$, then the barrier is not reached yet but I am not convinced that on $M < L$ $f$ would be constant in $M$. Furthermore, if it were constant, this would induce discontinuity at $M = L$ and I am not sure if Ito's lemma would be applicable in the first place then.

My question is how can I make this approach to pricing of barrier options work? If that is not possible, then I would like to know why and specifically what it is that I am missing which is blocking this path.

$\endgroup$
  • $\begingroup$ I would start by thinking through your assumptions about $f_{MM}$ and $f_{MS}$, they look way too casual to me. $\endgroup$ – Ivan Apr 1 '18 at 21:38
  • $\begingroup$ @Ivan Thanks for your input but that is actually not an assumption. Since $[M,M]$ and $[M,S]$ are both zero (not by assumption but as a fact), it does not matter what those derivatives are. $\endgroup$ – Calculon Apr 1 '18 at 21:44
  • 4
    $\begingroup$ The valuation function does not explicitly depend on the running maximum $M_t$ as long as $M_t < L$ but if that condition is not satisfied then the option worthless anyways. So you compute the current value conditional on no prior breach of the barrier and this value does not depend on $M_t$ but only $S_t$. The option price satisfies the standard Black-Scholes PDE but with an additional upper boundary condition. Barrier options are thus often referred to as weakly path-dependent. This is different from e.g. an Asian option which has a strong path dependence. $\endgroup$ – LocalVolatility Apr 3 '18 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.