Hedging error in a stochastic volatility model

Question

I would like to find how much error I make when I hedge a call option using Black Scholes model in a market which is actually governed by a stochastic volatility process such as $$dS_t = rS_tdt + \sigma_t S_tdW_t^Q$$ where $(\sigma_t)_{t\geq 0}$ is some stochastic process and $(W_t^Q)_{t\geq 0}$ is Brownian motion under the risk neutral measure. Just for the sake of clarity the Black Scholes model is given by $$dS_t = rS_tdt + \sigma_{BS} S_tdW_t^Q$$ where $\sigma_{BS}$ is some known constant.

First I want to determine how my hedging portfolio evolves. I denote this by $V$. There are three things I need to impose on this portfolio. The number of shares in it is given by the Black Scholes delta (so I calculate it based on $\sigma_{BS}$), the portfolio is self-financing and its value at inception has to be equal to the Black Scholes price of the call option. Then it must hold that $$dV_t = \Delta_{BS} dS_t + (V_t - \Delta_{BS} S_t)rdt \qquad V_0 = C_0$$ This then yields $$dV_t = rV_tdt + \Delta_{BS}\sigma_t S_tdW_t^Q \tag{1}$$

On the other hand the price of the call option under the Black Scholes model obeys the SDE $$dC_t = \underbrace{\left(\theta_{BS} + \Delta_{BS}rS_t + \frac{1}{2}\Gamma_{BS}\sigma_{BS}^2S_t^2\right)dt}_{rC_tdt} + \Delta_{BS}\sigma_{BS}S_tdW_t^Q \tag{2}$$

I define the hedging error as $e_t = V_t - C_t$ and I am interested in $e_T$ where $T$ is the expiry of the call option.

I can subtract $(2)$ from $(1)$ but the terms with Brownian motion do not vanish and according to the notes I am looking at I should get something like $\int_0^T\Gamma_{BS}(\sigma_{BS}^2 - \sigma_t^2)\ldots dt$. I am not sure if I defined the objects the right way. Some help would be appreciated.

More details on my reasoning: Let me phrase my question in a different way. Perhaps I am doing something wrong while converting my interpretation into mathematical statements. Just for the sake of argument forget the fact that the Black Scholes model exists and we are given that the market is governed by an SV model. At time $0$ I sell a call option (strike $K$, maturity $T$) in this market and I want to hedge it. Now instead of hedging correctly, I hold $\Delta^{wrong}$ number of stocks at each time where $$\Delta^{wrong}_t = \frac{\log\left(\frac{S_t}{K}\right) + \left(r + \frac{1}{2}\sigma_{wrong}^2\right)(T-t)}{\sigma_{wrong}\sqrt{T-t}}$$ with $\sigma_{wrong}$ being some constant positive number that I pick.

With the money I get from the sale, $V_0$, I set up my porfolio by buying $\Delta^{wrong}_0$ number of shares and borrowing/lending $V_0 - \Delta^{wrong}_0S_0$ in the money markets. Since I want my portfolio to be self-financing, I rebalance it such that

$$dV_t = \Delta_{wrong} dS_t + (V_t - \Delta_{wrong} S_t)rdt$$

Under the SV model, this becomes $$dV_t = rV_tdt + \Delta_{wrong}\sigma_tS_tdW_t$$

If instead of choosing some constant $\sigma_{wrong}$ in $\Delta^{wrong}_t$, I had made the right hedging decisions, it would follow that at maturity $V_T = \max(S_T-K,0)$. But since I hedged incorrectly, it won't be true that $V_T = \max(S_T-K,0)$ and that difference is the hedging error in my interpretation.

Answer 1

Let's assume that at time $t$ you become long an option, which you wish to price and risk-manage under the BS framework. The delta-hedged portfolio at time $t$ reads $$ \Pi_t = (V^{BS}(t) - \Delta_{BS} S_t) - \frac{(V^{BS}_t - \Delta_{BS} S_t )}{B_t} B_t \tag{0}$$ where the first portion on the RHS denotes your long option & short delta position and the second portion figures the cash balance required to fund the former position (risk-free money-market account). At time $t$, the value of the hedging portfolio - or the replication error - is therefore zero $e_t = \Pi_t = 0$.

Further assuming the stock pays no dividend, we have from the self-financing property \begin{align} d\Pi_t &= dV^{BS}_t - \Delta^{BS}dS_t - \frac{(V^{BS}_t - \Delta^{BS} S_t)}{B_t} dB_t \\ &= dV^{BS}_t - \Delta^{BS}dS_t - (V^{BS}_t - \Delta^{BS} S_t)r dt \tag{1} \end{align} Now, from Itô formula (and under our modelling assumption of a 1D Markov diffusion model in $S_t$) $$ dV^{BS}_t = \theta^{BS} dt + \Delta^{BS} dS_t + \frac{1}{2} \Gamma^{BS} d\langle S \rangle_t $$ such that $(1)$ is equivalent to $$ d\Pi_t = \theta^{BS} dt + \frac{1}{2} \Gamma^{BS} d\langle S \rangle_t - (V^{BS}_t - \Delta^{BS} S_t) r dt \tag{2} $$

