1
$\begingroup$

Suppose I have the formula for computing $\mathbb E^P\big[\int_0^T v\,dt\big]$ for the variance process $v$ in the real world measure $P$. Can I set it to the VIX$^2$ price and solve for the variance risk premium? My concern is that VIX is a traded asset and its risk premium is zero and the equality does not hold. However, would it be correct to argue that VIX$^2$ is not traded and thus the procedure is correct?

I will put the above question in specific terms. Assume the variance $v$ undergoes the process $$dv = a\,dt+b\,dB$$ where $B$ is the standard Brownian motion. The transformation between the real-world and risk-neutral measures is $a_P = a_Q-\lambda_v b$ where $\lambda_v$ is the market price of variance risk, and subscripts $Q$ and $P$ denotes the risk-neutral and real-world measures, respectively. We than set $$f[a_Q,b]:=\frac1T\mathbb E^Q\Big[\int_0^T d\langle\ln S\rangle_t\Big]$$ where $f[u,v]$ denotes a function $f$ of functions $u$ and $v$.

From the vanilla option market, I calibrate the functions $a_Q$ and $b$. Now I set $f[a_Q-\lambda_vb,b]=$VIX$^2$ where VIX denotes the price TRADED on the market of a fresh start VIX of maturity $T$, to compute $\lambda_v$.

Is this correct? My rationale is that although $VIX$ is traded and thus evaluated in the risk-neutral measure $Q$, VIX$^2$ is not and therefore is evaluated in the real-world measure $P$.

Edit: I now think this is wrong because the variance swap is traded and thus should be valued in the risk-neutral measure $Q$ and we cannot obtain the risk premium this way. The correct way to estimate this is to set $$f[a_P,b]=\frac1T\int_0^T d\langle\ln S\rangle_t$$ for the realized $S_t$ to solve for $\lambda_v$.

As a matter of fact, if we have a mean reversion form for the variance process, for example, $a=-\kappa(v-v_\infty)$, which generates an asymptotic stationary variance $v_\infty$ at long time, we can simply set $$v_\infty[a_P,b]=\lim_{T\to\infty}\frac1T\int_0^T d\langle\ln S\rangle_t$$ where $S$ is the realized stock price, to compute $\lambda_v$.

Improve this question
$\endgroup$
11
  • $\begingroup$ "My concern is that VIX is a traded asset" -- no? It's given by a formula that is a function of traded asset prices, but it is not traded itself. $\endgroup$
    – user217285
    Commented Aug 1, 2018 at 2:52
  • $\begingroup$ @Nitin: I agree with your last sentence. But what do you mean by the question mark following your "no"? Also, what is your conclusion? Do you agree with my rationale as stated in my last sentence that my setup is correct? $\endgroup$
    – Hans
    Commented Aug 1, 2018 at 6:02
  • 1
    $\begingroup$ It's not entirely clear to me what you're trying to achieve. What is the variance process $(v_t)_{t \geq 0}$ which you are talking about? Is this in the framework of a particular model? What do you define as the variance risk premium? $\endgroup$
    – Quantuple
    Commented Aug 1, 2018 at 7:55
  • 1
    $\begingroup$ So assuming $VIX^2$ represents the fair strike of a fresh-start, idealised (i.e. continuous returns sampling) variance swap maturing in 30 days then indeed: $VIX^2 = \frac{1}{T} \Bbb{E}^\Bbb{Q} \left[ \int_0^T d\langle \ln S \rangle_t \right]$ with $T=30/DCC$ years. And the idea would be to write that: $VRP = VIX^2 - \frac{1}{T} \Bbb{E}^P \left[ \int_0^T d\langle \ln S \rangle_t \right] = \frac{1}{T} \int_0^T \left( \Bbb{E}^Q[v_t] - \Bbb{E}^P[v_t] \right) dt$ assuming that $d\ln S_t = \cdot dt + v_t dW_t$. Did I understand that well? $\endgroup$
    – Quantuple
    Commented Aug 1, 2018 at 12:50
  • $\begingroup$ @Quantuple: Your understanding is mostly correct but with some difference to my intended argument. I have edited my question to let it stand on a more specific ground. Please review. Thank you. $\endgroup$
    – Hans
    Commented Aug 1, 2018 at 17:22

0

Reset to default

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.