# Tag Info

## Hot answers tagged stochastic-processes

8

I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}$$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since  ...

2

It is a Wiener integral as your integrand is a deterministic function of time. It is known that the Wiener integral is stationary gaussian process with independent increments. So $z(t) \sim \mathcal N\left(0, \int_0^te^{-2k(t-s) }~ds\right)$ and $(z(t)-z(s)) \amalg z(u), \ \forall u,s,t \in \mathbb R_+ \text{ such that }u\leq s, s\leq t$ or alternatively ...

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