# quadratic variations of Brownian motion squared

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 !

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

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).
• 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 $(B_{t}-B_{s})^{2}-(t-s)\neq B_{t}^{2}-t-(B_{s}^{2}-s)$ 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