0
$\begingroup$

I have a question about the following derivation in this pdf (sample chapter from Bergomi - Stochastic Volatility Modeling). He derives the PnL for a delta hedged position as

$$PnL = -[P(t+\delta,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$ where $r$ is the interest rate, $q$ the repo rate including dividend yield and $P(t,S)$ denotes the price of the option with underlying $S$ at time $t$. He chooses $\Delta = \frac{\partial P}{\partial S}$. Then he wants to expand the PnL in powers of $\delta S$ and $\delta t$ just looking at $\delta t, \delta S$ and $\delta S \delta S$ terms. His result is

$$ PnL = -(\frac{dP}{dt}- rP +(r-q)S \frac{dP}{dS})\delta t - \frac{1}{2}S^2\frac{d^2P}{dS^2}(\frac{\delta S}{S})^2$$

I'm not sure how he gets that one. If I expand $P$ in the terms mentioned above I find

$$P(t+\delta,S+\delta S) = P(t,S) + \frac{dP}{dt}\delta t + \frac{dP}{dS}\delta S + \frac{1}{2}*\frac{d^2P}{dS^2}(\delta S)^2$$

Doing this for the three terms of $P$ above I dont get the correct result. So how do you get his formula

$\endgroup$
  • $\begingroup$ Well there you go for the $P(t,S) - P(t+\delta,S+\delta S)$ par of the $PnL$ but now you have to add the remaining part (see your first equation), with $\Delta = \partial P/\partial S$. It's as easy as that $\endgroup$ – Quantuple Nov 18 '16 at 15:36
  • $\begingroup$ I'm not even sure this question should remain open actually. $\endgroup$ – Quantuple Nov 18 '16 at 15:53
5
$\begingroup$

$\require{cancel}$ $$\text{PnL} = -[P(t+\delta t,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$ Assuming a pure diffusion, at the order 1 as $\delta t \to 0$ $$P(t+\delta,S+\delta S) = P(t,S) + \frac{\partial P}{\partial t}\delta t + \frac{\partial P}{\partial S}\delta S + \frac{1}{2}\frac{\partial^2P}{\partial S^2}(\delta S)^2$$ Plugging that back in the first equation and using the identity $\Delta = \frac{\partial P}{\partial S}$ gives: \begin{align} \text{PnL} &= -\frac{\partial P}{\partial t}\delta t \cancel{- \frac{\partial P}{\partial S}\delta S} - \frac{1}{2} S^2 \frac{\partial^2P}{\partial S^2}\left(\frac{\delta S}{S}\right)^2 + rP\delta t + \cancel{\Delta\delta S} - (r - q) S \Delta \delta t \\ &= -\left( \frac{\partial P}{\partial t} - rP + (r - q) S \frac{\partial P}{\partial S} \right) \delta t + \frac{1}{2} S^2 \frac{\partial^2P}{\partial S^2}\left(\frac{\delta S}{S}\right)^2 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.