I am trying to work out the PnL of continuous delta hedging. I saw This link to an answer here, however, I obtained a different answer without resorting to Black Scholes, which I will outline below.

Lets assume a constant $\Gamma$. We take the time interval $[0,1]$, and divide it into the discrete points $i/n$ for $i=0,\ldots, n$. We enter a position in some option at time $0$, and initially hedge by shorting $\Delta_0$ of the underlying, where $\Delta_0$ is the delta of the option at time $0$. Denote by $P$ the spot price of the underlying. At time $ 1/n$, the delta of the option moves to $\Delta_0 + \Gamma (P_{1/n} - P_0)$, so we short an additional $\Gamma (P_{1/n} - P_0)$ of the underlying, this time at price $P_{1/n}$, so we spend $\Gamma P_{1/n} (P_{1/n} - P_0)$. We continue this process until time $ 1$, at which time we leave the position by shorting $-\Delta_1$ of the underlying. Our total spending amounts to

$$ \Delta_0 P_0 + \sum_{i=1}^{n-1} \Gamma P_{i/n} ( P_{i/n} - P_{(i-1)/n}) - \Delta_1P_1.$$

Now, $\Delta_1 = \Delta_0 + \Gamma (P_1-P_0)$ so that \begin{align} \Delta_0 P_0 - \Delta_1P_1 &= \Delta_0P_0 - (\Delta_0 +\Gamma (P_1- P_0))(P_0 + (P_1-P_0))\\ &= - \Delta_0 (P_1-P_0) -P_0\Gamma(P_1 - P_0) - \Gamma (P_1-P_0)^2\\ &= -\Gamma (P_1-P_0)^2 -(\Delta_0 + P_0 \Gamma)(P_1-P_0) \end{align}

Letting $n$ become large in our expression, and multiplying by $-1$ to obtain PnL we have

$$ PnL = \Gamma (P_1-P_0)^2 + (\Delta_0 + P_0 \Gamma)(P_1-P_0) + \int_0^1 \Gamma P (t) \mathrm{d} P(t). $$

So is this something that you have seen before? Or did I make some mistakes along the way? And does this somehow lead to what is in the link? As it stands I see no obvious connection to any implied or realized volatility? Thanks

Edit: I think maybe the quantity being evaluated in the link I added measures something else. I am specifically trying to evaluate the additional PnL obtained when hedging to stay delta neutral during the holding time of an option.



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