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This question comes from the book of Sheldon Natenberg's book "Option Volatility and Pricing: Advanced Trading Strategies" 2nd.

In chapter 8 "Dynamic Hedging" page 129, it says: In theory, if we ignore interest, the sum of all these small profits (the unhedged amounts in Figure 8-3) should approximately equal the value of the option. and states the equation as

Option theoretical value = {.} + ... {.} where {.} is unhedged amount.

My question is how to understand what he says and how does this come out? What he says means the optional value = sum(gamma)? because the unhedged amount seems to be gamma

Hope someone who read the book answers this, thanks so much!!

yles will
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  • To be precise you should say 'option value = sum(gamma trading P&L)'. Each of those little profits and losses are not equal to $\Gamma$ but rather $\frac{1}{2}\Gamma \cdot dS^2$ where dS is the (random) change in stock price. With this minor change in wording, the statement you made is correct. HTH. – nbbo2 Nov 01 '21 at 17:53
  • Yep, thanks so much! I agree with you but I am still confused about how the sum of little profits equals option value. By using the formula it means $\int \frac{1}{2}\Gamma (dS)^2=option value$ or makes it like this using Ito $\int \frac{1}{2}\Gamma \sigma^2 S^2 dt=option value$, which means $\int \frac{1}{2}\Gamma \sigma^2 S^2 dt = SN(d_1) - Ke^{-r(T-t)}N(d_2)$. I just don't get how this formula come out, could you explain this? – yles will Nov 02 '21 at 01:56

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