I have a question for you. Obviously, by looking at the sample paths of a Poisson process with parameter $\lambda >0$, this process does not have a hölder-continuous version.
But why? I have the feeling that the conditions for the Kolmogorov-Chentsov theorem are not hurt. This is part of my homework and it puzzles me the whole weekend already.
A simple version of that theorem states:
" Let $X_t$ be a real-valued process. For every $T >0$ there exists $\alpha, \beta, C > 0$ with
$E[|X_t - X_s|^\alpha] \leq C|t - s|^{1 + \beta}$ on $[0,T]$
Then there exists a version of $X_t$ with hölder-continuous paths."
Assuming $t > s$ and $\alpha \geq 1$, we can use Jensen and find
$E[|X_t - X_s|^\alpha] \geq \lambda^\alpha (t - s)^\alpha$
which goes to zero, so the conditions are fulfilled (?). For $\alpha < 1$, I could calculate the fractional moments of the Poisson distribution via fractional Bell polynomials - but those are even negative, depending on $\lambda$ and $\alpha$. So they could be fulfilling the prerequisites for a given $\beta$ as well.
Where am I wrong? I have to be somewhere, but I don't find out where.
Since the Poisson process has stationary increments, i.e.
$$\mathbb{E}(|X_t-X_s|^{\alpha}) = \mathbb{E}(|X_{|t-s|}|^{\alpha}), \qquad s,t \geq 0,$$
it suffices to consider the case $s=0$. By definition $X_t$ is Poisson distributed with parameter $\lambda t$, and therefore
$$\mathbb{E}(|X_t|^{\alpha}) \geq \mathbb{E}(1 \cdot 1_{\{X_t=1\}}) = \mathbb{P}(X_t = 1) = e^{-\lambda t} (\lambda t).$$
This gives
$$\frac{1}{t} \mathbb{E}(|X_t|^{\alpha}) \geq e^{-\lambda t} \lambda \xrightarrow[]{t \to 0} \lambda.$$
Hence,
$$0 < c:= \inf_{t \in (0,1]} \frac{1}{t} \mathbb{E}(|X_t|^{\alpha})$$
which in turn implies
$$\mathbb{E}(|X_t|^{\alpha}) = t \left( \frac{1}{t} \mathbb{E}(|X_t|^{\alpha}) \right) \geq c t$$
for small $t$. In particular, this shows that an estimate of the form $\mathbb{E}(|X_t|^{\alpha}) \leq C t^{1+\beta}$ cannot hold for any $\beta>0$.