The volatility $\sigma$ of an Ito process $dS_t = r S_t dt + \sigma S_t dW_t$ is not the square root of its variance.

But you often hear that "volatility = standard deviation".

What's going on here?

  • $\begingroup$ I don't think it makes sense to say the variance of the ito process since variance is for a random variable and that's different from a stochastic process. Maybe you are talking about quadratic variation? $\endgroup$
    – Slade
    Jun 15, 2019 at 15:24
  • $\begingroup$ A stochastic process $S_t$ has variance for a given time point $t$. $\ $\endgroup$
    – Banjor
    Jun 15, 2019 at 15:30
  • $\begingroup$ You mean $\sigma^2 S_t^2$ as the instantaneous variance of $dS_t$? Or the variance of the solution of this SDE (the value of the process)? Or the instantaneous variance of $\frac{dS_t}{S_t}$? The statement you have vol=std dev=sqrt of var, is true for each one of them individually, and if you assume $\sigma$ is constant then you can easily establish how they are linked. $\endgroup$ Jun 15, 2019 at 15:46

1 Answer 1


$S_t$ is log-normal, so indeed its variance will be different.

$\sigma ^2$ is, however, the variance of the returns $\log S_t$ per unit of time since

$$\log S_t \sim N\left(\log S_0 + \left(r-\frac 12 \sigma ^ 2\right) t, \sigma ^2 t\right)$$


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