Ito isometry and the covariance of an Ito process

Question

Let $(B_t)_{t \geq 0}$ et $(W_t)_{t \geq 0}$ be two independent Brownian motions and let $f: \mathbb{R} \rightarrow \mathbb{R}$ a deterministic function of time. We define the following process: \begin{equation} X_t = \int_0^t \frac{1}{\sqrt{1 + f(u)^2}} dB_u + \int_0^t \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u. \end{equation}

We want to prove that

1. $\forall s,t \geq 0 \; X_{t+s} - X_t \sim N(0,s)$;
2. $\forall s,t \geq 0 \; E(X_t) = 0 \text{ and } Cov(X_t, X_s) = min \{ s,t \}$.

Regarding (2), the Ito integrals are each of null expectation, hence $E(X_t) = 0$ must be true. Now, without loss of generality, assume $s < t$

\begin{align} COV(X_t, X_s) = \int_0^t \frac{1}{\sqrt{1 + f(u)^2}} \frac{1}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du \; + \\ \int_0^t \frac{f(u)}{\sqrt{1 + f(u)^2}} \frac{f(u)}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du + \\ \int_0^t \frac{2}{\sqrt{1 + f(u)^2}} \frac{f(u)}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du \\ COV(X_t, X_s) = \int_0^s \frac{1 + f(u)^2 + 2f(u)}{1 + f(u)^2} du = \int_0^s du + 2 \int_0^s \frac{f(u)}{1 + f(u)^2}du \end{align} Clearly, I did something wrong, but I don't see where. It looks like I applied the Ito isometrie correctly here. Anyone sees the problem?

Answer 1

Recall that for any deterministic function $g,$ Ito's integral follows a normal distribution: $$\int_0^t g(u) dW_u \sim N\left(0,\int_0^t g^2(u) du\right).$$ Therefore, since $$X_{t+s} - X_t = \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u + \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u,$$ each integral follows a normal distribution and they are independent, so $X_{t+s}-X_t$ follows a normal distribution with mean $0$ and variance $$\mathbb{E}\left( \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u \right)^2 + \mathbb{E}\left( \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u \right)^2 = \mathbb{E}\left( \int_t^{t+s}du \right) = s.$$