Market price of volatility risk

Market price of volatility risk

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Anna Taurogenireva · External communityPost link
External question — Quantitative Finance Stack Exchange Author: Anna Taurogenireva Original post: https://quant.stackexchange.com/questions/16138 License: CC BY-SA 3.0 — https://creativecommons.org/licenses/by-sa/3.0/ Adaptation: HTML converted to plain text; contact email addresses removed. Reading Gatheral's The volatility surface , page 7. The model they are talking about is $$\begin{align}dS_t&=\mu_tS_tdt+\sqrt{\nu_t}S_tdZ_1\\d\nu_t&=\alpha(S_t,\nu_t,t)dt+\eta\beta(S_t,\nu_t,t)\sqrt{v_t}dZ_2\\\left[Z_1,Z_2\right]&=\rho dt\end{align}$$ where $S_t$ is a stock price, and $\nu_t$ stochastic volatility. There is an option on $S$ with price $V(S_t,\nu,t)$, and another asset $V_1$ depending on the volatility. A portfolio $\Pi=V-\Delta V-\Delta_1 V_1$, with $\Delta$ and $\Delta_1$ chosen such that this evolves like the money market account $$dB_t=r_tB_tdt$$ with $r_t$ assumed deterministic. [skipping the long equations] They get a big differential operator applied to $V$ and applied to $V_1$ most both equal to some function which they write in the form $-(\alpha-\phi\beta\sqrt{\nu})$, for some function $\phi$. So far so good. And they call $\phi$ the market risk of volatility risk. With the choices of $\Delta$ and $\Delta_1$ made, they form a portfolio $\Pi_1=V-\Delta S$ from where $$d\Pi_1-r\Pi_1=\beta\sqrt{\nu}\frac{\partial V}{\partial \nu}\left(\phi dt+dZ_2\right)$$ So far so good, too. Now they say "defining the risk neutral drift as" $$\alpha'=\alpha-\beta\sqrt{\nu}\phi$$ the equation for $\nu$ becomes $$d\nu_t=\alpha'dt+\beta\sqrt{\nu}dZ_2$$. Here is what I don't understand. Why are they free to define the risk neutral drift? Is the model is complete and arbitrage free the risk neutral drift is no something one chooses, isn't it? It comes out of the unique risk free measure, isn't it? I am a beginner in this things. I could be very wrong. Alternatively, why is the risk-free drift equal to $\alpha-\beta\sqrt{\nu}\phi$? My possible answer. But I am not sure. Probably my confusion is that I don't know well the definition of risk free. Is it that risk free just means to get the portfolio, $\Pi_1$ in this case, to have drift $r$, as it happens with the stock in the Black-Scholes model? We see that $$d\Pi_1=r\Pi_1dt+\beta\sqrt{\nu}\frac{\partial V}{\partial \nu}d Z_2'$$ with $Z_2':=\int_{0}^{t}\phi+Z_2$.
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user16891 · External communityPost link
External answer — Quantitative Finance Stack Exchange Author: user16891 Original post: https://quant.stackexchange.com/a/19159 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. The stock price and variance follow the processes in this bivariate system of stochastic differential equations under the historical measure $\mathbb{P}$ also called the physical measure. For pricing purposes,however, we need the processes for under the risk-neutral measure $\mathbb{Q}$ .This is done by modifying each SDE separately by an application of Girsanov’s theorem.The risk-neutral process for the variance is defineded by introducing a function $\phi(S_t,v_t,t)$ into the drift of $dv_t$ as follows $$dv_t=[\alpha(S_t,v_t,t)-\phi(S_t,v_t,t)]dt+\eta\,\beta(S_t,v_t,t)\sqrt{v_t}dW_2(t),$$ where $$W_2(t)=Z_2(t)+\frac{\phi(S_t,v_t,t)}{\eta\,\beta(S_t,v_t,t)\sqrt{v_t}}t$$ The risk-neutral process for the stock price is $$dS_t=rS_{t}dt+\sqrt{v_t}S_tdW_1(t)$$ where $$W_1(t)=Z_1(t)+\frac{\mu_t-r}{\sqrt{v_t}}t.