Instantaneous Variance of an Itô Process

Summary

The note explains how to obtain the instantaneous variance of an Itô process with drift coefficient f and diffusion coefficient g. Over a small time interval, the expected increment is the drift contribution, f times dt. Subtracting that mean leaves the random diffusion term, g times the Brownian increment. Squaring and taking expectations gives variance g squared times dt, because the squared Brownian increment has expectation dt.

The same result follows from Itô multiplication rules: terms involving dt squared or dt times a Brownian increment vanish at this order, while the square of the Brownian increment contributes dt. Thus the variance rate per unit time is g squared. This is a local, infinitesimal result for the stated Itô model; the note does not address finite-horizon variance, changing coefficients, or the distinction between variance and standard deviation. The question’s suggested square-root expression is not derived as written; the derivation supports variance of the increment equal to g squared dt.

Key ideas

• For an Itô process, the drift term determines the conditional mean of an infinitesimal increment.
• Removing the drift leaves the diffusion contribution to the increment’s variance.
• The Brownian increment has variance dt, so the process increment has variance g squared times dt.
• The instantaneous variance rate is g squared under the stated model.

Tags

# Why the variance of a process is $\left( \frac{dS_T^2}{dt}\right)^2$?


# Why the variance of a process is $\left( \frac{dS_T^2}{dt}\right)^2$?












Consider an Ito process $dS_t = f(t,S_t) dt + g(t,S_t)dW_t $

What is the reason that we can compute the variance as: $\sqrt{VaR(S_t)} = \frac{(dS_t)^2}{dt}$

## Answer by Magic is in the chain (score 2, accepted)

https://quant.stackexchange.com/a/46114

Because instantaneous variance can be written as follows:

$V \left[ dS_t\right]=E\left[ \left( dS_t -E\left[dS_t\right] \right)^2\right]$

$V \left[ dS_t\right]=E\left[ \left( dS_t -f \, dt \right)^2\right]$

$V \left[ dS_t\right]=E\left[ \left( g \, dW_t \right)^2\right]=g^2dt$

Which is the same thing as:

$V \left[ dS_t\right]=E\left[ dS_t dS_t\right]=g^2dt$

Where I used the familiar drill $dtdt=0,dtdW=0, \text{ and } dWdW=dt$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.