Distribution of stochastic integral 
Assume that
$\mathrm{d}S = \sigma \, \mathrm{d}W$
with initial level $S(0)$ and where $\mathrm{d}W$ is usual Brownian motion. Now
$$A(T) = \frac{1}{T} \int_0^T S(t) \, \mathrm{d}t.$$
What is the distribution of $A(T)$?

Solution:
$$S(t) = S(0) \, \sigma \, W(t) \sim \mathcal{N}(0, \sqrt{S(0)}\sigma^2t)$$
so
$$A(T) = \frac{1}{T} \int_0^T S(0) \, \sigma \, W(t) \, \mathrm{d}t$$
.... and now I'm confused. Any help appreciated.
 A: Your "solution" (to the SDE for $S(t)$) is wrong: it should be $S(t) = S(0) + \sigma W(t)$.
In any case, your problem reduces to computing the distribution of $\int_0^T W(t)\,dt$. One way to do so is by writing it as a double integral and using Fubini's theorem (which continues to hold in the stochastic context) to exchange the order of integration:
  \begin{align*}
  \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u)
  \end{align*}
The term $\int_0^T u\, dW(u)$ is a standard Ito integral, so its distribution is normal with mean $0$ and variance $\int_0^T u^2\, du = \frac{T^3}{3}$. The final answer should be easy to compute from this.
(Another way to rearrange the integral would be to use an "integration by parts" formula for $d[t W(t)]$.)
A: Since $S(t)=S(0)+\sigma W(t)$, $A(T)=S(0)+\frac1T\sigma C(T)$ where $C(T)=\int\limits_0^TW(t)\mathrm dt$. Since $C(T)$ is a barycenter of the centered normal family $(W(t))_{0\leqslant t\leqslant T}$, $C(T)$ is centered normal. Furthermore, $E[C(T)^2]=\int\limits_0^T\int\limits_0^TE[W(t)W(s)]\mathrm dt\mathrm ds=\int\limits_0^T\int\limits_0^T\min(t,s)\mathrm dt\mathrm ds=\frac13T^3$.
Thus, $A(T)$ is normal with mean $S(0)$ and variance $\frac13\sigma^2T$.
A: Let $\mathcal{F}$ be the natural filtration of $W$. Define
$$\begin{align} B &= \int_0^T W_tdt\\
 Y_t &= \mathbb{E}[B|\mathcal{F}_t]\\
                   &= \int_0^tW_udu + \mathbb{E}\left[\int_t^TW_udu\bigg|\mathcal{F}_t\right]\\
       &= \int_0^tW_udu + (T-t)W_t.
\end{align}$$
Then, by Ito's formula
$$\begin{align}dY_t &= W_tdt+TdW_t-tdW_t-W_tdt\\
              &= (T-t)dW_t.\end{align}$$
In particular the martingale $Y$ is a deterministic time change of Brownian motion and hence a Gaussian process. So $Y_u$ is normally distributed for each $u\in[0,T]$. Note that $B = Y_T$.
As $Y$ is a martingale, $\mathbb{E}[B]=0$.
By the Ito Isometry,
$$\begin{align}\text{var}(B) &= \int_0^T(T-t)^2dt \\
&= \frac{T^3}3.\end{align}$$
$A(T) = S(0) + \sigma B /T$, so
$$A(T) \sim N\left(S(0), \frac{\sigma^2T}{3}\right).$$  
