Sunday, January 4, 2015

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)?



Answer



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.


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