Existence of a Continuous Modification of Fractional Brownian Motion

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?

• 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.. – JustSomeGuy Mar 20 '15 at 21:59
• So how would you prove $X_t = t^H X_1$ for $H<1$? – saz Mar 21 '15 at 7:06
• @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.. – JustSomeGuy Mar 21 '15 at 14:30
• 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.) – saz Mar 21 '15 at 14:34
• 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? – JustSomeGuy Mar 21 '15 at 14:40

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.