TITLE: Integral of Brownian motion is Gaussian?
QUESTION [36 upvotes]: Let $(W_t)$ be a standard Brownian motion, so that $W_t \sim N(0,t)$. I'm trying to show that the random variable defined by $Z_t = \int_0^t W_s \ ds$ is a Gaussian random variable, but have not gotten very far.
I tried approximating the integral by a Riemann sum: choose $\delta, M$ such that $M\delta = t$, then the integral is approximated by
$$
\sum_{k=0}^{M-1} (W_{(k+1)\delta} - W_{k\delta}  )\delta = \delta \sum\limits_{k=0}^{M-1} X_k
$$
where using standard properties of the Brownian motion, the $X_k$'s are independent identically distributed $N(0, \delta)$ random variables. So I find that $Z_t$ is approximated by a random variable with distribution $ N(0, M\delta^3) = N(0,t\delta^2) $. Now letting $ \delta \to 0$, I find the variance of $Z_t$ is also $0$, which does not make sense to me.
Any help is appreciated!

REPLY [17 votes]: I just found out that we can use the following fact:
If $f:[0,T] \rightarrow [0,T]$ is continuous and deterministic, then
\begin{equation}
\int_{0}^T \bigg( \int_{0}^T f(s,t) \,dW_s \bigg) \,dt =  \int_{0}^T \bigg( \int_{0}^T f(s,t) \, dt \bigg) \,dW_s.
\end{equation}
Hence (I suppose that it works for piecewise continuous functions),
\begin{eqnarray}
\int_{0}^T W_t \,dt & = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dW_s \,dt \\
& = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dt  \,dW_s\\
& = &  \int_0^T T-s \,dW_s\\
& \sim & N \bigg( 0, \int_{0}^T (T-s)^2 \,ds \bigg).
\end{eqnarray}