I have just started learning stochastic calculus and my professor posed the following as exercises to help understand how we construct the Ito Integral.

Let $B$ be a standard Brownian motion. Fix $t>0,n>0,\delta = t/n $. Let $V_j = B_{j\delta}$ and $\Delta_j = V_{j+1} - V_j $. Evaluate the limits of the following as $n ->\infty$:

  1. $I_1(n)=\sum_j{V_{j+1}}*\Delta_j $
  2. $I_2(n)=\sum_j{\frac{1}{2}(V_{j+1}+V_j)}*\Delta_j $

I am pretty stuck on this so any hints or help would be great. I am interested in understand how to prove these. Thanks!

  • $\begingroup$ Probably you mean $V_j = B_{j \delta}$...? And how did you define/construct the Itô integral? $\endgroup$ – saz Jun 11 '15 at 6:06
  • $\begingroup$ This is what I mean, sorry I am new to using Latex type writing. And we defined the Ito integral but taking the limit of $I_1(n)$ but replacing $V_{j+1}$ with $V_j$. $\endgroup$ – EE_13 Jun 11 '15 at 15:06
  • $\begingroup$ Basically the heart of this question is getting at why we took the left endpoint instead of the right endpoint or a weighted sum of the two. Do you have some intuition as to why we did this? $\endgroup$ – EE_13 Jun 11 '15 at 15:10

By the definition of the Itô integral, we know that

$$\sum_{j=1}^n V_j \cdot \Delta_j \to \int_0^t B_s \, dB_s. \tag{1}$$

Note that we can write

$$I_1(n) = \sum_{j=1}^n V_{j+1} \Delta_j = \sum_{j=1}^n \underbrace{(V_{j+1}-V_j)}_{\Delta_j} \Delta_j + \sum_{j=1}^n V_j \Delta_j. \tag{2}$$

By $(1)$, the second term at the right-hand side converges to $\int_0^t B_s \, dB_s$ in $L^2$, so we just have to find the limit of the first term. To this end, recall the following theorem on quadratic variation of Brownian motion:

Theorem: Let $(B_t)_{t \geq 0}$ a Brownian motion. Then $$\sum_{j=1}^n (B_{t \frac{j+1}{n}}-B_{t \frac{j}{n}})^2$$ converges in $L^2$ to $t$.

Using this in $(2)$, we find that

$$I_1(n) \stackrel{n \to \infty}{\to} t + \int_0^t B_s \, dB_s. \tag{3}$$

The argumentation for the second one is very similar; I leave it to you.

Remark: Since you asked for the reason for the choice of the left endpoint: If we use the left endpoint, then it is not difficult to see that the "discrete" stochastic integral

$$\sum_{j=1}^n \phi(t_{j-1}) (B_{t_{j+1}}-B_{t_j})$$

of some function $\phi$ is a martingale. The martingale property is preserved if we take the limit and this means that any Itô integral is a (local) martingale. This turns out to be quite useful. In contrast, if we take a e.g. the right endpoint or a weighted sum, then the result is in general not a martingale (as $(3)$ shows).

  • $\begingroup$ Very clear explanation of the choice of the left endpoint, thanks! $\endgroup$ – EE_13 Jun 11 '15 at 18:02
  • $\begingroup$ If we were to consider the midpoint instead what would be the way to go about this... i.e. if we want to evaluate the limit of $I_3(n)=\sum_j{V_{j+\frac{1}{2}}}*\Delta_j $ $\endgroup$ – EE_13 Jun 11 '15 at 18:13
  • 1
    $\begingroup$ @user3692250 For fixed $\alpha \in [0,1]$ set $t_j^* := \frac{j}{n} t + \alpha \frac{1}{n}$. Then one can show that $$\sum_j B(t_j^*) \Delta_j$$ converges to $$\int_0^t B_s \, dB_s + \alpha t.$$ (left endpoint: $\alpha=0$, midpoint: $\alpha = \frac{1}{2}$, right endpoint: $\alpha=1$) $\endgroup$ – saz Jun 11 '15 at 18:18

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