We now use 2 important informations. First, that the price function $V^{BS}(t,S)$ by definition verifies the BS PDE $$ \theta^{BS} + rS\Delta^{BS} + \frac{1}{2} \Gamma^{BS} \sigma_{BS}^2 S^2 - rV^{BS} = 0 $$ such that $(2)$ may be rewritten as $$ d\Pi_t = \frac{1}{2} \Gamma^{BS} \left( d\langle S \rangle_t - \sigma_{BS}^2 S^2 dt \right) \tag{3}$$ Second, the fact that the market actually behaves in a stochastic fashion, different than what is postulated by the model $$ dS_t = \cdot + \sigma_t S_t dW_t $$ so that $(3)$ becomes $$ d\Pi_t = \frac{1}{2} \Gamma^{BS} S_t^2 \left( \sigma_t^2 - \sigma_{BS}^2 \right) dt $$ The total hedging error up to maturity then writes: \begin{align} e_T &= e_t + \int_t^T d\Pi_u \\ &= \frac{1}{2} \int_t^T \Gamma^{BS}(u,S_u) S_u^2 \left( \sigma_u^2 - \sigma_{BS}^2 \right) du \end{align}

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Here is an answer that follows up to your edits. We are now pricing with a Stochastic Volatility model while using a "wrong" Delta, e.g. that given by some Black-Scholes model. We have

\begin{align} \Pi_t &= (V(t) - \Delta_{BS} S_t) - \frac{(V_t - \Delta_{BS} S_t )}{B_t} B_t \\ d\Pi_t &= dV_t - \Delta^{BS}dS_t - (V_t - \Delta^{BS} S_t) rdt \\ &= \left( {\color{blue}{\frac{\partial V}{\partial t} dt}} + \frac{\partial V}{\partial S} dS_t + \frac{\partial V}{\partial v} dv_t + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} d\langle S \rangle_t + \frac{1}{2} \frac{\partial^2 V}{\partial v^2 } d\langle v \rangle_t + \frac{\partial^2 V}{\partial S \partial v } d\langle S, v \rangle_t \right) - \Delta^{BS}dS_t - ( {\color{blue}{V_t}} - \Delta^{BS} S_t) {\color{blue}{rdt}} \end{align} since now the modelling framework is 2D Markov in $(S_t, v_t=\sigma_t^2)$ (hence requiring bivariate Itô to express $dV_t$). From there, you can replace some of the terms (in blue) by using the pricing PDE verified by the function $V(t,S,v)$.

But this PDE is of course model-specific. Consider the Heston case. You'd have: $$ {\color{blue}{\frac{\partial V}{\partial t}}} + \frac{\partial V}{\partial S_t} r S_t + \frac{\partial V}{\partial v_t} \left( \kappa (\theta - v_t) \right) + \frac{1}{2}\frac{\partial^2 V}{\partial S_t^2} v_t S_t^2 + \frac{1}{2}\frac{\partial^2 V}{\partial v_t^2} \xi^2 v_t + \frac{\partial^2 V}{\partial v_t \partial S_t} \rho \xi v_t S_t - {\color{blue}{r V_t}} = 0 $$ such that \begin{align} d\Pi_t &= \left( \frac{\partial V}{\partial S} - \Delta_{BS} \right) (dS_t - rS_t dt) \\ &- \frac{\partial V}{\partial v} \left( \kappa (\theta - v_t) \right) dt - \frac{1}{2}\frac{\partial^2 V}{\partial S^2} v_t S_t^2 dt - \frac{1}{2}\frac{\partial^2 V}{\partial v^2} \xi^2 v_t dt - \frac{\partial^2 V}{\partial v \partial S} \rho \xi v_t S_t dt \\ & + \frac{\partial V}{\partial v} dv_t + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} d\langle S \rangle_t + \frac{1}{2} \frac{\partial^2 V}{\partial v^2 } d\langle v \rangle_t + \frac{\partial^2 V}{\partial S \partial v } d\langle S, v \rangle_t \end{align} and finally \begin{align} d\Pi_t &= \left( \frac{\partial V}{\partial S} - \Delta_{BS} \right) (dS_t - rS_t dt) \\ &+ \frac{\partial V}{\partial v} \left( dv_t - \kappa (\theta - v_t) dt \right) \\ &+ \frac{1}{2} \frac{\partial^2 V}{\partial S^2} \left( d\langle S \rangle_t - v_t S_t^2 dt \right) \\ &+ \frac{1}{2} \frac{\partial^2 V}{\partial v^2 } \left( d\langle v \rangle_t - \xi^2 v_t dt \right) \\ &+ \frac{\partial^2 V}{\partial S \partial v }\left( d\langle S, v \rangle_t - \rho \xi v_t S_t dt \right) \end{align} Again to conclude you write that $e_T = \int_t^T d\Pi_u$. Note the additional terms due to (i) the fact that you've delta-hedged using the "wrong" delta; (ii) volatility moves (here instantaneous variance) are stochastic and potentially correlated to spot moves. Finally note that if $v_t$ is not stochastic but equal to $\sigma^2$ and if you use the right Delta then you end up on the same result as above.