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Why Floating-Strike Lookback Options Gain Gamma After Inception

Article Quant Q&A · Author: Archetupon

Summary

The discussion explains why a floating-strike lookback call can have zero gamma at inception under a homogeneous underlying model, yet develop nonzero gamma later. Its payoff depends on the terminal price relative to the minimum price reached over the option’s life. At inception, scaling the underlying scales the entire price path and the option value proportionally, making the value linear in the spot price.

After inception, the running minimum is already known and does not scale with a change in current spot. The option value therefore depends on both spot and the recorded minimum, so it is no longer linear in spot. Gamma should be computed as the second derivative with respect to spot while holding the running minimum fixed; the suggested finite-difference method does the same. This distinction addresses the reported hedge behavior, though the zero-gamma-at-inception result assumes a degree-one homogeneous model, such as geometric Brownian motion.

Key ideas

  • At inception, a homogeneous underlying model makes the lookback call value linear in spot and its gamma zero.
  • After inception, the running minimum is fixed history, so the option value is generally nonlinear in current spot.
  • The option price should be treated as a function of spot, running minimum, and time.
  • When estimating gamma by finite differences, vary spot while holding the running minimum constant.

Tags

Full text
# Floating Strike Lookback Delta Risk


# Floating Strike Lookback Delta Risk












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.

## Answer by Antoine Conze (score 4, accepted)

https://quant.stackexchange.com/a/37267

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 $m$.

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