One of the properties of the Wiener process is that
$$ W_{s}-W_{t} \sim \sqrt{s-t} N(0,1) $$
where $N(0,1)$ is the standard normal distribution and the relation between the Wiener process with a random, stochastic variable $\xi(t)$ is such that
$$ W_{t} = W(t) = \int_{t_{0}}^{t}\xi(t^{\prime})dt^{\prime} $$
My question is, how do I infer the relationship of $\xi(t)$ from $W(t)$? Naively, one would think that
$$ W_{s}-W_{t} = \int_{t_{0}}^{s}\xi(t^{\prime})dt^{\prime} - \int_{t_{0}}^{t}\xi(t^{\prime})dt^{\prime} \sim \sqrt{s-t} N(0,1) $$
I would first take the derivative with respect to $s$
$$ \partial_{s}\left( \int_{t_{0}}^{s}\xi(t^{\prime})dt^{\prime} - \int_{t_{0}}^{t}\xi(t^{\prime})dt^{\prime}\right) = \xi(s)= \partial_{s}(\sqrt{t-s} N(0,1)) $$
since $\partial_{s}\left(\int_{t_{0}}^{t}\xi(t^{\prime})dt^{\prime}\right) = 0$. I will then take a derivative with respect to $t$ from the original equation, such that
$$ \partial_{t}\left( \int_{t_{0}}^{s}\xi(t^{\prime})dt^{\prime} - \int_{t_{0}}^{t}\xi(t^{\prime})dt^{\prime}\right) = \xi(t)= \partial_{t}(\sqrt{t-s} N(0,1)) $$
whereby similarly, $\partial_{t}\left(\int_{t_{0}}^{s}\xi(t^{\prime})dt^{\prime}\right) = 0$. Subtracting the two equations, I have that
$$ \xi(s) - \xi(t) = \partial_{s}(\sqrt{t-s} N(0,1)) - \partial_{t}(\sqrt{t-s} N(0,1)) $$
Is this correct? I am actually quite confused and not sure what to do with the right-hand side. How does one evaluate $\partial_{s,t}(\sqrt{t-s} N(0,1))$?
Moreover, what should I do with $N(0,1)$? It is not a variable, it is just saying that it is a standard normal distribution with zero mean and unit variance. But if I need to calculate the correlation function/mean, namely
$$ \left<\left(\xi(s) - \xi(t)\right)\left(\xi(u) - \xi(v)\right)\right> = \left<\left(\partial_{s}(\sqrt{t-s} N(0,1)) - \partial_{t}(\sqrt{t-s} N(0,1)\right)\left(\partial_{u}(\sqrt{u-v} N(0,1)) - \partial_{v}(\sqrt{u-v} N(0,1)\right)\right> $$
what should I do and how should I proceed? I think I'm not understanding $N(0,1)$ and I'm taking it a bit too literally.