Let $W_t$ be a Wiener process. In the demonstration provided below (which is right), it is said, knowing $s<t$, that $W_s$ and $W_t - W_s$ are independent because increments of Wiener processes are independent.
My question is : as $W_s$ and $W_t$ can be considered respectively as $W_s-W_0$ and $W_t-W_0$ (as $W_0=0$), why can't we use the same argument (that they are independent because increments of Wiener processes are independent) and say that they are uncorrelated.
Here is the computation of the covariance of $W_s$ and $W_t$ where $W_t$ is a Wiener process in which we use this property.
The covariance between $W_s$ and $W_t$ is given by:
$\text{Cov}(W_s, W_t) = \mathbb{E}[(W_s - \mathbb{E}[W_s])(W_t - \mathbb{E}[W_t])]$
Since Wiener processes have mean zero at all points, $\mathbb{E}[W_s] = \mathbb{E}[W_t] = 0$, and this simplifies to:
$\text{Cov}(W_s, W_t) = \mathbb{E}[W_s W_t]$
Now, let's consider the joint distribution of $W_s$ and $W_t$. Since Wiener processes have independent and normally distributed increments, we have:
$W_t - W_s \sim \mathcal{N}(0, t - s)$
Now, let's express $W_t$ in terms of $W_s$ and the increment $W_t - W_s$:
$W_t = W_s + (W_t - W_s)$
Now, we can write the covariance:
$\text{Cov}(W_s, W_t) = \mathbb{E}[W_s(W_s + (W_t - W_s))]$
$\text{Cov}(W_s, W_t) = \mathbb{E}[W_s^2 + W_s(W_t - W_s)]$
$\text{Cov}(W_s, W_t) = \mathbb{E}[W_s^2] + \mathbb{E}[W_s(W_t - W_s)]$
$\text{Cov}(W_s, W_t) = s + \mathbb{E}[W_s(W_t - W_s)]$
Now, $W_s$ and $W_t - W_s$ are independent because increments of Wiener processes are independent, so the expectation of their product is the product of their expectations:
$\text{Cov}(W_s, W_t) = s + \mathbb{E}[W_s] \mathbb{E}[W_t - W_s]$
$\text{Cov}(W_s, W_t) = s + 0$
Therefore, we have:
$\text{Cov}(W_s, W_t) = s$
This is the same result as: $\text{Cov}(W_s, W_t) = min(s, t)$, but for $(s < t)$.