Is the process $X_t = \sqrt{t} \xi$ a Wiener process if $\xi \sim N(0,1)$

Question

I have the following question from my course, Let $\xi$ be an $\mathcal{N}(0,1)$ random variable and define $X_t = \sqrt{t} \xi$. Is $X_t$ a Wiener process?

I understand the proof can be completed by showing that the variance of an increment is not equal to the minimum of the time step. However, I am confused with the following proof:

We consider $t_0 < t_1 < \dots < t_k$ and see that the random variables $X_{t_{1}}-X_{t_{0}} = \sqrt{t_{1}}\xi$ and $X_{t_{2}}-X_{t_{1}} = (\sqrt{t_{2}}-\sqrt{t_{1}})\xi$ are not independent, hence the process is not Wiener.

I know this is probably a simple question, but why are these random variables not independent?

Answer 1

Let $X = a \xi$ and $Y = b \xi$ for $a, b \in \mathbb R$ and $\xi \sim N(0,1)$. Here are four approaches:

1. If $X$ and $Y$ are independent, $\mathbb E[XY] = \mathbb E[X]\, \mathbb E[Y]$. With $X$ and $Y$ as above, $$\mathbb E[XY] = a b \,\mathbb E[\xi^2] = ab \neq 0 = \mathbb E[X]\, \mathbb E[Y] $$ So, unless $a = 0$ or $b=0$, the two variables are not independent.

2. If $X$ and $Y$ are independent, then $\mathbb E[X \mid Y] = \mathbb E[X]$, a constant almost surely. With $X = a \xi$ and $Y = b \xi$ we have $$ \mathbb E[X \mid Y] = \mathbb E [ a \xi \mid b \xi] = \mathbb E \Big[ \frac{a}b b \xi \mid b \xi \Big] = \frac{a}b Y $$ which is not constant almost surely.

3. Two random variables are independent if and only if their joint moment-generating function (MGF) factorizes: $$ \mathbb E e^{s X + t Y} = \mathbb E e^{s X} \mathbb E e^{t Y}. $$ In other words, the two are independent iff we can write $\mathbb E e^{s X + t Y} = \phi(s) \psi(t)$ for some function $\phi$ and $\psi$. For $X$ and $Y$ are above, $$ \mathbb E e^{s X + tY} = \mathbb E^{(a s + t b) \xi} = \exp\Big( \frac{(as+tb)^2}{2} \Big) = f(s) g(t) \, e^{ab st} $$ which clearly cannot be factorized as the product of a function in $s$ and a function in $t$.

4. $X$ and $Y$ are independent iff $\mathbb P(X \le t, Y \le s) = f(s) g(t)$ for some function $f$ and $g$. With $X$ and $Y$, we have $$ \mathbb P(X \le t, Y \le s) = \mathbb P\Big( \xi \le \min\{\frac{t}a, \frac{s}b\}\Big) = \Phi\Big(\min\{\frac{t}a, \frac{s}b\}\Big) $$ where $\Phi$ is the CDF of $N(0,1)$. The above clearly does not factorize.

One can most likely come up with more ways ... .

Answer 2

You are correct thet it's not a Wiener process because it's increments are not independent.

Let $ D_{t_1, t_2} := X_{t_2} - X_{t_1} = (\sqrt{t_2} - \sqrt{t_1}) \xi$

It's easy to prove this, since $D_{t_1}$ and $D_{t_2}$ are both scaled versions of the same random variable $\xi$. So $$P(D_{t_1, t_2}| D_{t_3, t_4}) = P(D_{t_1, t_2}| \xi) = I_{\{D_{t_1, t_2} = (\sqrt{t_2} - \sqrt{t_1}) \xi\}} \ne P(D_{t_1, t_2}) \sim N(0, (\sqrt{t_2} - \sqrt{t_1})^2)$$

You can also say that because they are scaled versions of the same variable, they generate the same $\sigma$ algebra