(locally) square integrable process 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.
 A: 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.
A: Begin with 
$$E [e^{a W_s} ] = e^{\frac{a^2s}{2}}.$$ 
This is true for all complex-valued $a$. Now assume $a$ is real. 
Set $X_s^a = e^{a W_s}$. Note  that $(X_s^a)^2 =X_s^{2a}$. 
Therefore for $T<\infty$
$$ E [\int_0^T (X_s^a)^2 ds ] =\int_0^T E [ e^{2aW_s} ]ds = \int_0^T  e^{2a^2 s} ds=\frac{1}{2a^2} (e^{2a^2T} -1)<\infty.$$
This implies $ \int_0^T X_s^2 ds <\infty$ a.s. and answers 2. (specifically:  a.s. the integral is finite for all rational $t$, which implies that a.s. the integral is finite for all $t$).   As for 1. take the limit $T\to\infty$ and apply monotone convergence to obtain that for  any $a$ (including the trivial case $a=0$), $E [ \int_0^\infty (X_s^a)^2 ds]=\infty$. 
Think what would happen if $a$ were  purely imaginary. 
