First of all, I keep on finding different answers to this question: Is Brownian motion (BM) process the same as Wiener process?or the Standard BM is Wiener?
My second question is about the integral of a BM. If a BM process is the input of an integral system, what is the autocorrelation of the output?
$Z=\int_{0}^{t}B(s)ds,t>=0 $
Is Z a stationary process? Is Z ergodic?
First Question
In mathematics, the Wiener process is a continuous-time stochastic process named in honor of Norbert Wiener. It is often called standard Brownian motion ,$\,B(0)=0\,$, after Robert Brown.
Second Question
Let $\{W_t\}_{t\ge 0}$, be a Wiener process (Standard Brownian motion). By application of Ito's lemma, we have $$ d(tW_t) = W_t dt + tdW_t+\underbrace{d[t\,,\,W_t]}_{0}\tag 1 $$ thus $$ \int_0^t W_s ds = tW_t -\int_0^t sdW_s \tag{2}= \int_0^t (t-s)dW_s $$ Thus ,by definition of Ito's integral, we can say $Z_t=\int_0^t W_s ds=\int_0^t (t-s)dW_s$ is a normal process, such that $$ \mathbb{E}\left[\int_0^t W_s ds\right] = 0\tag 3 $$ on the other hand \begin{align} & \text{Cov}\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]\tag 4 \end{align} then \begin{align} & \text{Cov}\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}\mathbb{E}\,[W_uW_v]\,\,du\,dv\tag 5 \end{align} Since $\mathbb{E}\,[W_uW_v]=\min \{\,u\,,v \}$ therefor \begin{align} & \text{Cov}\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}\min \{\,u\,,v \}\,\,du\,dv\tag 6 \end{align} For the case $s<t$
\begin{align} \int_{0}^{s}\int_{0}^{t}\min \{\,u\,,v \}\,\,du\,dv=\int_{0}^{s}\int_{0}^{s}\min \{\,u\,,v \}\,\,du\,dv+\int_{0}^{s}\int_{s}^{t}\min \{\,u\,,v \}\,\,du\,dv\tag 7 \end{align} we immediately have \begin{align} \int_{0}^{s}\int_{0}^{t}\min \{\,u\,,v \}\,\,du\,dv=\frac{1}{3}s^3+\frac{1}{2}(t-s)s^2\tag 8 \end{align} Following the same steps as described above, for the case $s > t$ we can also show \begin{align} \int_{0}^{s}\int_{0}^{t}\min \{\,u\,,v \}\,\,du\,dv=\frac{1}{3}t^3+\frac{1}{2}(s-t)s^2\tag 9 \end{align} Thus, \begin{align} \text{Cov}\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\frac{1}{3}\min\{s^3\,\,,t^3\}+\frac{1}{2}|t-s|\min\{s^2\,\,,t^2\}\tag {10} \end{align} $\text{Cov}(Z_t,Z_s)$ is not the function of $t-s$ thus $Z_t$ is not a stationary process. it is clear $$\text{Var}(Z_t)=\text{Cov}(Z_t,Z_t)=\frac 13 t^3$$ thus, the autocorrelation between times $s$ and $t$ is $$R(s,t)=\frac{\min\{s^3\,\,,t^3\}+\frac{3}{2}|t-s|\min\{s^2\,\,,t^2\}}{ts\sqrt{ts}}\tag {11}$$
