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

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?

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:

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

• Why the down-vote? If something is wrong with the answer I gave, please explain what it is. – Jan Stuller 2 days ago
• I think that you have not answered the authors question. – Martinus Maximus 2 days ago
• The "added value" in my answer is to: (i) point the OP to the existence of Ito Isometry and the Martingale Property of Ito Integrals, which the OP seems unaware of. Indeed, once these concepts are known, computing the Var and Expectation of $I(t)$ is trivial. I additionally sketch how Ito Isometry and the zero-expectation property can be derived from the definition of Ito Integrals... these mechanical "proofs" are often omitted in lecture notes and pages like Wiki, but they are non-trivial if you're encountering the subject for the first time. – Jan Stuller 2 days ago
• You are not proving that the stochastic integral I(t) is normal distributed. In fact this would not be true if the integrand is not deterministic, think about $\int_0^t B_s d B_s$. – Martinus Maximus 2 days ago
• @MartinusMaximus: yes, if the integrand is stochastic or indeed linked to the integrator (i.e. $B(s)$, the "last recorded value of the realized Brownian"), it's a little bit more difficult to prove the martingality condition or zero expectation. Also, $I(t)$ is only guaranteed to be Normally distributed if the integrand is non-stochastic. But the OP states that $g(s)$ is a bounded, continuous function and from the notation it follows that it's a function of "$s$", i.e. a deterministic function, therefore I believe my sketches work ok... – Jan Stuller 2 days ago

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 $$\operatorname{law}(\int_0^t g(s) d B_s) = \mathcal{N}(0,\int_0^t g^2(s) d s).$$

• You quote Ito Isometry without proving it or sketching how it works. Ito Isometry can be used directly to show that the Variance of the Ito Integral is $\int_0^t g^2(s) d s$, as shown on Wiki. Indeed, quoting the Wiki page, >"One of its main applications is to enable the computation of variances for random variables that are given as Itô integrals": so you might as well have written "By Ito Isometry, the variance is $\int_0^t g^2(s) d s$". – Jan Stuller 2 days ago
• Yes, but why is the limit Normal-distributed? – Martinus Maximus 2 days ago
• To be more clear: We used the Itô-Isometry not to calculate the expectation or variance. We used it to show convergence of $\int_0^t g_n(s) d B_s \to \int_0^t g(s)$ in $L^2(\mathbb{P})$ (and thus in probability and distribution). The variance and mean are determined by the characteristic function. – Martinus Maximus 2 days ago