Statistical property of the difference between wiener process and the integral of the wiener process It is known that Wiener increments are stationary and normally distributed:
$$
W_{s}-W_{t} \sim N(0,s-t)
$$
Is the difference between a Wiener process and its integral also normally distributed? Namely, is
$$
W_{s}-\int_{0}^{s}W_{t}dt
$$
normally distributed ($t<s$)? If so, how do I show it? And what are its statistical properties(mean, variance)?
Edit: Here is my attempt
Based on this post, it seems that if
$$
X_{s}=\int_{0}^{s}W_tdt
$$
then $X_{s} \sim N(0,\frac{s^{3}}{3})$. Since the sum of two independent random variables is simply the sum of their means and variances, I have that
$$
W_s - X_s \sim N\left(0,s+\frac{s^{3}}{3}\right).
$$
Is this correct? I'm assuming $W_s$ and $X_s$ are independent.
 A: For all $n\in\mathbb N^*$, $W_s-\frac{s}{n}\sum_{k=1}^nW_{ks/n}$ is gaussian, so its almost sure limit $W_s-\int_0^sW_t\,dt$ is gaussian.
Then the mean is
$$
\mathbb E[W_s]-\int_0^s\mathbb E[W_t]\,dt=0,
$$
and the variance is
$$
\begin{align}
\mathbb E\left[\left(W_s-\int_0^sW_t\,dt\right)^2\right]&=\mathbb E[W_s^2]-2\int_0^s\mathbb E[W_sW_t]\,dt+\int_0^s\int_0^{s'}\mathbb E[W_tW_{t'}]\,dt\,dt'\\
&=s-s^2+\frac{s^3}{3}.
\end{align}$$
A: We transform the integral $\int_0^sW_tdt$
$$\begin{align}
\int_0^sW_tdt &= \int_0^s\left(\int_0^t dW_u  \right)dt  \\
&= \iint_{0\le u \le t \le s} dW_u \cdot dt  \\
&= \iint_{0\le u \le t \le s}  dt \cdot  dW_u \\
 &= \int_0^s\left(\int_u^s dt  \right)dW_u  \\
&= \int_0^s(s-u)dW_u  \\
\end{align}$$
Then
$$\begin{align}
\color{red}{W_s - \int_0^sW_tdt} &=\int_0^sdW_u - \int_0^s(s-u)dW_u \\
&=\color{red}{\int_0^s (1-s+u) dW_u} \\
\end{align}$$
We deduce easily that
$$W_s - \int_0^sW_tdt \sim\mathcal{N} \left(0, \int_0^s(1-s+u)^2du  \right) = \color{red}{\mathcal{N} \left(0, s-s^2 +\frac{s^3}{3}  \right)}$$
