# why is the BNS model the way it is

what I am puzzled about is, why dont we instead of having

$$dX_t = \sqrt{V_t} dB_t - (\frac{1}{2} V_t^2-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber$$

we just have

$$dX_t = V_t dB_t - (\frac{1}{2} V_t-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber$$

where $$dV_t = -\lambda V_t dt + dZ_{\lambda t}\nonumber$$

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?

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could you give a link to some paper on the model ? –  Probilitator Feb 19 at 21:06
Could you please post a link to a free paper where BNS is defined. A paper where we can find your equation. –  Richard Feb 20 at 12:38
@Richard I found it here: economics.ouls.ox.ac.uk/13781/1/read.pdf –  Probilitator Feb 20 at 14:11
why would you need to model the jump in vol to impact equity same level? You have another random process to move the equities. If you have to random processes that have similar effects, you will have a calibration nightmare –  adam Feb 22 at 7:33

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.

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where is the mean reversion in the geometric Brownian motion ? –  Probilitator Feb 21 at 8:39
Just a typo - of course it does not habe mr. –  Richard Feb 21 at 9:20
I always hat the idea to work with the squareroot process of variance. As far as I remember Itô gives you some terms with $\sqrt{V}^{-1}$ - this tends to be numerically unstable –  Probilitator Feb 21 at 11:34
Of course the first derivative of a square-root is proportional to $1/\sqrt(...)$. But which application of Ito do you mean - an application to $f(X_t)$ or $f(V_t)$ for some $C2$ function $f$? –  Richard Feb 21 at 14:09
I meant applying Itô to $\sqrt{V}$ - to get the dynamics of $\sqrt{V}$ –  Probilitator Feb 21 at 14:18

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 $dX_t$.

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