0
$\begingroup$

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.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.