Problem with particular proof regarding infinite total variation of Brownian motion

Question

I have some problems with a proof from the last page of this pdf: Brownian motion has infinite total variation.

Could we say that variance is exactly $\frac{c_{1}}{n}$ for some constant $c_{1}$?

How Var of $V_{n}$ is calculated? Is it just a sum of variances of every absolute increment? That is every two absolute increments are independent, just like increments of B-motion withouth moduli?

And most of all how to apply Tchebychev’s inequality? I see that some kind of strange squaring took place before Tchebychev could be used, but that's that. I do not see how to get greater equal instead of less equal, and other things.

Answer 1

• Since the increments $X_{t_n}-X_{t_{n-1}},\ldots,X_{t_1}-X_{t_0}$ are independent, we know that $|X_{t_n}-X_{t_{n-1}}|,\ldots,|X_{t_1}-X_{t_0}|$ are independent. Therefore, the variance of the sum $$V_n = \sum_{j=1}^n |X_{t_j}-X_{t_{j-1}}|$$ is indeed, by Bienaymé's formula, given by the sum of variances: $$\text{var} \, (V_n) = \sum_{j=1}^n \text{var} \, (|X_{t_j}-X_{t_{j-1}}|).$$ We can even calculate the variance explicitly, but the important point for the proof is that the variance is uniformly bounded, i.e. $$\sup_{n \in \mathbb{N}} \text{var} \, (V_n)<\infty.$$
• Tschebysheff's inequality states that $$\mathbb{P}(|X-\mathbb{E}X| \geq r) \leq \frac{1}{r^2} \text{var} \, (X)$$ for any $X \in L^2$ and $r>0$. This implies $$\mathbb{P}(|X-\mathbb{E}(X)|<r) \geq 1- \frac{\text{var}(X)}{r^2} \tag{1}.$$ Since $V_n \geq 0$, we have $$\begin{align*} \mathbb{P}\left(V_n- c_0 \sqrt{n} \geq - \frac{1}{2} c_0 \sqrt{n} \right) &= \mathbb{P}\left(-V_n+c_0 \sqrt{n} \leq \frac{1}{2} c_0 \sqrt{n} \right) \\ &\stackrel{V_n \geq 0}{=} \mathbb{P}\left( |-V_n+c_0 \sqrt{n}| \leq \frac{1}{2} c_0 \sqrt{n} \right). \end{align*}$$ Applying $(1)$ with $r = \frac{1}{2} c_0 \sqrt{n}$ and $X = - V_n$ gives the claimed inequality.