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

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


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

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


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


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

Thanks a lot!

  • $\begingroup$ 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. $\endgroup$ – saz Apr 27 at 19:38
  • $\begingroup$ @saz Yes that's a mistake, I meant the former $\endgroup$ – 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 $$ X_3 = \frac{W(t_{j+1}) - W(t_j)}{\sqrt{t_{j+1} - t_j}}. $$

  • $\begingroup$ Thanks a lot! We only wrote the headline 'Ito integral' last time and didn't have time to define it, so this task seems to be overkill. I added some questions I still have above $\endgroup$ – user668718 Apr 28 at 7:53
  • $\begingroup$ I have edited the answer to address those questions. $\endgroup$ – Jason Swanson Apr 30 at 14:42

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