# Lookback option with floating strike: boundary condition

I am trying to make sense of one of the boundary conditions of a look-back option with floating strike.

Some notation first: let $v(t,x,y)$ denote the price at time $t$ of the option under the assumption that $S(t) = x$ and $Y(t) = y$, such that $Y(t) = max_{0\leq u \leq t} S(u)$, $S(t)$ being the geometric Brownian Motion asset price.

When computing the discounted value of the option, we get the usual Black-Scholes Merton equation satisfying: $$v_t(t,x,y) + rxv_x(t,x,y) + \frac{1}{2}\sigma^2x^2v_{xx}(t,x,y) = rv(t,x,y)$$

The new feature in this derivation is that we obtain, additionally, the term $$e^{-rt}v_y(t,S(t),Y(t))dY(t)$$

which must be $0$ so that $d\left(e^{-rt}v(t,S(t),Y(t))\right)$ is a martingale.

The book I am using as reference (Stochastic Calculus for Finance, by Shreve) states that the term $dY(t)$ is $0$ on the flat spots (which is obvious), and that "the times when $S(t) = Y(t)$ , the term $e^{-rt}v_y(t,S(t),Y(t))dY(t)$ must also be $0$ because $dY(t)$ is positive".

Could somebody explain better this last statement? I really do not see the reason why this is true. This should conclude that one of the boundary conditions is that $v_y(t,y,y) = 0$.

• I think this is the intuitive idea: Let say $S(t) = 100$ and $Y(t) = 100$. Now $S(t)$ goes to $101$. What happens to your option price? $Y(t)$ also goes to $101$, so it should leave your option price unchanged. Jun 2 '14 at 20:11

As mentioned in the question, we want $e^{-rt}v_y(t,x,y)dY_t$ to be equal to $0$ in order to have a martingale.
When $Y_t$ is a flat line, the term $dY_t$ is equal to $0$ (this was already stated in the question). Now, when $S_t$ increases over the current maximum, the variable $Y_t$ increases together with it, making the term $dY_t$ different than $0$. This means that in order to obtain a martingale, $v_y(t,x,y)$ must be $0$, and because $S_t = Y_t$, we must have $v_y(t,y,y) = 0$, yielding the desired boundary condition.