# Mathematical underpinnings of the square root of time rule

Often when I am reading about options pricing (and/or options greeks) the square root of time continually comes up. What the mathematical justification for why this keeps on turning up?

• When you add two normal distributions with standard deviations alpha and beta, you get a normal distribution with standard deviation sqrt(alpha^2 + beta^2). The other answers are probably correct, but, in my opinion, unnecessarily verbose.
– user59
Sep 20, 2016 at 16:17
• @barrycarter in fact the normality assumption is not required at all. Plus what you say is not true if the normal variables are correlated. See Alex C answer which is spot on IMHO. Sep 20, 2016 at 20:34

For any process with independent increments, by the very fact of statistical independence the variance of $x_{t3}-x_{t1}$ is going to be the sum of the variances of $x_{t2}-x_{t1}$ and $x_{t3}-x_{t2}$ for $t1\leq t2 \leq t3$. Many processes have independent increments, including ABM, GBM, Poisson, etc. Then if you add a homogeneity assumption (the parameters of the process do not change over time) you get a proportionality of the variance to the length of the time interval and thus a proportionality of the standard deviation to $\sqrt t$.

The reason is that in many common models including geometric Brownian motion, the variance of the logarithmic returns is proportional to time. Thus, their standard deviation/volatility is proportional to the square root of time.

Consider for example the class of Levy models where $X$ is is the logarithmic stock price process such that $S_t = S_0 e^{X_t}$. The cumulant generating function of $X_t$ takes the form

$$\psi_{X_t}(\omega) = \psi_{X_1}(\omega) t,$$

where $\psi_{X_1}(\omega)$ is the characteristic exponent which is independent of $t$. You obtain the $n$-th cumulant as the $n$-th derivative of the cumulant generating function with respect to the transform parameter evaluated at zero, i.e.

$$c_n(X_t) = \frac{1}{\mathrm{i}^n} \frac{\partial^n \psi_{X_t}}{\partial \omega^n}(0).$$

You immediately see that the cumulants are proportional to $t$. Since the second cumulant corresponds to the variance, this confirms that the standard deviation of logarithmic returns is proportional to the square root of time.

As mentioned the above setup includes the geometric Brownian motion setting with

$$\psi_{X_t}(\omega) = \left( \mathrm{i} \omega \left( r - \frac{1}{2} \sigma^2 \right) - \frac{1}{2} \sigma^2 \omega^2 \right) t.$$

You get

$$c_2 \left( X_t \right) = \sigma^2 t$$

as expected.