# Gaussian Ito Integrals

In this problem, we will show that the Ito integral of a deterministic function is a Gaussian Random Variable.

Let $$\phi$$ be deterministic elementary functions. In other words there exists a  sequence of  real numbers $$\{c_k : k=1,2,\dots,N\}$$ so that

$\sum_{k=1}^\infty c_k^2 < \infty$

and there exists a partition

$0=t_0 < t_1< t_2 <\cdots<t_N=T$

so that

$\phi(t) = \sum_{k=1}^N c_k \mathbf{1}_{[t_{k-1},t_k)}(t)$

1. Show that if $$W(t)$$ is a standard brownian motion then the Ito integral
$\int_0^T \phi(t) dW(t)$
is a Gaussian random variable with mean zero and variance
$\int_0^T \phi(t)^2 dt$
2. * Let $$f\colon [0,T] \rightarrow \mathbf R$$ be a deterministic function such that
$\int_0^T f(t)^2 dt < \infty$
Then it can be shown that there exists a sequence of  deterministic elementary functions $$\phi_n$$ as above such that
$\int_0^T (f(t)-\phi_n(t))^2 dt \rightarrow 0\qquad\text{as}\qquad n \rightarrow \infty$
Assuming this fact, let $$\psi_n$$ be the characteristic function  of the random variable
$\int_0^T \phi_n(t) dW(t)$
Show that for all $$\lambda \in \mathbf R$$, show that
$\lim_{n \rightarrow \infty} \psi_n(\lambda) = \exp \Big( -\frac{\lambda^2}2 \big( \int_0^T f(t)^2 dt \big) \Big)$
Then use the the convergence result here to conclude that
$\int_0^T f(t) dW(t)$
is a Gaussian Random Variable with mean zero and variance
$\int_0^T f(t)^2 dt$
by identifying the limit of the characteristic functions above.

Note: When Probabilistic say the “characteristic function” of a random distribution they just mean the Fourier transform of the random variable. See here.