$$ To summarize, the risk-neutral process is \begin{align} &dS_t=rS_{t}dt+\sqrt{v_t}S_tdW_1(t)\\ &dv_t=[\underbrace{\alpha(S_t,v_t,t)-\phi(S_t,v_t,t)}_{\alpha^*(S_t,v_t,t)}]dt+\eta\,\beta(S_t,v_t,t)\sqrt{v_t}dW_2(t) \end{align} where $$\mathbb{E^Q}[dW_1(t)dW_2(t)]=\rho\,dt$$ Now by application of delta hedging argument, we have $$\frac{\partial V}{\partial t} +\frac{1}{2}v\,S^2\frac{\partial^2 V}{\partial S^2}+\rho\,\eta\,\,v\,S \frac{\partial^2 V}{\partial v\,\partial S} + \frac{1}{2}\eta^2v\frac{\partial^2 V}{\partial v^2} + rS \frac{\partial V}{\partial S}-rV=-\alpha^* \frac{\partial V}{\partial v}$$ Note that,when $\phi=0$ we have $\alpha^*=\alpha$ so that these parameters under the physical and risk-neutral measures are the same.we set $\phi=0$ , because when we estimate the risk-neutral parameters to price options we do not need to estimate $\phi$ . Estimation of $\phi$ is the subject of its own research. Let $M$ denote the number of underlying traded assets in the model excluding the risk free asset, and let $R$ denote the number of random sources. The model is arbitrage free if and only if $M\leq R$ . The model is complete if and only $M\geq R$ The model is complete and arbitrage free if and only if $M=R$ . In the stochastic volatility model we have $M=1$ and $R=2$ thus model is arbitrage free.In other words we can say,the risk-neutral measure is not unique.
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Quoted from Forex.com.bd-Editorial External answer — Quantitative Finance Stack Exchange Author: user16891 Source score (net votes, not local likes): 6 Original post: https://quant.stackexchange.com/a/19159 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. The stock price and variance follow the processes in this bivariate system of stochastic differential equations under the historical measure $\mathbb{P}$ also called the physical measure. For pricing purposes,however, we need the processes for under the risk-neutral measure $\mathbb{Q}$ .This is done by modifying each SDE separately by an application of Girsanov’s theorem.The risk-neutral process for the variance is defineded by introducing a function $\phi(S_t,v_t,t)$ into the drift of $dv_t$ as follows $$dv_t=[\alpha(S_t,v_t,t)-\phi(S_t,v_t,t)]dt+\eta\,\beta(S_t,v_t,t)\sqrt{v_t}dW_2(t),$$ where $$W_2(t)=Z_2(t)+\frac{\phi(S_t,v_t,t)}{\eta\,\beta(S_t,v_t,t)\sqrt{v_t}}t$$ The risk-neutral process for the stock price is $$dS_t=rS_{t}dt+\sqrt{v_t}S_tdW_1(t)$$ where $$W_1(t)=Z_1(t)+\frac{\mu_t-r}{\sqrt{v_t}}t.$$ To summarize, the risk-neutral process is \begin{align} &dS_t=rS_{t}dt+\sqrt{v_t}S_tdW_1(t)\\ &dv_t=[\underbrace{\alpha(S_t,v_t,t)-\phi(S_t,v_t,t)}_{\alpha^*(S_t,v_t,t)}]dt+\eta\,\beta(S_t,v_t,t)\sqrt{v_t}dW_2(t) \end{align} where $$\mathbb{E^Q}[dW_1(t)dW_2(t)]=\rho\,dt$$ Now by application of delta hedging argument, we have $$\frac{\partial V}{\partial t} +\frac{1}{2}v\,S^2\frac{\partial^2 V}{\partial S^2}+\rho\,\eta\,\,v\,S \frac{\partial^2 V}{\partial v\,\partial S} + \frac{1}{2}\eta^2v\frac{\partial^2 V}{\partial v^2} + rS \frac{\partial V}{\partial S}-rV=-\alpha^* \frac{\partial V}{\partial v}$$ Note that,when $\phi=0$ we have $\alpha^*=\alpha$ so that these parameters under the physical and risk-neutral measures are the same.we set $\phi=0$ , because when we estimate the risk-neutral parameters to price options we do not need to estimate $\phi$ . Estimation of $\phi$ is the subject of its own research. Let $M$ denote the number of underlying traded assets in the model excluding the risk free asset, and let $R$ denote the number of random sources. The model is arbitrage free if and only if $M\leq R$ . The model is complete if and only $M\geq R$ The model is complete and arbitrage free if and only if $M=R$ . In the stochastic volatility model we have $M=1$ and $R=2$ thus model is arbitrage free.In other words we can say,the risk-neutral measure is not unique.

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