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When we buy a call and continuously delta hedge using some implied volatility $\sigma_i$, what is the formula for our aggregate profit given that the actual realized volatility is $\sigma_r$?

Say $S_0 = 1000, \sigma_i = 0.25, \mu = 0.10$, and the call has expiry a year from now.

How does the formula look like in terms of $\sigma_r$? What happens if $\sigma_r =0$? $> \sigma_i$? $< \sigma_i$?

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    $\begingroup$ See this answer quant.stackexchange.com/questions/33205/… for the daily P&L. For the overall P&L you integrate this from 0 to T. $\endgroup$ – noob2 Mar 30 '17 at 17:57
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    $\begingroup$ Chapter 12 "How to Delta Hedge" in Wilmott's "Paul Wilmott on Quantitative Finance" discusses this problem in detail. In particular it shows how your hedging p&l differs when using either the implied or the estimated (true) volatility for computing the hedge ratio. $\endgroup$ – LocalVolatility Mar 30 '17 at 19:46
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    $\begingroup$ Possible duplicate of : quant.stackexchange.com/questions/33345/…. @LocalVolatility maybe you should write this as an answer so we can indeed mark these as duplicates? $\endgroup$ – Quantuple Mar 31 '17 at 7:45
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This is a slightly extended version of my comment that summarizes the main result of the reference that I provided.

This problem is discussed in detail in Chapter 12 of Wilmott (2006), which is based on the paper Ahmad and Wilmott (2005). See also the related Question 9 in Carr (2005).

In your case, you are selling and delta hedging the option using the implied volatility $\sigma_{(i)}$ while the actual volatility of the underlying asset is $\sigma_{(r)}$. Your portfolio profit and loss is given by the sum of value changes of (i) the derivative and (ii) the hedging position, both using the implied volatility. We have

\begin{equation} \mathrm{d} \Pi_t = \mathrm{d}V_t^{(i)} - \Delta_t^{(i)} \mathrm{d}S_t - r \left( V_t^{(i)} - \Delta_t^{(i)} S_t \right) \mathrm{d}t. \end{equation}

Since

\begin{equation} \mathrm{d}V_t^{(i)} = \frac{\partial V^{(i)}}{\partial t} \mathrm{d}t + \underbrace{\frac{\partial V^{(i)}}{\partial S}}_{= \Delta^{(i)}} \mathrm{d}S_t + \frac{1}{2} \underbrace{\frac{\partial^2 V^{(i)}}{\partial S}}_{=\Gamma^{(i)}} \mathrm{d} \langle S \rangle_t, \end{equation}

we get

\begin{equation} \mathrm{d}\Pi_t = \left( \frac{\partial V^{(i)}}{\partial t} + \frac{1}{2} \sigma_{(r)}^2 S_t^2 \frac{\partial^2 V^{(i)}}{\partial S^2} - r \left( V_t^{(i)} - \Delta_t^{(i)} S_t \right) \right) \mathrm{d}t. \end{equation}

Now we use that $V^{(i)}$ satisfies the Black-Scholes PDE

\begin{equation} \frac{\partial V^{(i)}}{\partial t} + r S_t \underbrace{\frac{\partial V^{(i)}}{\partial S}}_{=\Delta^{(i)}} + \frac{1}{2} \sigma_{(i)}^2 S_t^2 \underbrace{\frac{\partial^2 V^{(i)}}{\partial S^2}}_{=\Gamma^{(i)}} - r V^{(i)} = 0 \end{equation}

to obtain

\begin{equation} \mathrm{d}\Pi_t = \frac{1}{2} \left( \sigma_{(r)}^2 - \sigma_{(i)}^2 \right) S_t^2 \Gamma^{(i)} \mathrm{d}t. \end{equation}

I.e. over each short interval, your profit and loss is proportional to the difference in realized to implied variance times the current gamma (computed using the implied volatility). While $\mathrm{d}\Pi_t$ is deterministic, the overall hedging profit and loss over the lifetime of the option is path-dependent. Its absolute value is higher for paths that fluctuate around the strike (where the gamma is higher).

References

Ahmad, Riaz and Paul Wilmott (2005) "Which Free Lunch Would You Like Today, Sir? Delta Hedging, Volatility Arbitrage and Optimal Portfolios," Wilmott Magazine, available here

Carr, Peter (2005) "FAQs in Option Pricing Theory", Working Paper, available here

Wilmott, Paul (2006) Paul Wilmott on Quantitative Finance, Vol. 1: Wiley, 2nd Edition

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