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

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.

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.
• Thanks for your answer KeSchn. Is there a way to prove the same using Ito's lemma and moment generating functions since this is what was asked? I edited my question to reflect this. Do you have any idea how to do that? Is it possible to show that $E[e^{uX_{t}}]=1+\frac{1}{2}u^{2}\int_{0}^{t}\delta _{s}^{2}E[e^{uX_{s}}]ds$ Apr 11 '20 at 17:11