Integration of a deterministic function w.r.t. a Brownian motion

Question

Help me solve this problem:

Let $W_t$ be a Brownian motion and suppose $X_t = \int_{0}^{t}\delta _{s}dW_{s}$ where $\delta _{s}$ is a deterministic function. Then show that $X_t$ is a Gaussian process with mean, $m(t) = 0$ and covariance function $\rho (s,t)=\int_{0}^{min(s,t)}\delta _{s}^{2} ds$.

Edit: I am looking for a specific approach which utilizes Ito's Lemma and moment generating functions.

Answer 1

These are key properties of the Ito integral. Let $(X_t)$ be a cadlag, adapted process with $\int_0^t \mathbb{E}[X_s^2]\mathrm{d}s<\infty$. Then, the Ito integral $$I_t=\int_0^t X_s\mathrm{d}B_s$$ is well-defined, a martingale and satisfies the Ito isometry.

1. The martingale property tells you that $I_t$ has a constant mean and since $I_0=0$, we obtain $\mathbb{E}[I_t]=0$ for all $t$. This holds if $X_s$ is a stochastic process (not necessarily a deterministic function).
2. Question (ii) follows from the zero mean property: \begin{align*} \mathbb{C}\mathrm{ov}(I_t,I_s)&=\mathbb{E}[I_tI_s] \\ &= \mathbb{E}\left[\int_0^t X_u\mathrm{d}B_u \int_0^s X_\tau\mathrm{d}B_\tau\right] \\ &= \mathbb{E}\left[\int_0^t \int_0^s X_uX_\tau \mathrm{d}B_u \mathrm{d}B_\tau \right] \\ &= \mathbb{E}\left[\int_0^{t\wedge s} X_u^2 \mathrm{d}u \right]. \end{align*} This again holds if $X_s$ is a stochastic process.
3. For this point, we need that $X_s$ is a deterministic function. Recall that the Ito integral $I_t$ is defined as a limit (in mean squared sense): You take simple processes $X_s^n$ to approximate the integrand $X_s$ and define the integral of simple processes with respect to Brownian motion, i.e. \begin{align*} I_t^n=\int_0^t X_s^n\mathrm{d}B_s := \sum_{i=1}^n C_{i-1}(B_{s_i}-B_{s_{i-1}})+C_n(B_t-B_{s_n}). \end{align*} If $X_s$ is deterministic, then the $C_i$ are constants and $I_t^n$ is a sum of normally distributed random variables (the increments of Brownian motions). Thus, the Ito integral $I_t$ is just the limit of normally distributed sums and thus, Gaussian itself. If $X_s$ is any stochasic process, then the $C_i$ are random variables and $I_t^n$ is not necessarily normally distributed.