Why is $\int_0^t g(s)dB_s \sim \mathcal{N}\left(0, \int_0^t g^2(s) ds\right)$

Question

I've found in different references that $\int_0^t g(s)dB_s$ is a Wiener integral, where $g(s)$ is a bounded, continuous function and $dB_s$ is the standard Brownian Motion. Also, that $\int_0^t g(s)dB_s \sim \mathcal{N}(0, \int_0^t g^2(s) ds)$. Why is this?

My attempt so far, is to approximate it as a series of normal random variables. Since the increments $B_{t_{i+1}}-B_{t_i}$ have a distribution $\mathcal{N}(0, t_{i+1}-t_i)$ and they are pairwise independent. Then, thinking it as a Riemann integral $\sum_{i=0}^{\infty} g_i(s) (B_{t_{i+1}}-B_{t_i}) \sim \mathcal{N}(0, \sum_{i=0}^{\infty} g_i^2(s)(t_{i+1}-t_i))$ for some $s\in [t_i, t_{i+1}]$. My question is, is it posible to take the limit $\lim_{t_{i+1} \to t_i} \sum_{i=0}^{\infty} g_i^2(s) (t_{i+1}-t_i) = \int_0^t g^2(s)ds$ in this case? Why?

Answer 1

We can "derive" the fact that Ito Integral $I(t)$ is normally distributed with mean zero and variance $\int_{s=0}^{s=t}g(s)^2ds$ using two fundamental properties of Ito Integrals:

(i) The martingale property

(ii) Ito Isometry

Sketch of why $\mathbb{E}[I(t)]=0$ and why $I(t) \sim N$:

The integral $I(t):=\int_{s=0}^{s=t}g(s)dB_s$ is defined as:

$$I(t):=\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\left(B_{t_{i+1}}-B_{t_i}\right)$$

The limit above is in probability. Notice that the integrator $B(t)$ is (by definition of Ito Integral) forward-looking, and thus at any time $t_i$, the difference $\Delta(B_{t_i}):=B_{t_{i+1}}-B_{t_i}$ is independent from the integrand $g(t_i)$. Therefore:

$$\mathbb{E}[I(t)]=\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\Delta(B_{t_i})\right]=\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\mathbb{E}\left[\Delta(B_{t_i})\right]=0$$

Because each $\Delta(B_{t_i})$ is normally distributed with mean zero (by definition of Brownian motion), the sum of all the $\Delta(B_{t_i})$ terms is also normally distributed, so indeed $I(t)$ is normally distributed.

Sketch of Ito Isometry: because $\mathbb{E}[I(t)]=0$, clearly the variance of $I(t)$ must equal to $\mathbb{E}[I(t)^2]$. Now:

$$\mathbb{E}\left[I(t)^2\right]=\mathbb{E}\left[\left(\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\Delta(B_{t_i})\right)^2\right]=\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(B_{t_i})^2\right]$$

The above is true because all the cross-terms of the type $g(t_i)g(t_j)\Delta(B_{t_i})\Delta(B_{t_j})$ have expectation equal to zero whenever $j\neq i$.

The expectation $\mathbb{E}[\Delta(B_{t_i})^2]=t_{i+1}-t_i$ (by definition of Brownian motion), and so one can see that:

$$\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(B_{t_i})^2\right]\rightarrow\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(t_i)\rightarrow \int_{s=0}^{s=t}g(s)ds$$

Answer 2

We consider the general case that $g: [0,t] \longrightarrow \mathbb{R}$ is a square integrable function.

Choose a sequence of simple/elementary functions $(g_n)_{n\in\mathbb{N}}$ that converge in $L^2([0,t])$ to $g$. Then, by Ito's Isometry one has $$ \mathbb{E}\left[ \left(\int_0^t g(s) d B_s - \int_0^t g_n(s) d B_s \right)^2 \right] = \mathbb{E}\left[ \left(\int_0^t (g(s)-g_n(s)) d B_s\right)^2\right] = \|g_n-g\|_{L^2}^2 \to 0, $$ as $n\to\infty$. This implies that $\left(\int_0^t g_n(s) d B_s\right)_{n\in\mathbb{N}}$ converges in $\int_0^t g(s) d B_s$ in distribution.

Next, calculate the characteristic function of $\int_0^t g_n(s) d B_s$. As you have seen, $\int_0^t g_n(s) d B_s$ is simply a sum of independent Normal-distributed variables and you should see that this is the characteristic function of a $\mathcal{N}(0,\int_0^t g_n^2(s) d s)$-distributed random variable. Use $L^2$-convergence of $(g_n)_{n\in\mathbb{N}}$ to $g$ to show that the characteristic function of $\int_0^t g_n(s) d B_s$ converges point-wise to the characteristic function of a $\mathcal{N}(0,\int_0^t g^2(s)ds)$-distributed random variable. This implies $$ \operatorname{law}(\int_0^t g_i(s) d B_s) \Longrightarrow \mathcal{N}(0,\int_0^t g^2(s) d s) $$ as $n\to\infty$. Since the limit in distribution is unique one has $$ \operatorname{law}(\int_0^t g(s) d B_s) = \mathcal{N}(0,\int_0^t g^2(s) d s). $$