This is a follow up on this question on quant SE:

The question mentions for a Brownian motion : $X_t = X_0 + \int_0^t\mu ds + \int_0^t\sigma dW_t $ , the quadratic variation is calculated as

$dX_t dX_t = \sigma^2 dW_t dW_t = \sigma^2 dt $

I cannot understand how is the differential with time ($\mu ds $) eliminated from the equation. When I square the differential form of the equation:

$(dX_t)^2 = (\mu dt + \sigma dW_t)^2 = \mu^2 dtdt + \sigma^2 dW_tdW_t + \mu \sigma dt dW_t$. From here on, I'm finding it difficult to reduce it to above form.

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    $\begingroup$ The rule is that $(dt)^2$ and $(dt)(dW)$ can be replaced by zero, while $(dW)^2$ is replaced by $dt$. $\endgroup$
    – nbbo2
    Commented May 10, 2021 at 13:54
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    $\begingroup$ Guys, I understand the reason why somebody would consider the way the question was phrased as "basic" and "not meeting the Quant SE guidelines", but underneath the surface, the question is non-trivial. For example, I reference the formal proof in my answer below, but if someone asked me to replicate it, I wouldn't be able to do it, without a thorough preparation (and basically memorizing it). Therefore, I do think this a non-trivial question that deserves to stay open, as others might find it useful. I personally found useful trying to answer it rigorously, to refresh knowledge of the subject. $\endgroup$ Commented May 10, 2021 at 16:40
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    $\begingroup$ @JanStuller I think the argument is not that it's elementary, but that it's basic in the sense that you can find it in every single book on brownian motion or stochastic calculus. Not sure why you would replicate these calculations here. $\endgroup$
    – LazyCat
    Commented May 10, 2021 at 18:45
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    $\begingroup$ @LazyCat: you have a valid point, but only to a certain extent: for example, it took me a long time to find a rigorous proof (that I link below): most textbooks scratch the surface only and "hand-wave" the argument, using the usual "rule" of $dW_t^2=dt$. I haven't really found a text-book that would prove the Brownian motion quadratic variation as a limit a.s. or in probability. $\endgroup$ Commented May 10, 2021 at 19:07

1 Answer 1


In all honesty, Quadratic Variation for Stochastic Processes is an advanced topic, and computing it rigorously from first principles is a graduate-level probability question.

Part 1: Quadratic Variation: Informal "proof"

First, how is Quadratic Variation Defined? For a stochastic process $X_t$, the quadratic variation, denoted $<X_t>$, is defined as (loosely speaking, I provide the rigorous definition at the end):

$$<X_t>=\lim_{n \to \infty} \left(\sum_{i=1}^{i=n}(X_i-X_{i-1})^2\right)$$

So in words, the quadratic variation expresses the sum of square differences, as the mesh-size gets finer and finer. The limit is in the probability sense (see end of this post).

Now, we have:


Usually, $X_0:=0$, so ignoring the $X_0$ term, we have:

$dX_i^2=\mu^2di^2+\sigma^2dW_i^2+2\mu di \sigma dW_i$

Notice that as $n \to \infty$, $di \to 0$, so $di^2 \to 0$ even faster. So ignoring the $di^2$ term (and all other $di$-terms of order higher than 1), we can focus on the $dW_i^2$ term. Let's try and compute its expectation & variance:



As $n \to \infty$, $di^2 \to 0$ faster than $di \to 0$, so the Variance converges to zero. That is basically what is meant when somebody writes $dW_t^2=dt$

Notice the above is not mathematically rigorous, it's just "hand-waving" to get an intuitive understanding. The rigorous proof is below:

Quadratic Variation: "Rigorous" proof

Formally, Quadratic Variation for a Wiener process $W_t$ is defined as below:

$\forall \epsilon > 0$:

$$\left<W\right>_t:=\lim_{n \to \infty} \mathbb{P}\left(\left|\sum_{i=1}^{i=n}\left(W_{t_i}-W_{t_{i-1}}\right)^2-t\right|>\epsilon\right)=0$$

In words: the probability that the Quadratic variation converges to "$t$", goes to $1$, as the mesh size gets infinitely fine.

The proof is pretty technical, the best one I know is in these lecture notes: but the proof stretches over 5 pages (and they prove convergence almost surely, which is a stronger convergence that "in probability", so it implies convergence in probability).

  • $\begingroup$ Thanks Jan! This makes it a lot clearer now. $\endgroup$
    – Jay Mangal
    Commented May 10, 2021 at 18:56
  • $\begingroup$ Is it really PhD level in America? We had this in the bachelor and master Econometrics on the math courses. $\endgroup$
    – simsalabim
    Commented May 10, 2021 at 20:15
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    $\begingroup$ @simsalabim: the link to the proof in my answer is to a course 6.265 ("advanced stochastic processes") at MIT: in the USA, stand-alone Masters degrees in classical disciplines, such as maths or probability, don't really exists: instead, if people want to do a graduate course after their bachelors, they do a PhD, which usually takes 5 years, with the first two years having exams (the first 2 years is 60% exam, 40% "research" , the last 3 years is 100% research). The course 6.265 would typically be taken by 2nd year PhD candidates. $\endgroup$ Commented May 11, 2021 at 5:21

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