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 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..