We are given a process $\left(X_t\right)_{t\geq 0} = \left(e^{aW_t^2}\right)_{t\ge0}$, where $W_t$ is Wiener process, $a > 0$. Check for which $a$:
1) $\mathbb{E}\int_{0}^{\infty} X_s^2 ds <\infty$
2) $\forall t<\infty, \int_{0}^t X_s^2 ds < \infty$ a.s.
As for the first problem, I think that by Fubini and inequality $e^x \ge 1+x$, for any $a$ there is
$\mathbb{E}\int_{0}^{\infty} e^{2aW_s^2}ds \ge \mathbb{E}\int_{0}^{\infty} 1+2aW_s^2 ds= \int_{0}^{\infty} \mathbb{E}\left(1 + 2aW_s^2\right)ds = \int_{0}^{\infty}\left ( 1 + 2as \right)ds= \infty$
But I don't know how to do the second part. I'm quite new to this topic and I don't know what I'm supposed to do. Should I calculate it are just try to bound. Thanks for help.
The Brownian Motion $(W_t)_{t \geq 0}$ has (almost surely) continuous sample paths. Consequently, we have by the extrem value theorem
$$M(T,\omega) := \sup_{t \leq T} |W_t(\omega)|<\infty$$
for all $T \geq 0$ and (almost) all $\omega \in \Omega$. This implies $$X_s^2(\omega) = e^{2a W_s(\omega)^2} \leq e^{2a M(T,\omega)}$$ for all $s \in [0,T]$. Thus, $$\int_0^T X_s(\omega)^2 \, ds \leq T e^{2a M(T,\omega)} < \infty.$$ Since this holds for arbitrary $T \geq 0$ and almost all $\omega$, this finishes the proof.