# 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: $$\sqrt{VaR(S_t)} = \frac{(dS_t)^2}{dt}$$

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 $$dtdt=0,dtdW=0, \text{ and } dWdW=dt$$.

• $E\left[ \left( g \, dW_t \right)^2\right]=g^2dt$. Why does this hold? thx – econmajorr Jun 15 at 15:25
• This becomes $E\left[ g^2 dt \right]$ which is conditionally constant/known, so its expected value is its amazing self! – Magic is in the chain Jun 15 at 15:34
• This is Ito's isometry: $\mathbb{E}[(\int g dW_t)^2] = \mathbb{E}[\int g^2 dt]$ – byouness Jun 17 at 19:45