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While trying to compute $\int_0^TB_t^2\ dB_t$, $B$ being the standard Brownian motion, I got stuck at showing the following. $$\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{t_i})^2 \rightarrow \int_0^TB_t\ dt \quad \text{in} \quad L^2$$ as $mesh(\pi) \rightarrow 0$ where $\pi: 0 = t_0 < t_1 < \ldots < t_n = T$.

Since I have that $$\sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i) \rightarrow \int_0^TB_t\ dt \quad \text{in} \quad L^2$$ I figured if I can show $$\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{t_i})^2 \rightarrow \sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i) \quad \text{in} \quad L^2$$ I will have the result that I want. It is easy to see that $$E\left[\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{{t_i}})^2\right] = E\left[\sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i)\right]$$

But when I involve square of the difference of the sums to show convergence in $L^2$, things get out of hand (my hand at least). Could someone tell me if I am on the right track up until now? If yes, how do I proceed from this point on? I would appreciate any help/hint.

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1 Answer 1

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Steps:

  1. It suffices to show that $$\left\| \sum_{i=0}^{n-1} B_{t_i} \bigg( (B_{t_{i+1}}-B_{t_i})^2 - (t_{i+1}-t_i) \bigg) \right\|_{L^2} \to 0.$$
  2. Recall that $M_t := B_t^2-t$ is a martingale (with respect to the canonical filtration). For brevity of notation set $\Delta_i := B_{t_{i+1}}-B_{t_i}$ and $\Delta t_i := t_{i+1}-t_i$. Show that $$\mathbb{E}(B_{t_i} B_{t_j} \cdot (\Delta B_i^2 - \Delta t_i)(\Delta B_j^2- \Delta t_j))=0$$ whenever $i \neq j$.
  3. Conclude from step 2 that $$\left\| \sum_{i=0}^{n-1} B_{t_i} \bigg( (B_{t_{i+1}}-B_{t_i})^2 - (t_{i+1}-t_i) \bigg) \right\|_{L^2}^2 = \sum_{i=0}^{n-1} \mathbb{E}( B_{t_i}^2 (\Delta B_i^2-\Delta t_i)^2.).$$
  4. Use the independence of the increments and the fact that $\Delta B_i \sim B_{t_{i+1}-t_i} \sim \sqrt{t_{i+1}-t_i} B_1$ to conclude that the expression at the right-hand side (in step 3) converges to $0$.
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  • $\begingroup$ You are a life saver :) If you don't mind I made a minor change in your last step of the proof. $$\sum_{i=0}^{n-1}E[B_{t_i}^2(\Delta B_i^2-\Delta t_i)^2] = 2E\left[\sum_{i=0}^{n-1}B_i^2(t_{i+1}-t_i)^2\right] \leq 2 \ mesh(\pi)E\left[\sum_{i=0}^{n-1}B_i^2(t_{i+1}-t_i)\right]$$ $\sum_{i=0}^{n-1}B_i^2(t_{i+1}-t_i)$ converges to $\int_0^T B_t^2 \ dt$ in $L^2$. I know that the expectation of this integral is $\frac{T^2}{2}$ (I computed that using Fubini before). Hence $mesh(\pi)E\left[\sum_{i=0}^{n-1}B_i^2(t_{i+1}-t_i)\right]$ goes to $0$ as the mesh goes to $0$. Is this also OK you think? $\endgroup$
    – Calculon
    Commented Mar 21, 2015 at 17:28
  • $\begingroup$ @Calculon How do you prove the first "=" in your last comment? $\endgroup$
    – saz
    Commented Mar 21, 2015 at 17:39
  • $\begingroup$ $$E[B_{t_i}^2(\Delta B_i^2-\Delta t_i)^2] = E[E[B_{t_i}^2(\Delta B_i^2-\Delta t_i)^2\mid\mathcal{F}_{t_i}]] = E[B_{t_i}^2E[(\Delta B_i^2-\Delta t_i)^2]]$$ Then I compute $E[(\Delta B_i^2-\Delta t_i)^2]] = 2(t_{i+1}-t_i)^2$. $\endgroup$
    – Calculon
    Commented Mar 21, 2015 at 17:43
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    $\begingroup$ @Calculon I see, thanks; I was confused about the factor "$2$", but you are totally right. Concerning the remaining part of the argumentation: Using that $\sum B_{t_i}^2 (t_{i+1}-t_i)$ converges to $\int_0^T B_t^2 \, dt$ is correct, but rather overkill. Note that we can calculate the expectation $$\mathbb{E} \left[ \sum_{i} B_{t_i}^2 (t_{i+1}-t_i) \right]$$ explicitly. Then it it easy to see that this sum can be bounded by $T^2$. $\endgroup$
    – saz
    Commented Mar 21, 2015 at 18:44
  • $\begingroup$ @user893458 What exactly do you mean by "compute"? $\endgroup$
    – saz
    Commented Mar 29, 2017 at 18:59

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