# Showing if sum converges in $L^2$ (Brownian motion)

Consider a probability space $$(\Omega, \mathcal F, P)$$ and a Brownian motion $$(W(t),t\ge 0)$$.

Let $$T>0$$, $$t_j^n=jT/n$$ and

$$\xi_j^n=\frac{1}{3}t_{j+1}^n+\frac{2}{3}t_j^n, j=0,\ldots,n-1$$

Does

$$\sum_{j=0}^{n-1}W(\xi_j^n)(W(t_{j+1}^n)-W(t_j^n))$$ converge in $$L^2$$?

I'm looking for any hints. I just used all the definitions and arrived at:

$$\sum_{j=0}^{n-1}W(\xi_j^n)(W(t_{j+1}^n)-W(t_j^n))=\sum_{j=0}^{n-1}(\frac{1}{3}\frac{(j+1)T}{n}+\frac{2}{3}\frac{jT}{n})(W(\frac{(j+1)T}{n})-W(\frac{jT}{n}))=\sum_{j=0}^{n-1}\frac{T}{n}(j+1/3)-(W(\frac{(j+1)T}{n})-W(\frac{jT}{n}))$$

What could I do next?

Questions: \begin{align*} E|J_n - T/3|^2 &= \sum_{j=0}^{n-1} \text{Var}((W(\xi_j) - W(t_j))(W(t_{j+1}) - W(t_j)))\\ \end{align*}

So for this part you use $$E(X^2)=Var(X)+E(X)$$? And because of independence you can write $$Var(\sum...)=\sum Var(...)$$. Why does each term have mean $$0$$? I can see that $$E(W(t_{j+1})-W(t_j))=0$$ but what about $$\frac{T}{3n}$$?

Which of the formulas in the pdf do you mean?

And you showed that

$$\sum_{j=0}^{n-1}W(\xi_j^n)(W(t_{j+1}^n)-W(t_j^n))$$

converges in $$L^2$$ for the exponent $$n=1$$, does this also work for arbitrary $$n$$?

Thanks a lot!

• How do you arrive at the first "=" in your calculation? Do you consider $\sum_j W(\xi_j^n) (W(t_{j+1}^n)-W(t_j^n))$ or $\sum_j \xi_j^n (W(t_{j+1}^n)-W(t_j^n))$...? Your computiations suggest the latter but the statement of your problem the former. – saz Apr 27 at 19:38
• @saz Yes that's a mistake, I meant the former – user668718 Apr 28 at 7:54

Let $$I_n = \sum_{j=0}^{n-1} W(\xi_j)(W(t_{j+1}) - W(t_j)).$$ Then $$I_n = \sum_{j=0}^{n-1} W(t_j)(W(t_{j+1}) - W(t_j)) + J_n,$$ where $$J_n = \sum_{j=0}^{n-1} (W(\xi_j) - W(t_j))(W(t_{j+1}) - W(t_j)).$$ Since the left-endpoint Riemann sums converge in $$L^2$$ to the Ito integral $$\int_0^T W(t)\,dW(t)$$, it remains only to show the convergence in $$L^2$$ of $$J_n$$.
In fact, $$J_n\to T/3$$ in $$L^2$$. To see this, write $$J_n - T/3 = \sum_{j=0}^{n-1} \left({ (W(\xi_j) - W(t_j))(W(t_{j+1}) - W(t_j)) - \frac T{3n} }\right).$$ Each term in this sum has mean $$0$$ and distinct terms are independent. Thus, \begin{align*} E|J_n - T/3|^2 &= \sum_{j=0}^{n-1} \text{Var}((W(\xi_j) - W(t_j))(W(t_{j+1}) - W(t_j)))\\ &\le \sum_{j=0}^{n-1} E[(W(\xi_j) - W(t_j))^2(W(t_{j+1}) - W(t_j))^2]. \end{align*} You can use the formula in this note (http://math.swansonsite.com/instructional/prodgaus.pdf) to show that each term in the above sum is equal to $$5T^2/(9n^2)$$. Thus, $$E|J_n - T/3|^2 \le \frac{5T^2}{9n} \to 0$$ as $$n\to\infty$$, and this shows that $$I_n \to \int_0^T W(t)\,dW(t) + \frac13T\qquad\text{in L^2}.$$
The following addresses some additional questions that were posed. First note that I have everywhere omitted the superscript $$n$$ for notational simplicity, but quantities such as $$t_j$$ and $$\xi_j$$ still depend on $$n$$, of course. All I did was to simplify the notation.
Now, regarding the fact that each term in the sum for $$J_n-T/3$$ has mean $$0$$, note that \begin{align*} E[(W(\xi_j) &- W(t_j))(W(t_{j+1}) - W(t_j))]\\ &= E[(W(\xi_j) - W(t_j))(W(t_{j+1}) - W(\xi_j) + W(\xi_j) - W(t_j))]\\ &= E[(W(\xi_j) - W(t_j))(W(t_{j+1}) - W(\xi_j))] + E[(W(\xi_j) - W(t_j))^2]. \end{align*} Since $$W(\xi_j) - W(t_j)$$ and $$W(t_{j+1}) - W(\xi_j)$$ are independent and mean $$0$$, this gives $$E[(W(\xi_j) - W(t_j))(W(t_{j+1}) - W(t_j))] = E[(W(\xi_j) - W(t_j))^2] = \xi_j - t_j = \frac T{3n}.$$ Finally, the needed formula in the linked note is on the first page and says that if $$X_1,\ldots,X_4$$ are jointly Gaussian with mean $$0$$ and variance $$1$$, then $$E[X_1X_2X_3X_4] = \rho_{12}\rho_{34} + \rho_{13}\rho_{24} + \rho_{14}\rho_{23}.$$ In the special case that $$X_1=X_2$$ and $$X_3=X_4$$, this simplifies to $$E[X_1^2X_3^2] = 1 + 2\rho_{13}^2.$$ I am applying this with $$X_1 = \frac{W(\xi_j) - W(t_j)}{\sqrt{\xi_j - t_j}},$$ and $$X_3 = \frac{W(t_{j+1}) - W(t_j)}{\sqrt{t_{j+1} - t_j}}.$$