I'm trying to refresh my memories about stochastic processes. We know that Brownian motion has as quadratic variation equals to t. What is the quadratic variation of the Brownian motion squared ? Usually for this I would just use Ito's formula and pick out whatever is in front of the dWt, except in that case it doesn't work.

Is there a straightforward way to compute this ? thanks !

  • 1
    $\begingroup$ Why doesn't it work using Itô's formula? $\endgroup$
    – saz
    Nov 19, 2014 at 7:47

2 Answers 2


There are several definitions for the quadratic variation:

  1. For a "nice" process $(X_t)_{t \geq 0}$ ("nice" means semimartingale), the quadratic variation is defined by $$[X]_t := X_t^2-X_0^2 -2 \int_0^t X_{s-} \, dX_s.$$ For $X_t := B_t^2$, we know from Itô's formula that $$d(B_s^2)= 2 B_s \, dB_s+ \, ds$$ and $$B_t^4 = 4 \int_0^t B_s^3 \, dB_s + 6 \int_0^t B_s^2 \, ds.$$ Combining these two equalities yields $$\begin{align*} [B^2]_t &= B_t^4 - 2 \int_0^t B_s^2 \, d(B_s^2) = 4 \int_0^t B_s^2 \, ds. \end{align*}$$
  2. The quadratic variation of a process $(X_t)_{t \geq 0}$ is the limit (in probability) of the sums $$S_{\Pi}(X) := \sum_{j=1}^n (X_{t_j}-X_{t_{j-1}})^2$$ as the mesh size $|\Pi|$ of the partition $\Pi=\{0=t_0< \ldots < t_n \leq t\}$ tends to $0$. For $X_t := B_t^2$, we have $$\begin{align*} S_{\Pi}(B^2) &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2)^2 = \sum_{j=1}^n (B_{t_j}+B_{t_{j+1}})^2 \cdot (B_{t_j}-B_{t_{j-1}})^2 \\ &= \sum_{j=1}^n (B_{t_j}+B_{t_{j+1}})^2 \bigg[(B_{t_j}-B_{t_{j-1}})^2-(t_j-t_{j-1}) \bigg] + \sum_{j=1}^n (B_{t_j}+B_{t_{j+1}})^2 (t_j-t_{j-1}) \\ &= I_1+I_2. \end{align*}$$ Recall that $$\sum_{j=1}^n g(t_j) \cdot (t_j-t_{j-1}) \to \int_0^t g(s) \, ds \qquad \text{as} \, n \to \infty$$ for any continuous function $g$. Using the continuity of $g$, it is therefore not difficult to see that $$\sum_{j=1}^n (g(t_j)+g(t_{j+1}))^2 (t_j-t_{j-1}) \to 4\int_0^t g(s)^2 \, ds.$$ Applying this to the continuous sample paths of the Brownian motion, we find that $$I_2 \to 4 \int_0^t B_s^2 \, ds.$$ It remains to prove that $I_1$ converges to $0$. This follows from some straightforward calculations (Hint: Show that the $L^2$-norm converges to $0$ using that $B_t^2-t$ is a martingale).
  • $\begingroup$ I am struggling to prove that $I_{1}$ converges in $L^{2}$ to $0$. Particularly, why would the hint that $B_{t}^{2}-t$ is a martingale help us given that $(B_{t}-B_{s})^{2}-(t-s)\neq B_{t}^{2}-t-(B_{s}^{2}-s)$ $\endgroup$
    – MinaThuma
    Dec 9, 2020 at 12:04

As mentionned by saz from Itô's lemma applied to $Y_t=B^2_t$ you get :

$dY_t=finite_variation_ter.dt + 2.B_tdB_t$

so the quadratic variation is (using Itô's isometry) is equal to :

$$<Y>_t=4\int_0^t B_s^2ds$$

Please note that this is a stochastic process.

Best regards


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .