Let $Z$ be a standard normal random variable (with mean zero and unit variance). For all $t ≥ 0$, let $X_t =\sqrt t Z$
First I need to find the distribution function which I have done by:
$$ F_t(x)= P(X_t \le x) = P(\sqrt t Z \le x) = P\left(Z \le \frac{x}{\sqrt t}\right) = F_Z\left(\frac{x}{\sqrt t}\right) $$ and therefore $X_t$ is found by $\int \frac{x}{\sqrt t} =2x \sqrt t$
I then need to show that $X_t ∼ N(0;t)$ which is basically me proving that $X_t$ is a normally distributed random variable with zero mean and variance $t$.
I'm not sure how to do this because I feel like I dont have enough information, would anyone mind helping me?
Further from that I need to prove that $X_t$ is Brownian motion, I can do this by proving three properties:
1) Every increment $X_{t+s} X_s $is normally distributed with mean $0$ and variance $\sigma^2t$ with $\sigma$ a fixed parameter.
2) For every pair of disjoint time intervals $[t_1, t_2], [t_3, t_4]$ with $t_1 < t_2 ≤ t_3 < t_4$, the increments ${X_t}_4 - {X_t}_3$ and ${X_t}_2 - {X_t}_1$ are independent random variables with distributions given in (1).
3) $X_0 = 0$ and $X_t$ is continuous at $t = 0$.
However I can't work out the exact link between these properties and my problem, any help would be greatly appreciated!
The variance of $X_t$ is just $t \mbox{Var}(Z)=t$. Also, normal variables are closed under scalar multiplication, so $X_t$ is definitely normal with mean 0 and variance $t$. You can confirm this as well by differentiating $F(x/\sqrt{t})$ and getting back a normal density with variance $t$.
However, looking closer at $X_t$, you'll realize that once you draw $Z$, it's deterministic. E.g. if you know $X_1$, then $X_t = \sqrt{t}X_1$
This suggests that intervals are not independent. For example $P(X_2-X_1<0|X_1-X_0)=P(X_2-X_1<0|X_1)$, and so if $X_1-X_0\geq 1$ (thus $X_1\geq 1$), then $P(X_2-X_1< 0|X_1\geq 1)=0$.
So $X_t$ is not a Brownian motion.