# Ito isometry and the covariance of an Ito process

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?

• Indeed, your expression is same as what I got after applying Ito isometry. I think the problem might have an error. – Idonknow Dec 7 '19 at 7:02
• It looks like there is indeed a problem. – Stéphane Dec 8 '19 at 2:31

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 $$\mathbb{E}\left( \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u \right)^2 + \mathbb{E}\left( \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u \right)^2 = \mathbb{E}\left( \int_t^{t+s}du \right) = s.$$