path dependency and dollar gamma

Question

On a previous question on this website, a user derived the following PnL of a delta-hedged option: $$P\&L_{[0,T]} = \int_0^T \frac{1}{2} \underbrace{\Gamma(t,S_t,\sigma^2_{t,\text{impl.}})S_t^2}_{\text{Gamma dollar}}( \sigma^2_{t,\text{real.}} - \sigma^2_{t,\text{impl.}}) dt$$

Indeed, taking the derivative with respect to time, we obtain that the daily PnL of a position is therefore $$\frac{1}{2} \Gamma(t,S_t,\sigma^2_{t,\text{impl.}})S_t^2( \sigma^2_{t,\text{real.}} - \sigma^2_{t,\text{impl.}}),$$ and hence the daily profit and loss is weighted down by a factor of the gamma dollar.

Therefore, at expiration, the PnL of the position might be positive or negative, independent of the final realized volatility versus implied volatility, since given the value at time $t=T$, it may be the case that we can be dragged down by all preceeding values from time $t=0$ to $t=T-\epsilon$ dependent on the gamma dollar.

This, however, leaves me somewhat disgruntled, because I feel like I am missing something important. When learning about options, we are told that the act of "delta-hedging" removes "path-dependency", i.e. makes it a pure volatility play. This, however, is clearly false, because we have the dollar gamma term there! Why are we taught this? It is not even close to negligible.

Secondly, because my intuition is so bad in this circumstance, I can't exactly think of a time where this happens. Let us say that I sell an options contract at an implied volatility of 45%, and subsequently the stock realizes a volatility of 42%, and also, assume I'm continuously delta-hedging. By the above, it is impossible to tell whether I am profitable or not, based on this information alone. Under what circumstances am I profitable? Must the stock be realizing a 42% volatility continuously throughout it, i.e. there cannot be a type of discontinuous jump -- is that what this is saying, in mathematical terms? (i.e. low vol -> really high vol suddenly? avg. realized may be 42%, but I may be losing due to that spike?)

Answer 1

Lorenzo Bergomi's book was very instructive here; as follows is me stealing his great intellectual contribution. The first chapter solves many of my quarrels, and in particular Bergomi shows that, given some relatively "elementary" assumptions, $$P\&L = -\frac{S^2}{2} \frac{d^2 P}{d S^2} \left( \frac{\delta S}{S} - \frac{\hat{\sigma}^2 \delta t}{S^2} \right)$$ is the P&L of a short delta-hedged option position. He goes on to state, quite explicitly, that the "P&L will be positive or negative depending upon whether $\frac{\delta S^2}{S^2}$ is larger or smaller than $\hat{\sigma}^2 \delta t.$"

As I initially thought, the entire point of delta-hedging is to remove the exposure from the underlying movement; it does do this, attempting to remove the linear term $\delta S$, as one expects. Indeed, Bergomi mentions how, if delta hedging is exactly continuous (i.e. $\delta t \to 0$), the sum of the terms in our expression "vanishes with probability $1$."

More poignant, however, is the fact that our processes are not exactly lognormal; nor is hedging on a continuous basis actually palpable. Thus, although in theory it might be possible to hedge continuously in that manner, such that the sum of terms vanishes, in practice it is not. There is further instruction in his first chapter.

On that note, if anyone would like to contribute a more intuitive explanation, with less discussion of the internal machinery, that would possibly be even more apt.

Answer 2

What you have in your portfolio is an incorrectly valued instrument. It is misleading to say it is an option, because the option that we understand is that which is priced correctly. When the PnL above is positive, you overvalued it and if it is negative, you undervalued it.

If you don't price it correctly, you are of course going to be exposed to randomness.

You can say an option is characterized by the contract, so why the problem in saying this is an option, but I would say an option is characterized by the mapping from $S(t)->C(t,S(t))$. This is a different mapping, so you should not expect it to behave like an option.