Observations/doubts on Brownian motion as limit of a random walk

Question

I quote Kuo (2006). While quoting, I will write in $\color{red}{\text{bold red}}$ all my $\color{red}{\text{observations/doubts}}$.

Consider a random walk starting at $0$ with jumps $h$ and $-h$ equally likely at times $\{\delta,2\delta,\ldots\}$, where $h$ and $\delta$ are positive numbers. Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of i.i.d. random variables with $$\mathbb{P}\{X_j=h\}=\mathbb{P}\{X_j=-h\}=\displaystyle{\frac{1}{2}}\tag{1}$$ Let $Y_{\delta,h}(0)=0$ and put $$Y_{\delta,h}(n\delta)=X_1+X_2+\ldots+X_n\tag{2}$$ [...] $$Y_{\delta,h}(t)=\displaystyle{\frac{(n+1)\delta-t}{\delta}}Y_{\delta,h}(n\delta)+\displaystyle{\frac{t-n\delta}{\delta}}Y_{\delta,h}((n+1)\delta)\tag{3}$$
[At this point, we can ask ourselves] which is the limit of the random walk $Y_{\delta,h}$ as $\delta,h\rightarrow0$? To find out the answer, compute the following limits of the characteristic function of $Y_{\delta,h}(t)$: $$\lim\limits_{\delta,h\to0}E\exp\bigg[i\lambda Y_{\delta,h}(t)\bigg]\tag{4}$$ where $\lambda\in\mathbb{R}$ is fixed.
For the heuristic derivation, let $t=n\delta$. Then [after some passages]: $$E\exp\bigg[i\lambda Y_{\delta,h}(t)\bigg]=(\cos(\lambda h))^{\frac{t}{\delta}}$$ Obviously, for fixed $\lambda$ and $t$, the limit of $\exp\bigg[i\lambda Y_{\delta,h}(t)\bigg]$ does not exist when $\delta$ and $h$ tend to $0$ independently. Thus, in order for the limit to exist we must impose a certain relationship between $\delta$ and $h$. $\color{red}{\text{(what does it mean that the limit does not exist when }\delta\text{ and } h \text{ tend to }0\text{ independently?}}$
$\color{red}{\text{In which sense "independently"?}}$$\color{red}{\text{Does that mean that we must impose a relationship between $\delta$ and $h$ such that}}$
$\color{red}{\text{whenever $\delta$ tends to $0$, so does $h$, and viceversa?}}$ $\color{red}{\text{For example, I am not mistaken, $h=\frac{1}{\delta}$ would not be a suitable}}$
$\color{red}{\text{relation. Is my interpretation correct?)}}$
If $\delta$ and $h$ are related by $h^2=\delta$, [...] $$\lim\limits_{\delta\to 0}E\exp\bigg[i\lambda Y_{\delta,h}(t)\bigg]=e^{-\frac{1}{2}t\lambda^2},\text{ }\lambda\in\mathbb{R}\tag{5}$$ Hence, assuming $h^2=\delta$, for each $t\geq0$ the limit $$B(t)=\lim\limits_{\delta\to0}Y_{\delta,h}(t)\tag{6}$$ exists in distribution.
$\color{red}{\text{(Does the limit of $Y_{\delta,h}(t)$ exist in distribution since we have shown that the limit of}}$$\color{red}{\text{the characteristic function of $Y_{\delta,h}(t)$ exists and, since we know that characteristic function}}$ $\color{red}{\text{of a random variable completely defines the probability distribution of}}$ $\color{red}{\text{that random variable, so it will completely define the probability distribution of}}$ $\color{red}{\lim\limits_{\delta\to 0}Y_{\delta,h}(t)\text{?)}}$
Moreover, we have: $$E e^{i\lambda B(t)}=e^{-\frac{1}{2}t\lambda^2},\text{ }\lambda\in\mathbb{R}\tag{7}$$ $\color{red}{\text{(Hence, by (5), we have that }\lim\limits_{\delta\to 0}E\exp\bigg[i\lambda Y_{\delta,h}(t)\bigg]=E \exp\bigg[i\lambda\lim\limits_{\delta\to0}Y_{\delta,h}(t)\bigg]}$ $\color{red}{\text{ I think that this is an application of Lebesgue's Dominated Convergence Theorem with }}$ $\color{red}{|Y_{\delta,h}(n\delta)|\leq n\delta h \text{, each } n\text{. Right?)}}$
[...] The absolute value of the slope of $Y_{\delta,h}$ in each step is $\frac{h}{\delta}=\frac{1}{\sqrt{\delta}}\to\infty$ as $\delta\to0$. Thus it is plausible that every path $B(t)$ is nowhere differentialbe. In fact, if we let $\delta=|t-s|$, then $$|B(t)-B(s)|\approx\frac{1}{\sqrt{\delta}}|t-s|=|t-s|^{\frac{1}{2}}\tag{8}$$ $\color{red}{\text{(Why does } (8)\text{ hold true? And why is that plausible that every path $B(t)$ is nowehere}}$ $\color{red}{\text{ differentiable as a consequence of }(8)\text{?)}}$
[...] The stochastic process $B(t)$ has independent increments, namely, for any $0\leq t_1<t_2<\ldots<t_n$, the random variables $$B(t_1), B(t_2)-B(t_1),\ldots, B(t_n)-B(t_{n-1})\tag{9}$$ are independent.
$\color{red}{\text{Why are increments in }(9) \text{ independent? How is it possible to show that fact?}}$

Answer 1

1. "Independently" is a bit vague, to be a bit more precise you could say "in succession" (in either order). The point is that they have to go to zero together. Not only that but the rate must be controlled. The control is exactly to set a scaling relationship between them such that the variance after a period of time $t$ (thus after $t/\delta$ jumps) remains a finite nonzero number (namely $h^2 t/\delta$, at least when $t$ is divisible by $\delta$). The fact that it is $h^2$ and not $h$ is a little bit counterintuitive, and is closely related to the fact that Brownian motion has no drift. If you considered a random walk with drift, the length scale for the drift and the length scale for the diffusion would need to scale differently in order for both to contribute in the limit of many small steps.

2. Knowing the characteristic function converges implies convergence in distribution (this is not totally obvious a priori but it is a standard theorem).

3. This statement here basically follows from continuity of characteristic functions with convergence in distribution; the characteristic function of the distributional limit is the limit of the characteristic functions.

4. (8) is basically true because a typical value of a Gaussian deviates from the mean by an amount on the order of the standard deviation, meaning that there is a low probability of a deviation by at most $\varepsilon \sigma$ for $0<\varepsilon \ll 1$ and a low probability of a deviation by more than $M \sigma$ for $M \gg 1$. This means that the slope of a secant line between $(t,B(t))$ and $(s,B(s))$ scales like $|t-s|^{-1/2}$ as $t \to s$, which blows up. The discrete analogy is also a reasonable intuitive explanation, though it may give a "misleadingly discrete" picture on what Brownian motion actually looks like.

5. What constitutes a convenient way to prove independence of increments depends on what you want to use to start developing the theory. But just about any way of doing this can be looked up in many other references. In this development it intuitively holds because the discrete analogue of this property holds for the underlying random walk, but that's not a proof of course.