Let $f(t), 0 \leq t \leq T$ be a deterministic function with $f(t) = \sum_{i=1}^na_{i-1}1_[t_{i=1}, t_i)(t)$ with $0 \leq t_0<t_1<...<t_{n-1} = T$. Show that the stochastic integral $I_t(f) = \int_0^tf(s)dW_s$ is a Gaussian process.

Following the method in this: Distribution of stochastic integral, I can show that each $I_t(f)$ is normally distributed for all $t$.

To show that $I_t(f)$ is Gaussian, I need to show that for all $0 \leq t_1 < ... < t_n$ and any $a_1, ..., a_n \in \mathbb{R}$, $a_1I_{t_1}(f) + .. + a_n I_{t_n}(f)$ is Gaussian with mean 0. This would be true if the $I_{t_i}$ were jointly normally distributed, because then any linear combination of them is also gaussian. Is this the right way of going about it, or is there a simpler way?


1 Answer 1


Your approach makes sense. Consider two times $t_1 < t_2$. We are interested in the joint moment generating function

\begin{eqnarray} \Psi \left( u_1, u_2 \right) & = & \mathbb{E} \left[ \exp \left\{ u_1 \int_0^{t_1} f(u) \mathrm{d}W_u + u_2 \int_0^{t_2} f(u) \mathrm{d}W_u \right\} \right]\\ & = & \mathbb{E} \left[ \exp \left\{ \left( u_1 + u_2 \right) \int_0^{t_1} f(u) \mathrm{d}W_u \right\} \right] \mathbb{E} \left[ \exp \left\{ u_2 \int_{t_1}^{t_2} f(u) \mathrm{d}W_u \right\} \right]. \end{eqnarray}

Here, we used the independence of the integral over $\left[ 0, t_1 \right]$ and $\left[ t_1, t_2 \right]$ in the second step to factor the expectation.

We know that for any $s_1 < s_2$

\begin{equation} \mathbb{E} \left[ \exp \left\{ u \int_{s_1}^{s_2} f(u) \mathrm{d}W_u - \frac{1}{2} u^2 \int_{s_1}^{s_2} f^2(u) \mathrm{d}u \right\} \right] = 1 \end{equation}

and thus

\begin{equation} \mathbb{E} \left[ \exp \left\{ u \int_{s_1}^{s_2} f(u) \mathrm{d}W_u \right\} \right] = \exp \left\{ \frac{1}{2} u^2 \int_{s_1}^{s_2} f^2(u) \mathrm{d}u \right\}, \end{equation}

which is the moment generating function of a normal random variable with

\begin{equation} \mathcal{N} \left( 0, \int_{s_1}^{s_2} f^2(u) \mathrm{d}u \right). \end{equation}

This holds for any $s_1 < s_2$ and $u$ and thus also for

  1. $s_1 = 0$, $s_2 = t_1$ and $u = u_1 + u_2$ and
  2. $s_1 = t_1$ and $s_2 = t_2$ and $u = u_2$.

We get

\begin{eqnarray} \Psi \left( u_1, u_2 \right) & = & \exp \left\{ \frac{1}{2} \left[ \left( u_1 + u_2 \right)^2 \int_0^{t_1} f^2(u) \mathrm{d}u + u_2^2 \int_{t_1}^{t_2} f^2(u) \mathrm{d}u \right] \right\} \nonumber\\ & = & \exp \left\{ \frac{1}{2} \left[ \left( u_1^2 + 2 u_1 u_2 \right) \int_0^{t_1} f^2(u) \mathrm{d}u + u_2^2 \int_0^{t_2} f^2(u) \mathrm{d}u \right] \right\} \end{eqnarray}

But this is just the moment generating function of a bi-variate normal random variable with

\begin{equation} \mathcal{N}_2 \left( \left( \begin{array}{c} 0\\ 0 \end{array} \right), \left( \begin{array}{c c} \int_0^{t_1} f^2(u) \mathrm{d}u & \int_0^{t_1} f^2(u) \mathrm{d}u\\ \int_0^{t_1} f^2(u) \mathrm{d}u & \int_{t_1}^{t_2} f^2(u) \mathrm{d}u \end{array} \right) \right). \end{equation}


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