Why Delta Hedging Does Not Guarantee Profit in a Market Crash
Summary
The document examines a long call hedged by shorting stock in an amount equal to the option’s initial delta. It considers a sudden underlying-price decline alongside an increase in implied volatility, and asks whether the hedge must make money. A Taylor expansion is presented to separate the option’s local exposure to price changes, curvature through gamma, and volatility sensitivity through vega. Since the initial stock hedge offsets the first-order price term, the question centers on whether positive gamma and vega ensure that the remaining portfolio change is positive.
The document does not include an answer, proof, or empirical example. Its setup is useful for studying local option sensitivities, but the expansion is an approximation and does not establish a guaranteed outcome for a large jump. The result depends on the option’s repricing, the magnitude and path of the price and volatility moves, and model assumptions; delta hedging alone does not imply certain profit under crash conditions.
Key ideas
- An initial delta hedge offsets the option’s first-order sensitivity to a small underlying-price move.
- Gamma and vega describe curvature exposure and sensitivity to implied volatility.
- The question considers a large downward jump together with a rise in volatility.
- A local Taylor approximation does not prove that a hedged position is always profitable after a large move.
- The document poses the issue but supplies no resolution or evidence.
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Full text
# Why is a Delta-hedged option always profitable even in case of a sharp drop of value of the underlying?
# Why is a Delta-hedged option always profitable even in case of a sharp drop of value of the underlying?
I am trying to understand the following concept on a practical level.
Given a Delta-hedged long call position, so holding a portfolio
$$ Port_0 = C(S_0, \sigma) - \Delta_C(S_0) \, S_0 $$
If there is a market crash (so, a jump), and the value of the underlying $S$ drops by say, $30\%$ ($dS = S - S_0 = -0.3 \, S_0$), while the volatility $\sigma$ increases, would the above portfolio $Port$ end up making money?
I understand that the answer is yes. Taylor decomposition for the option implies:
$$ C(S, \sigma) = C(S_0, \sigma) + \frac{\partial C}{\partial S}(S_0) dS + \frac{\partial^2 C}{2 \partial S^2}(S_0) dS^2 + \frac{\partial C}{\partial \sigma}(S_0) d \sigma + O(dS^3, d \sigma^2) = \\ \simeq C(S_0, \sigma) + \Delta_C(S_0) dS + \frac{1}{2} \Gamma_C(S_0) dS^2 + \nu_C(S_0) d \sigma $$
so the portfolio after the crash becomes:
$$ Port = C(S, \sigma) - \Delta_C(S_0) \, S =\\ = C(S_0, \sigma) + \frac{1}{2} \Gamma_C(S_0) dS^2 + \nu_C(S_0) d \sigma - \Delta_C(S_0) \, S_0 $$
Given that we know that $\Delta_C, \, \Gamma_C, \, \nu_C \geq 0$, how can it be proven that $Port > Port_0$ ? That is to say, why the $(dPort \, , dS)$ curve is a convex function, and always $>0$ (see figure)?
Intuitively, to get a grip of the issue, after the crash the shorted, underlying-stock part of the portfolio ($- \Delta_C(S_0) \, dS$) should realize a gain, while the call option should drift towards being out-of-money, even while considering a positive influx from the volatility term, so its value should diminish, $dC < 0$.
Why can we say with certainty that under these conditions ($dS < 0, d \sigma > 0$) the gain from the stock part of the portfolio dominates the loss in option value?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.