Construction on Ito Integral with Brownian Motion

Question

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!

Answer 1

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).