For a course on stochastic processes, I've been working on an exercise on fractional Brownian Motion. Showing that this process has a continuous modification is one of the final steps of the exercise, but one of the earlier steps of the exercise has left me wondering whether my proof is correct. I hope someone can relieve me of my worries.

Let $X = (X_t)_{t \geq 0}$ be a Gaussian, zero-mean stochastic process starting from 0, which has both stationary increments and is H-self similar for $H > 0$. In a series of exercises I have already shown that \begin{align*} \mathbb{E} \: X_s X_t = \tfrac{1}{2} \left( s^{2H} + t^{2H} - |t - s|^{2H} \right), \end{align*} and that we must have $H \leq 1$.

My proof that that for every $H \in (0, 1]$ the process $X$ has a continuous modification is as follows:

By stationary increments and self similarity, we can write \begin{align*} \mathbb{E} \: \left| X_s - X_t \right|^\alpha = \mathbb{E} \: \left| X_{|s - t|} \right|^\alpha = |s - t|^{\alpha H} \mathbb{E}|X_1|^\alpha, \end{align*} for any $\alpha > 0$. By taking $\alpha$ such that $\alpha H = 2$, we can take $\beta = 1$ and $K = \mathbb{E} \: |X_1|^\alpha$ and Kolmogorov's Continuity Criterion will be satisfied: \begin{align*} \mathbb{E} \: \left| X_s - X_t \right|^{\frac{2}{H}} = |s - t|^{2} \mathbb{E}|X_1|^\alpha \leq \mathbb{E} |X_1|^\alpha |s - t|^2, \end{align*} for any $s, t \geq 0$. Thus a continuous modification $X'$ exists.

Now I've seen similar statements floating around on the web, so I think there is at least some measure of legitimacy to my proof. What worries me is an earlier exercise, in which I was tasked to show that for $H = 1$, we have $X_t = t Z$ a.s., for $Z \sim \mathcal{N}(0, 1)$.

I solved this exercise by claiming that $Z = X_1$, and showing that it had the same mean and covariance function. In addition, for the almost sure equality, I showed that $\mathbb{E} \: \left( X_t - t X_1 \right)^2 = 0$, which is an equivalent statement.

If however I can claim that $X_t = t^H X_1$ very easily, as I did in my proof above, isn't it immediately clear that $X_t = t X_1 = t Z$ a.s. as well?

Can you please verify my proof, and explain what it is exactly that I'm missing here?

  • $\begingroup$ I should add that nowhere in the exercise is $X$ referred to as fractional Brownian Motion. After some research however, I think it is fractional Brownian Motion in everything but name.. $\endgroup$ – JustSomeGuy Mar 20 '15 at 21:59
  • 1
    $\begingroup$ So how would you prove $X_t = t^H X_1$ for $H<1$? $\endgroup$ – saz Mar 21 '15 at 7:06
  • $\begingroup$ @saz, I thought about that shortly before, and I ran into problems with the covariance functions. While the covariance is $(st)^{H}$ for $sX_1 t X_1$, it is just the expression above for $X_s X_t$. You get similar issues with $\mathbb{E} \: |X_t - t^H X_1|^2$.. Is $X_t \neq t^H X_1$ for $H < 1$ then? That seems in direct contradiction with the self similarity property.. $\endgroup$ – JustSomeGuy Mar 21 '15 at 14:30
  • 1
    $\begingroup$ Note that for $H=\frac{1}{2}$ the process is a ("standard") Brownian motion. So, NO, we cannot expect $X_t = t^H X_1$ for $H<1$. (At least not almost surely. The identity holds in distribution.) $\endgroup$ – saz Mar 21 '15 at 14:34
  • $\begingroup$ Ah, I understand. The self similarity property ensures that $X_t$ and $t^H X_1$ have the same distribution, but a.s. equality is stronger, and only holds for $H = 1$? Why are the covariance functions not equal then? Also, I based my proof on statements such as math.stackexchange.com/questions/257904/…, is this incorrect as well then? $\endgroup$ – JustSomeGuy Mar 21 '15 at 14:40

I suppose I'll relieve my own worries then! Of course most credits should go to @Saz, who was tremendously helpful in the comments.

The problem here was basically that of modes of convergence, I think.

While $X_t = t^H X_1$ holds, it holds only in distribution. That is the reason that when we taking expectations of these variables on both sides, we get the same value.

The original variables are not equal however, and we only have a.s. equality for H = 1.

Looking back on it, this was a pretty silly question of course..

  • 1
    $\begingroup$ +1 for taking the time to write a nice answer to your question. $\endgroup$ – saz Apr 8 '15 at 19:37
  • $\begingroup$ Thanks @saz, its really nice to hear that from a veteran such as yourself! :) $\endgroup$ – JustSomeGuy Apr 8 '15 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.