# 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?

• We tried that in the past. In reality any possible profit will be lost between spreads and fees – Ariel Silahian Oct 26 '20 at 1:11
• In a transactions-free world it would be possible though, as you indirectly imply. But I fail to demonstrate that mathematically. – Giogre Oct 26 '20 at 1:18
• Locally (in the underlying space), you are reasoning seems correct and is solely driven by the fact that your portfolio is exposed to gamma (positive) and vega (positive). Yet the time effect is missing, which is negative. – Kermittfrog Oct 26 '20 at 9:32
• Excuse my ignorance, but isn't it sufficient to show that $0<d\Pi=dC-\Delta dS$ which is effectively $C_tdt+\Delta dS - \Delta dS + \frac{1}{2}\Gamma dS^2 + C_vdv$ which is $C_tdt + \frac{1}{2}\Gamma dS^2 + C_vdv$ which will be positive for most values of $dv$, no? – Kermittfrog Oct 26 '20 at 14:26
• Effectively, you own a gamma long portfolio, which is always bound to pay off. – Kermittfrog Oct 26 '20 at 14:54