# Is $(W_{2t}-W_{t})_{t \geqslant0}$ a brownian motion?

$W_{t}$ - Stochastic process ( Brownian motion). I need to check if $(W_{2t}-W_{t})_{t \geqslant0}$ is a Brownian motion.

I look at the independent increment property. I want to find contradiction for this property.

So $t>s\geqslant u$ for any $s$ and $u$ that are less, less and equal, respectively.

So I look at: $$cov(W_{2t} - W_t - W_{2s} - W_s, W_{2u} - W_u) = cov(W_{2t} - W_t,W_{2u} - W_u) - cov(W_{2s} - W_s,W_{2u} - W_u)$$

if I suppose that $t > 2u$ then I can say that my first covariance is $0$ due to independence of increments.

Then I have left only:

$$= 0 - cov(W_{2s} - W_s,W_{2u} - W_u)$$ which is definitely an overlapping intervals. So they have $cov \neq 0$, or simply if I put $s = u$ then $cov = s=u$

Is it correct?

One more question:

Can I check the $(W_{2t}-W_{t}) - (W_{2s}-W_{s})$ is $N(0,\text{smth})$ ? or not? Why not?

Why I am asking, I did write that $W_{2t}-W_{t}$ $N(0,t)$, logically $W_{2s}-W_{s}$ $N(0,s)$, so the difference is $N(0,t+s)$ (by difference on normal r v). BUT!

If I rearrange other way, which also seems to be ok, $(W_{2t}-W_{2s}) - (W_{t}-W_{s})$ I get $N(0,3t-3s)$.

This is nonsense: I got 2 different distributions. But, which law I break when I do such calculations?

And the last question: suppose I have stochastic process $(-W_{t})_{t \geqslant0}$.

The property that $W_{t}-W_{s}$ is $N(0,t-s)$ in our case holds:

$W_{s}-W_{t}$ has $-N(0,t-s)$ which is $N(0,t-s)$ as normal distribution is symmetric? Is it right?

I copy-pasted from wiki:

The results for the expectation and variance follow immediately from the definition that increments have a normal distribution, centered at zero. Thus $W_t = W_t-W_0 \sim N(0,t)$ Does this leads to the following. Each element of this stochastic process is normally distributed? I am getting confused as $W_t \sim N(0,t)$ from the previous formula.

• $(W_t)$ is a Brownian Motion, I'm assuming? – Shalop Oct 29 '15 at 20:49
• $Z_t$ and $Z_s$ are not independent if $\frac t s\in (.5,2)$, so their difference is not just a normal with difference of means and sum of variances. – A.S. Oct 29 '15 at 22:03
• I'm not really understanding your last question. Yes, it is true that each individual $W_t=W_t-W_0$ is normally distributed, simply by the definition of the Brownian Motion. – Shalop Oct 29 '15 at 22:03
• My last question is basically: Is $W_t$ a r.v. that is normally distributed? So can we treat the $(W_t)_{t\geqslant 0}$ as sequence of r.v. n.d.? – Ievgenii Oct 29 '15 at 22:14
• Yes, each $W_t$ is Normally distributed with variance $t-0=t$ and mean $0$. – Shalop Oct 29 '15 at 22:15

There is a slightly simpler solution. Let $Z_t:= W_{2t}-W_t$.

If $(Z_t)$ was a Brownian Motion, then it would be true that $cov(Z_t, Z_s) = s \wedge t$ for all $s,t \geq 0$. But, as one can check, this fails when $s=1$ and $t=\frac{3}{2}$, since $cov(Z_1,Z_{3/2})=1/2 \neq 1$.

In general, you should remember that a Gaussian Process with continuous paths is completely determined (in distribution) by its means and its covariances. For a Brownian Motion, the means are $0$ and covariances are $s \wedge t$.

To answer your second question: yes, it is true that $(W_{2t}-W_t)-(W_{2s}-W_s)$ is normally distributed. To prove this, assume $s<t$ without loss of generality.

When $2s\leq t$ note that $W_{2t}-W_t$, and $W_{2s}-W_s$ are independent Normals with variances $t$ and $s$ respectively, and hence their difference is Normal with variance $s+t$.

When $2s>t$, we can write $(W_{2t}-W_t)-(W_{2s}-W_s) = (W_{2t}-W_{2s})-(W_t-W_s)$ and by independence of increments we again see that this is a difference of two independent Normals with variances $2(t-s)$ and $t-s$, respectively, hence is Normal with variance $3(t-s)$.

So the different variances are a result of considering two separate cases, $2s \leq t$, and $2s>t$. Remember, increments are independent only when the intervals don't overlap.