What is the difference between standard deviation, volatility and quadratic variation?

As I know, volatility is the standard deviation of the log returns, so they are basically the same. (One of them is used in Mathematics, the other one in Finance.) I also know what is the quadratic variation of a stochastic process. In a lot of article I see, and in a lot of lesson I hear volatility is the quadratic variation of the stochastic process. Therefore the standard deviation is equal to quadratic variation? I wouldn't say this last sentence for sure, but that would be a logical conclusion, isn't it?

What is the difference? I think I have a big problem with the definitions.


1 Answer 1


Using only words and no equations:

Knowing the Variance (or standard deviation) of a Brownian Motion we can calculate the uncertainty in the future position of a particle. Knowing $\sigma^2$ and assuming the particle starts at $S_0$ we can say that $S_T$ will be in $[S_0-1.96 \sigma, S_0+1.96 \sigma]$ 95% of the time. In other words 95% of the trajectories that start at $S_0$ will end up in this interval at time $T$.

Volatility is another name given in Finance to the $\sigma$ which appears in the formulas for the GBM. It is the standard deviation of the logarithmic stock price changes. It is also the standard deviation of the underlying BM, but to find the stock price we then have to take the Exponential of this BM, giving a point on the GBM.

Now about the Quadratic Variation. It is often said that Quadratic Variation is a path based concept. When a process (not necessarily BM or GBM) goes from $S_0$ to $S_T$ it will follow a specific trajectory (or path), which we usually draw on the blackboard as a very jagged and bumpy curve. As the particle moves along this trajectory, we can compute the Quadratic Variation by summing (integrating) the squared movements in S. (So you can say that quadratic variation is another process $QV_t$ that is computed from the process $S_t$). When we come to time $t=T$ we will have found the total quadratic variation of this path $QV \triangleq QV_{[0,T]}$. Naturally if $S_t$ had taken a different path (a different curve drawn on the blackboard with a chalk of a different color) we would have found a different value for $QV_t$ at each $t$ and also for $QV$. The quadratic variation is measured along a specific path.

QV is a random variable (which we can observe experimentally by carrying out a measurement), $\sigma$ is a parameter (a value which we can select as we please for the purpose of calculation). A reasonable way to choose $\sigma^2$ for a BM is to run many experiments and take the average value of $QV$ that we observed.

(I am trying to give a non-technical description. If there are any inaccuracies or contradictions in what I wrote I would appreciate a correction. Thanks).

  • $\begingroup$ And if we would calculate the expected value of the quadratic variation at time $T$, then we would get the $sigma^2$ as result? Thank you for the answer, it was quite adequate and understandable. $\endgroup$
    – Kapes Mate
    Feb 29, 2020 at 12:52
  • $\begingroup$ Yes, because BM is a martingale (the mean change is zero) it is true. But in general Variance and SumOfSquares are different $\sum x_i^2 \ne \sum (x_i-\mu)^2$ and this $\mu$ would have to be taken into account (if non-zero). $\endgroup$
    – nbbo2
    Feb 29, 2020 at 14:38

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