# Why is the volatility of an Ito process not the square root of its variance?

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?

• 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? – Slade Jun 15 '19 at 15:24
• A stochastic process $S_t$ has variance for a given time point $t$. $\ – Banjor Jun 15 '19 at 15:30 • You mean$\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. – Magic is in the chain Jun 15 '19 at 15:46

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