Why VIX² Does Not Directly Reveal the Variance Risk Premium
Summary
The document examines whether a model’s expected integrated variance under the real-world probability measure can be equated to the market VIX-squared level to infer the variance risk premium. It describes a variance process whose drift differs between real-world and risk-neutral measures through a market price of variance risk. The proposed approach calibrates risk-neutral variance dynamics from vanilla option prices, then adjusts the drift and attempts to match the resulting expectation to VIX squared.
The author revises this reasoning, pointing out that variance swaps are traded and valued under the risk-neutral measure, so VIX squared cannot simply be treated as a real-world expectation because it is not itself a traded asset. The suggested alternative is to estimate the risk premium using realized variance under the real-world measure, and, with mean reversion, to compare long-run realized variance with the model’s stationary variance. These are proposed estimation ideas rather than demonstrated empirical results; the note does not address estimation error or provide validation.
Key ideas
- The variance drift is related across real-world and risk-neutral measures through a variance risk price.
- Option prices can be used to calibrate risk-neutral variance dynamics.
- VIX squared should not be equated directly with real-world expected variance to infer the risk premium.
- The proposed alternative uses realized variance under the real-world measure, including a long-run comparison for mean-reverting variance.
Tags
Full text
# Estimation of the variance risk premium via VIX
# Estimation of the variance risk premium via VIX
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 $\lambda_v$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.