The variance of Brownian motion Currently I'm learning about Brownian motion. In the lecture slides the following definition is given.

Definition: A Wiener process $W_{t}, t \geq 0,$ is a process with
$W_{0}=0$ and with increments $W_{t}-W_{s}$ that are Gaussian random
variables with mean $E\left\{W_{t}-W_{s}\right\}=0$ and variance
$\operatorname{Var}\left\{W_{t}-W_{s}\right\}=t-s .$ Non overlapping
increments $W_{t}-W_{s}$ and $W_{v}-W_{u}$ for any $0 \leq s<t \leq
> u<v$ are independent of each other.

Also the following properties are given:

\begin{array}{l} W_{t+1}-W_{t}=\Delta W_{t}, E\left\{\Delta
> W_{t}\right\}=0, E\left\{\Delta W_{t}^{2}\right\}=\Delta t \\
W_{t+1}=W_{+}+\sqrt{\Delta t} \cdot N(0,1), \quad W_{n}=0 \end{array}

Since the variance of a increment is t-s or Δt. I would actually think that $W_(t+1) = W_t +N(0,Δt)$. The expectation of t-s or Δt is 0. Is this the same as $W_(t+1) + \sqrt(Δt)N(0,1)$? If so, how is this converted?
 A: By definition, $W_t$ has Normally distributed independent increments with Variance proportional to the increment size, that is to say that $W(t-s)=W_t-W_s\sim N(0,t-s)$ for: $0<s<t$.
For any random variable, it is true that $Var(aX)=a^2Var(X)$
For $W_t$, we can use this property to say that $W(t-s)$ equals in distribution to $(\sqrt{t-s})W(1)$, because:
$Var((\sqrt{t-s})W(1))=(t-s)Var(W(1))=t-s$
We know that $W(1)\sim N(0,1)$, and from the above, we know that $Var((\sqrt{t-s})W(1))=t-s$, so we can conclude that $\sqrt{t-s})W(1)\sim N(0,t-s)$ and so has the same distribution as $W(t-s)$.
A: The notation on the second section of quotation and in your block of questions is not quite consistent, but the underlying answer is yes.
For a standard Wiener process, as stated in the first section of quotation, the change in a period of time $\Delta t$ is a random variable normally distributed with mean $0$ and variance $\Delta t$, i.e. standard deviation $\sqrt{\Delta t}$,
so is equal to $\sqrt{\Delta t}$ times a random variable normally distributed with mean $0$ and variance $1$
