Why Brownian Motion Has Quadratic Variation Equal to Time

# How is the formula of Quadratic Variation of Brownian Motion derived?


# How is the formula of Quadratic Variation of Brownian Motion derived?












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.

## Answer by Jan Stuller (score 4, accepted)

https://quant.stackexchange.com/a/63885

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:

$$(X_i-X_{i-1})^2=X_i^2-2X_iX_{i-1}+X_{i-1}^2$$

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:

$$\mathbb{E}[dW_i^2]=\mathbb{E}[W(di)^2]=\mathbb{E}[\left(\sqrt{di}W(1)\right)^2]=di\mathbb{E}[W(1)]=di$$

$$Var\left(dW_i^2\right)=\mathbb{E}[\left(\sqrt{di}W(1)\right)^4]-\mathbb{E}[\left(\sqrt{di}W(1)\right)^2]^2=di^2\mathbb{E}[W(1)^4]-di^2$$

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).