July 20th, 2017, 3:34 pm

Suppose you buy a call and delta hedge it (by selling the underlying) and put the proceeds of long call and short stock on a bank account.

Your initial portfolio is then

[$] \Pi = C - \Delta S + (\Delta S - C) = 0 [$]

If you only delta hedge and keep other variables/parameters constant, then the change in your portfolio value (marked-to-model) after a time dt is

[$] d \Pi = \Theta dt + \frac{1}{2} \Gamma dS^2 + r (\Delta S - C)dt [$]

Assuming you canwrite [$] dS^2 = \sigma^2 S^2 dt [$] (no jumps_, then the break-even move is that value of [$] dS^2 [$] such that [$] d\Pi =0 [$]. In other words when

[$] \Theta = - \frac{1}{2} \Gamma \sigma^2 S^2 - r (\Delta S - C) [$]

Lo and behold, this is the Black-Scholes equation... But not really, since the BS equation tells you, assuming you know the vol to start with, how to solve for your option price and greeks, and this equation tells you if you know your gamma and delta and theta (from a mis-speicfied Black Scholes model) what should be the vol / daily spot move to balance - which is obviously the vol you used initially for solving Black-Scholes. The next question should be so what happens if my daily move is not my implied volatility.

Anyway, I hope you're really confused now and decide to explore further what happens if you hedge with a mis-specified model.

To summarize, assumptions to arrive at the expression in the link (which oversimplifies things):

- you're only letting [$]S[$] vary and (hence) only delta hedging

- r = 0