# $L^2$-quadratic variation of $X_t = B_t-tB_1$ for Brownian motion $(B_t)_{t \geq 0}$

I have a process $$X_t = B_t - tB_1$$, $$B$$ is a BM, $$t\in[0,1]$$. I have calculated its quadratic variation to be $$t$$, i.e.: $$\lim_{n\to\infty}V_n^t=\lim_{n\to\infty}\sum_{t_i\in E_n, t_{i+1}\leq t}(B_{t_{i+1}}-t_{i+1}B_1 -B_{t_{i}}+t_{i}B_1)^2 = t.$$

How can I find a $$L^2$$ limit of $$V_n^t$$, if $$E_n=\{0, 2^{-n},...,1\}$$? I presume it goes to $$t$$, but how can I show this?

I can start with: $$\lim_{n\to\infty}E\left[\left|V_n^t-t\right|^2\right]$$ and now I have to show that this converges to $$0$$. But I don't see any way to do this directly. Is there some trick which takes into the account that we have a BM, and that means some Gaussian distribution?

• The key idea is to expand $|V_n^t - t|^2,$ but for this, you will have to use some general properties of varience and recall that the increments of Brownian motion are independent. – Will M. Nov 27 '18 at 23:36

The key point is that only the term $$B_t$$ contributes to the quadratic variation. For two stochastic process $$(Y_t)_{t \geq 0}$$ and $$(Z_t)_{t \geq 0}$$ set

$$V_n^t(Y,Z) := \sum_{t_i \in E_n, t_{i+1} \leq t} (Y_{t_{i+1}}-Y_{t_i})(Z_{t_{i+1}}-Z_{t_i}) \qquad V_n^t(Y) := V_n^t(Y,Y)$$

Since $$X_t = B_t-t B_1 =: B_t-A_t$$ we have

$$V_n^t(X) = V_n^t(B) -2 V_n^t(B,A) + V_n^t(A).$$

It is well-known that the Brownian motion $$(B_t)_{t \geq 0}$$ satisfies $$V_n^t(B) \to t$$ in $$L^2(\mathbb{P})$$, and therefore it just remains to show that the other two terms on the right-hand side converge in $$L^2(\mathbb{P})$$ to $$0$$ as $$n \to \infty$$. As

$$|V_n^t(B,A)| \leq |B_1| \sup_{|u-v| \leq 2^{-n}, u,v \leq t} |B_u-B_v| \underbrace{\sum_{t_i \in E_n, t_{i+1} \leq t} (t_{i+1}-t_i)}_{\leq t}$$

we find that $$\mathbb{E}(V_n^t(B,A)^2) \leq t^2 \mathbb{E} \left( |B_1|^2 \sup_{|u-v| \leq 2^{-n}, u,v \leq t}|B_u-B_v|^2 \right).$$

The sample path of Brownian motion are uniformly continuous on $$[0,t]$$, and therefore it is not difficult to see from the dominated convergence theorem that the right-hand side converges to $$0$$ as $$n \to \infty$$. For $$V_n^t(A)$$ we note that

$$|V_n^t(A)| = |B_1|^2 \sum_{t_i \in E_n, t_{i+1} \leq t} (t_{i+1}-t_i)^2 \leq 2^{-n}|B_1|^2 \underbrace{\sum_{t_i \in E_n, t_{i+1} \leq t} (t_{i+1}-t_i)}_{\leq t}$$

and so

$$\mathbb{E}(V_n^t(A)^2) \leq 2^{-2n} t^2 \mathbb{E}(|B_1|^4) \xrightarrow[]{n \to \infty} 0.$$