Why BNS Models Use Variance with a Square-Root Price Diffusion
Summary
The discussion asks why a Barndorff-Nielsen–Shephard style model uses the square root of a variance process as the diffusion coefficient for log price, instead of using the variance process directly. The proposed alternative is motivated by algebraic simplicity in an American put problem and by an intuition that volatility jumps might correspond more directly to price jumps.
The replies explain that the model treats the mean-reverting process as variance, while the diffusion coefficient in a price or log-price equation represents volatility, so the variance must be square-rooted. They distinguish variance from standard deviation through the interpretation of Brownian increments and refer to the use of variance in GARCH models. One response cautions that substituting variance for volatility changes the scale and interpretation of paths, and suggests transforming a square-root process with Itô’s formula as an alternative avenue. The exchange is conceptual and does not provide computations or a formal model comparison.
Key ideas
- The BNS specification models a mean-reverting variance process and uses its square root as the price diffusion scale.
- Variance and standard deviation have different interpretations and cannot be interchanged as diffusion coefficients without changing the model.
- The proposed unsquared specification is motivated by convenience for an American put problem.
- The replies offer intuition but no numerical comparison or formal analysis of the alternative model.
- An Itô transformation of a square-root process is suggested as a possible modeling route.
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Full text
# why is the BNS model the way it is
# why is the BNS model the way it is
what I am puzzled about is, why dont we instead of having
\begin{equation} dX_t = \sqrt{V_t} dB_t - (\frac{1}{2} V_t^2-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber \end{equation}
we just have
\begin{equation} dX_t = V_t dB_t - (\frac{1}{2} V_t-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber \end{equation}
where \begin{equation} dV_t = -\lambda V_t dt + dZ_{\lambda t}\nonumber \end{equation}
I have been working on American put problem for this. Without the squre root, I think some things can be simplified in a much nicer manner. Though I have not done any computation, without the square root, $V$ is 'in the same dimension' as the log price. The equation has a nice interpretation that a jump in 'volatility' correspond to a jump in price rather than a jump in 'volatility squared' correspond to a jump in price?
Paper: http://economics.ouls.ox.ac.uk/13781/1/read.pdf
## Answer by Richi Wa (score 1)
https://quant.stackexchange.com/a/10315
I would put it differently. Modelling variance in an additive way (an OU process is in some regard additive) is more natural than e.g. a gemetric Brownian motion model (which on the other hand does not model mean reversion). Volatility as it is a square-root is by no means additive.
Let $(B_t)_{t \ge 0}$ be Brownian motion then we have $$ VAR(B_t) = t = VAR(B_s-B_0)+VAR(B_t-B_s) = s+(t-s) = t. $$ This is true for the variance but by no means for volatility. Also think of GARCH modelling where $\sigma^2$ is modelled and not $\sigma$.
Finally if you model the variance then you have to take the square-root if you use it as a multiplier that represents volatility.
## Answer by Probilitator (score 0)
https://quant.stackexchange.com/a/10314
The second equation where you would be using variance instead of standard deviation won't provide "meaningful" paths.
The reason is: variance has no meaning/interpretation in space. If you consider a normal distribution of stock returns the standard deviation is actually a number that tells you the difference between the expected value and some quantile.
Variance on the other hand does not have such a direct interpretation. You could for example run an exponential Brownian motion with a $\sigma^2$ instead of $\sigma$ but this would equal a path with an underlying volatility of $\sigma^4$.
Instead of taking $\sqrt{V_t}$ one could try work with a square-root process. Thus applying Itô to $d(\sqrt{V_t})$. This way you might get rid of the $\sqrt{V_t}$ term in $dX_t$.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.