Show that $E[X_t^2]<\infty$ Show that $E[X_t^2]<\infty$, where
$$
X_t=e^{3W_t-\frac{3t}{2}}-3e^{W_t-\frac{t}{2}}\underbrace{\int_0^te^{2W_s-s}ds}_{A_t},\quad. t\geq0,
$$
where $t$ is a fixed number and $W_t$ is Brownian motion.
What I did was:


*

*By Itō's formula: $X_t=\int_0^t\left(3e^{3W_s-\frac{3s}{2}}-3e^{W_s-\frac{s}{2}}A_s\right)dW_s$

*Itō isometry:


\begin{align*}
E[X_t^2]&=E\left[\left(\int_0^t\left(3e^{3W_s-\frac{3s}{2}}-3e^{W_s-\frac{s}{2}}A_s\right)dW_s\right)^2\right]\\[1.ex]
&=E\left[\int_0^t\left(3e^{3W_s-\frac{3s}{2}}-3e^{W_s-\frac{s}{2}}A_s\right)^2ds\right]\\[1.ex]
&= ...\\[1.ex]
&= \int_0^t\left(9e^{-3s}\underbrace{E\left[e^{6W_s}\right]}_{=e^{18s}}-18e^{-2s}\underbrace{E\left[e^{4W_s}A_s\right]}_{(1)}+9e^{-s}\underbrace{E\left[e^{2W_s}A_s^2\right]}_{(2)}\right)ds
\end{align*}
My main problem is that I don't know how to make an estimation of $(1)$ and $(2)$. I've tried using Cauchy-Schwarz, but it does not help me. 
Any help would be appreciated. Thanks! 
 A: Let me remark that applying Itô's formula is really overkill (and not necessary at all). In my answer you will see that the expressions (1) and (2) defined in your question are in fact finite, so you can prove it this way - but as I said, I do not recommend to do so.

In order to show $X_t \in L^2$, it suffices to prove
$$Y_t := e^{3W_t-3t/2} \in L^2(\mathbb{P})$$
and
$$Z_t := 3 e^{W_t-t/2} A_t \in L^2(\mathbb{P}).$$
Recall that $W_t$ is a Gaussian random variable with mean $0$ and variance $t$. This implies $\mathbb{E}e^{\lambda W_t}< \infty$ for any $\lambda \in \mathbb{R}$; in fact, the exponential moments can be calculated explicitly:
$$\mathbb{E}e^{\lambda W_t} = \exp \left( \frac{\lambda^2}{2} t \right) \tag{1}$$
(see e.g. wikipedia). Therefore,
$$\mathbb{E}(Y_t^2) = e^{-3t} \mathbb{E}e^{6W_t} < \infty,$$
i.e. $Y_t \in L^2$. To show that $Z_t$ is square-integrable, we apply the Cauchy-Schwarz inequality
$$\mathbb{E}(Z_t^2) \leq 3 \sqrt{\mathbb{E}(e^{4W_t-2t})} \sqrt{\mathbb{E}(A_t^2)}.$$
As above, we have $\sqrt{\mathbb{E}(e^{4W_t-2t})}<\infty$. Consequently, it remains to show that $\mathbb{E}(A_t^2)<\infty$. Applying Jensen's inequality yields
$$\begin{align*} \mathbb{E}(A_t^2) &= \mathbb{E} \left( \left|  t \int_0^t e^{2W_s-s} \, \frac{ds}{t} \right|^2 \right) \\ &\leq t \mathbb{E} \left( \int_0^t (e^{2W_s-s})^2 \, ds \right) \\ &\leq t \int_0^t \mathbb{E}e^{4W_s} \, ds. \end{align*}$$
Using $(1)$ we get $\mathbb{E}(A_t^2)<\infty$. This finishes the proof.
