Exponential Martingales - Properties This question relates to the exponential martingale,
\begin{align}
Y(t) = \exp\left(-\int_{0}^{t} \lambda(s)\,dW(s) - \tfrac{1}{2} \int_{0}^{t} \lambda^2(s)\,ds \right)
\end{align}
and specifically how we prove the process 
\begin{align}
\left( -\lambda(t)\,Y(t) : t \in [0,T]\right) \in \mathcal{L}^2
\end{align}
i.e.
\begin{align}
\int_{0}^{T} \mathbb{E}\left[\left|\,-\lambda(t)\,Y(t)\,\right|^2 \right] dt < \infty
\end{align}
My attempt at a solution is as follows:
\begin{align}
\mathbb{E}\left[\left|\,-\lambda(t)\,Y(t)\,\right|^2 \right] &= \mathbb{E} \left[ \lambda(t)^2\,Y(t)^2 \right]\\
&= \lambda(t)^2 \mathbb{E}\left[Y(t)^2\right]
\end{align}
I believe the stochastic process $\int_{0}^{t} \lambda(s)\,dW(s)$ in $Y(t)$ is normally distributed with mean 0 and variance $\sigma^2_t = \int_{0}^{t} \lambda(s)^2 ds$.
$\textbf{Question 1}:$ Could someone explain how the variance of $\sigma^2_t = \int_{0}^{t} \lambda(s)^2 ds$ is obtained? I can't quite see it.
Assuming that is correct, then
\begin{align}
\mathbb{E}\left(Y(t)^2 \right) &= \frac{1}{\sqrt{2\pi}\sigma^2_t} \int_{-\infty}^{\infty} \exp\left(2x - \sigma^2_t - \tfrac{x^2}{2\sigma^2_t}\right) dx\\
&= \exp(\sigma^2_t) \frac{1}{\sqrt{2\pi}\sigma^2_t} \int_{-\infty}^{\infty} \exp\left(-\frac{(x - 2\sigma^2_t)^2}{2\sigma^2_t} \right)\,dx\\
&= \exp(\sigma^2_t) 
\end{align}
thus,
\begin{align}
\int_{0}^{T} \mathbb{E}\left[\left|\,-\lambda(t)\,Y(t)\,\right|^2 \right] \,dt &= \int_{0}^{T} \lambda(t)^2 \mathbb{E}\left[Y(t)^2\right] \,dt\\
&= \int_{0}^{T} \lambda(t)^2 \exp\left(\int_{0}^{t} \lambda(s)^2 \, ds \right)\,dt < \infty
\end{align}
$\textbf{Question 2}:$ Would anyone be able to confirm is this is correct or whether I'm wide of the mark? Are there any other ways to show this? 
All help is appreciated.
Many thanks,
John
 A: Let $\Pi$ be a partition of the interval $[0,t]$, i.e. $\Pi : 0 = t_0 <t_1 <\ldots<t_m = t$. Note that as $\text{mesh}(\Pi) \rightarrow 0$
$$S(\Pi) := \sum_{i=0}^{m-1}\lambda(t_i)(W_{t_{i+1}}-W_{t_i}) \rightarrow \int_0^t\lambda(s)dW_s$$
This convergence is in $L^2$. 
Next, you can see that $S(\Pi)$ is a linear combination of the increments of Brownian motion, which you know to be normal and independent of one another. Therefore, $S(\Pi)$ is itself normal. 
There is a result that says if $(X_n)$ is a sequence of normal r.v.'s and $X_n \rightarrow X$ in distribution, then $X$ is either normal or a.s. a constant. We had $L^2$ convergence above, which implies convergence in distribution. Hence, $\int_0^t\lambda(s)dW_s$ must be normal (it is clearly not a constant).
The fact that $\int_0^t\lambda(s)dW_s$ has zero mean follows from the martingale property of the stochastic integral with respect to an $L^2$ martingale, which in this case is BM.
To compute its variance you can use Ito isometry, that is
$$E\left(\int_0^t\lambda(s)dW_s\right)^2 = E\int_0^t\lambda^2(s)d[W]_s = \int_0^t\lambda^2(s)ds$$
