# Why square root of volatility in Heston model?  Why do we model it as sqrt root of v(t)? Is that because we don't want the volatility to go negative? If this is the case, can we model it as square of v(t)?

• In the Heston model you have the square root of variance, not volatility (like in Geometric Brownian motion, but here variance is stochastic too) - please clarify the question. Thank you. – vonjd Jul 18 '15 at 12:49

## 4 Answers

V(t) is the variance process of the stock price, not volatility process. Cox-Ingersoll-Ross demonstrated that that specific process can be non-negative under certain conditions, which is what you want for variance.

In this paper http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2626552 the authors compare the Heston model with volatility given by

$dV_t = \kappa_V(\bar{V}-V_t)dt+\sigma_V\sqrt{V_t}dW_t$

with the a model where the volatiltiy is given by

$dV_t = \kappa_V(\bar{V}-V_t)dt+\sigma_VV_tdW_t$.

They show that the latter is inverse gamma distributed and leads to a more stable volatility distribution with higher kurtosis and fater tails.

However, a quick read shows, that the inverse gamma model seems to be relatively unexplored.

• As you point out in each case V is the variance. Both models dont allow negativ variance. – Phun Jun 14 '16 at 5:57
• What do you mean by "more stable" distribution? Is it in the sense of stable distribution en.wikipedia.org/wiki/Stable_distribution or that the parameters in the calibration process do not vary much over frequent calibration to market data over time? – Hans Dec 25 '17 at 22:36

The reason is that Heston managed to solve the case with square root. The log-normal vol process leads to nasty properties. The 3/2 model is another case that have been solved.

## Edit

we assume $X_t$ follows the differential stochastic process $$d X(t)=\mu (t,{{X}_{t}})dt+\sigma (t,{{X}_{t}}) dW(t)$$ if $$\underset{{{X}_{t}}\to 0}{\mathop{\lim }}\,\,\mu (t,{{X}_{t}} )-\frac{1}{2}\frac{\partial }{\partial x}{{\sigma }^{2}}(t,{{X}_{t}})\geq 0$$ then $$P(\{\,t\in [0\,,\infty )|\,X(t\,,x )\leq 0\})=0$$

in the C.I.R Model ,we have $$\underset{{{v}_{t}}\to 0}{\mathop{\lim }}\,\,\left( \kappa (\theta -{{v}_{t}})-\frac{1}{2}\frac{\partial }{\partial v}{{(\sigma \sqrt{{{v}_{t}}})}^{2}} \right)=\underset{{{v}_{t}}\to 0}{\mathop{\lim }}\,\,\kappa (\theta -{{v}_{t}})-\frac{1}{2}{{\sigma }^{2}}=\kappa \theta -\frac{1}{2}{{\sigma }^{2}}$$ another property of the square-root process for instantaneous variance is the fact that it leads in many case of interest to close-form or semi-close form solution for the characteristic function. we are also able to drive a close-form solution based on hyper-geometric functions when the underlying follow as mean reverting process.

• I know those equations. I'm more interested in knowing why we model the volatility as sqrt root in the first place. – SmallChess Jul 17 '15 at 8:23
• @Student T in first glance I don't understand your Purpose – user16891 Jul 17 '15 at 9:31