Statistical property of the difference between wiener process and the integral of the wiener process

Question

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.

Answer 1

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

$$W_s - \int_0^sW_tdt \sim\mathcal{N} \left(0, \int_0^s(1-s+u)^2du \right) = \color{red}{\mathcal{N} \left(0, s-s^2 +\frac{s^3}{3} \right)}$$

Answer 2

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 $$ \begin{align} \mathbb E\left[\left(W_s-\int_0^sW_t\,dt\right)^2\right]&=\mathbb E[W_s^2]-2\int_0^s\mathbb E[W_sW_t]\,dt+\int_0^s\int_0^{s'}\mathbb E[W_tW_{t'}]\,dt\,dt'\\ &=s-s^2+\frac{s^3}{3}. \end{align}$$