why is the BNS model the way it is

Question

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 1

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 2

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 $dX_t$.