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

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

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