Why is $$\frac{ \left(\frac{ \Delta S}{S}\right)^2} {dt}$$ an estimator of volatility squared (as claimed by my book)?

As far as I understand it, we estimate volatitility squared as $$\frac{ Var( \text{Return})} {dt}$$ where $\text{Return} = \Delta S/S$.

So how does one justify the first formula? Is it the volatility of just a single change in price?

  • $\begingroup$ Yes, the most basic (but also highly inaccurate) estimator of variance takes just 1 term squared. To get more accurate estimates we would (of course) combine (average) several such squared terms, but at its most fundamental estimating variance of a 0-mean process could be in principle done with 1 term. $\endgroup$
    – nbbo2
    Commented Mar 22, 2017 at 1:43
  • $\begingroup$ Agree with @noob2. The result holds because for continuous (semi-)martingales there exists a direct link between quadratic (co)variation (sum of squared differences over a realisation of the stochastic process $(X_t)_{t \in [0,\tau]}$ for a fixed $\omega \in \Omega$) and (co)variance (variance of the the random variables $X_\tau$) for all $\tau > 0$, see this related question: quant.stackexchange.com/questions/31861/… $\endgroup$
    – Quantuple
    Commented Mar 22, 2017 at 8:22


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