I'm running through some delta hedging simulations of floating strike lookback call options (that is, I'm short the options) during a volatile (downside) period for the underlying and some very odd things have resulted.

First of all, these options have no Gamma, at least not by the standard understanding of gamma being $$\frac{\partial^2 C}{\partial S^2}$$

I've obtained Gamma values both by a finite difference approximation and by partially differentiating the analytical solution (messy!), and both come to 0. However, the delta values are very volatile, and the hedge is getting crushed. So, a few thoughts I had from drawing on analogies from vanilla calls.

1) If call deltas are known to underestimate the price increase in up moves and overestimate the loss in down moves, then being short delta in a down market should be good.

2) Again, there is no Gamma, at least by the "traditional" measure, or whatever you want to call it. However, in a volatile market, being short gamma (by being short either a call or put) is bad.

3) For vanillas, we know that Gamma is high for at-the-money options.

Now, this is where things get really weird in my head. Given that, as mentioned, I'm hedging short calls in a down market, the structure of the floating strike lookback option means that basically the strike is getting continually reset to the current value of the underlying. That is, the options are basically always at-the-money during this period. So, I have options with a volatile delta that are at-the-money. So in effect, the hedge is suffering for the same reason that a short call hedge might suffer by being short Gamma, yet I'm not short Gamma...!?

Can anyone clarify this for me, and perhaps lead me to some literature about hedging floating lookback options? It seems to be unfortunately sparse.

Thank you.


1 Answer 1


The floating strike lookback call options has zero gamma only on the day it is issued (and only assuming an homogeneous model for the underlying). Afterwards it has non zero gamma.

The payoff on maturity $T$ is: $$ \text{payoff} = S_T - \min \{S_u | u \in [0, T]\} $$ Now assume your model for the underlying is homogeneous with degree 1, that is when viewed from $t$, $S_u$ for $u \geq t$ is proportional to $S_t$. This is of course the case when the model is a geometric Brownian motion as in Black & Scholes.

On $t=0$, $$ E_0[\text{payoff}] = E_0[S_T] - E_0[\min \{S_u | u \in [0, T]\}] = S_0 E_0[S_T/S_0] - S_0 E_0[\min \{S_u/S_0 | u \in [0, T]\}] $$ Since $S_T/S_0$ and $S_u/S_0$ do not depend on $S_0$, $E_0[\text{payoff}]$ is linear in $S_0$ and the option has zero gamma.

On $t>0$, $$ E_t[\text{payoff}] = E_t[S_T] - E_t[\min \{S_u | u \in [0, T]\}] = E_t[S_T]- E_t[\min\{m_t, \min \{S_u | u \in [t, T]\}\}] $$ where the running minimum $m_t = \min\{S_u | u \in [0, t]\}$ is already known. Now $$ E_t[\text{payoff}] = S_t E_t[S_T/S_t] - S_t E_t[\min\{m_t/S_t, \min \{S_u/S_t | u \in [t, T]\}\}] $$ As you can see the second term on the RHS is no longer proportional to $S_t$ because $m_t/S_t$ depends on $S_t$. Therefore $E_t[\text{payoff}]$ is no longer linear in $S_t$ and the option has non zero gamma.

In practical terms that means that you should view the option price as a function of both $S_t$ and $m_t$, that is $C(S, m, t)$. When you compute the gamma you compute $\frac{\partial^2 C}{\partial S^2}(S, m, t)$. When you approximate gamma using finite differences you compute $$ (C(S+\epsilon, m, t) + C(S-\epsilon, m, t) - 2 C(S, m, t))/\epsilon^2 $$ that is you move $S$ by $\pm \epsilon$ but you do not change $m$.

  • $\begingroup$ Thank you, I understand now. In my case, of course it turns out I had simply messed up a formula...but it's nice to know I can develop logic that defends my mistakes :) $\endgroup$
    – Archetupon
    Dec 7, 2017 at 14:58

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