Show that for any centered Gaussian process $X$, there exists a non-decreasing function $f$ and a Wiener process $W$ such that $X(t)=W(f(t))$ is true.

Question

I am working a problem stated as follows:

Prove that for every centered Gaussian process $X$ with independent increments on $\mathbb{R}_{+}$, there is a non-decreasing non-random function $f:\mathbb{R}_{+}\longrightarrow\mathbb{R}_{+}$ such that $X$ has the same finite dimensional distribution as $Y$ defined by $Y(t)=W(f(t))$ for a Wiener Process $W$.

I do not really have any idea about this exercise. Where should I start with?

Thank you!

Answer 1

The key observation is that if the result holds for such a function $f$ then $\operatorname{Var}(X_t) = \operatorname{Var}(W(f(t)) = f(t)$. So our only choice is to define $f(t) = \operatorname{Var}(X_t)$.

Notice that for $t>s$, $$f(t) - f(s) = \mathbb{E}[X_t^2 - X_s^2] = \mathbb{E}[(X_t - X_s)^2 + 2X_s (X_t - X_s)] = \mathbb{E}[(X_t - X_s)^2] \geq 0$$ where the final equality holds by independence of increments so that $f$ is a deterministic, non-decreasing function.

Now fix times $0 \leq t_1 < \dots < t_n$. We want to check that $$(X_{t_1}, \dots, X_{t_n}) \stackrel{d}{=} (W(f(t_1)), \dots, W(f(t_n)))$$ Since these are both Gaussian vectors, it suffices to check that for $1 \leq i<j \leq n$ we have that $$\operatorname{Cov}(X_{t_i},X_{t_j}) = \operatorname{Cov}(W(f(t_i)),W(f(t_j))) = f(t_i) = \operatorname{Var}[X_{t_i}]$$ where the second to last equality holds since $f$ is increasing. This is again just a computation using independence of increments.

$$\mathbb{E}[X_{t_i}X_{t_j}] = \mathbb{E}[X_{t_i}(X_{t_j} - X_{t_i}) + X_{t_i}^2] = \operatorname{Var}(X_{t_i})$$