Moments and covariance
Because each W_t \sim N(0, t):
\mathbb{E}[W_t] = 0, \qquad \operatorname{Var}(W_t) = t.
The values at two different times are correlated through their shared history. For
s \le t, write W_t = W_s + (W_t - W_s):
the increment W_t - W_s is independent of W_s,
so
\operatorname{Cov}(W_s, W_t) = \operatorname{Var}(W_s) = s = \min(s, t).
In one tidy formula: \operatorname{Cov}(W_s, W_t) = \min(s, t). The
envelope below is \pm\sqrt{t} — one standard deviation of
W_t at each time. It is the widening cone the path almost always
lives inside, a direct picture of \operatorname{Var}(W_t) = t.
Self-similarity, derived property by property
Brownian motion is statistically self-similar: rescaling time and height in
the right proportion gives back the very same process. Fix a scale factor
c > 0 and define the rescaled process
B_t \;:=\; \frac{1}{\sqrt{c}}\, W_{ct} \;=\; c^{-1/2}\, W_{ct}.
The claim is that (B_t)_{t \ge 0} is itself a standard
Brownian motion. There is nothing to do but check the four defining properties one at a time —
each follows from the corresponding property of W.
(1) Starts at zero. Substituting t = 0,
B_0 = c^{-1/2}\, W_{c \cdot 0} = c^{-1/2}\, W_0 = c^{-1/2}\cdot 0 = 0,
since W_0 = 0.
(2) Independent increments. Take disjoint time intervals for
B, say [s_1, t_1] and
[s_2, t_2]. The corresponding increment of B
is
B_{t} - B_{s} = c^{-1/2}\big(W_{ct} - W_{cs}\big),
an increment of W over the interval
[cs, ct] (just scaled by the constant
c^{-1/2}). Multiplying by the time factor c
maps disjoint intervals to disjoint intervals, so the underlying
W-increments live over disjoint time spans and are independent by
W's independent-increments property; scaling by a constant does not
disturb independence.
(3) Gaussian increments with the right variance. For
s < t, write the increment and track its law step by step:
B_t - B_s = c^{-1/2}\big(W_{ct} - W_{cs}\big).
The bracket is an increment of W over
[cs, ct], whose length is ct - cs = c(t - s),
so by the Gaussian-increments property
W_{ct} - W_{cs} \sim N\big(0,\, c(t - s)\big).
Multiplying a N(0, \sigma^2) variable by a constant
a scales its variance by a^2. Here
a = c^{-1/2}, so a^2 = c^{-1} and
B_t - B_s = c^{-1/2}\big(W_{ct} - W_{cs}\big) \sim N\big(0,\; c^{-1}\cdot c(t - s)\big) = N\big(0,\, t - s\big).
The two factors of c cancel exactly — that cancellation is the whole
point — leaving the variance equal to the elapsed time t - s, just as
a standard Brownian motion demands.
(4) Continuous paths. The map t \mapsto B_t(\omega)
is c^{-1/2} times W evaluated at the
continuous time-change t \mapsto ct. A continuous function composed
with the continuous map t \mapsto ct and multiplied by a constant is
continuous, so almost every path of B is continuous because almost
every path of W is.
All four properties hold, so B is a standard Brownian motion.
Equivalently, \sqrt{c}\,W_t matches
W_{ct} in law: zoom in (or out) with the matching
\sqrt{c} rescaling of height against time, and the picture is
statistically indistinguishable — fractal noise.
For every scale factor c > 0, the rescaled process
B_t = c^{-1/2}\, W_{ct}
is again a standard Brownian motion. Equivalently, the two processes
(W_{ct})_{t \ge 0} and
(\sqrt{c}\, W_{t})_{t \ge 0} have the same law. Brownian motion is
therefore a random fractal: it looks the same, in distribution, at every
magnification, provided height is scaled by \sqrt{c} whenever time is
scaled by c.
Infinite total variation
The same roughness forces one more property, the decisive one for calculus: on any
interval the path has infinite total variation,
\sup_{\text{partitions}} \sum_i \left| W_{t_{i+1}} - W_{t_i} \right| = \infty.
The path is so wiggly that the total distance it travels is unbounded, even over a finite time.
That is precisely why the ordinary Riemann–Stieltjes integral
\int f\, dW cannot be defined the usual way — and the rescue is to
look one power up, at the quadratic variation.
Here is the heart of why \sum_i |W_{t_{i+1}} - W_{t_i}| diverges.
Cut [0, t] into n equal pieces of length
\Delta t = t/n. Each increment is
N(0, \Delta t), so its typical size is its standard deviation,
\mathbb{E}\big|W_{t_{i+1}} - W_{t_i}\big| = \sqrt{\tfrac{2}{\pi}}\,\sqrt{\Delta t} \;\propto\; \sqrt{\Delta t}.
A single increment is of order \sqrt{\Delta t} — much larger than
\Delta t for small \Delta t. Now add up
the n = t/\Delta t of them:
\mathbb{E}\!\left[\sum_{i} \big|W_{t_{i+1}} - W_{t_i}\big|\right] = n \cdot \sqrt{\tfrac{2}{\pi}}\,\sqrt{\Delta t} = \frac{t}{\Delta t}\cdot \sqrt{\tfrac{2}{\pi}}\,\sqrt{\Delta t} = \sqrt{\tfrac{2}{\pi}}\; \frac{t}{\sqrt{\Delta t}}.
As the partition refines, \Delta t \to 0 and the factor
1/\sqrt{\Delta t} \to \infty, so the expected sum diverges. The
increments are too big, individually of order \sqrt{\Delta t}, for
their absolute values to add up to anything finite. (Contrast the squared increments,
each of order \Delta t: those sum to a finite limit — that is the
quadratic variation.)
How big does the path get? The law of the iterated logarithm. Refining the
measurement of "how far out does W wander as
t \to \infty", the sharp answer is
\limsup_{t \to \infty} \frac{W_t}{\sqrt{2\, t \log \log t}} = 1 \quad\text{almost surely}.
So the envelope is not merely \sqrt{t} but a hair larger, by a
factor \sqrt{2 \log \log t} — the path repeatedly grazes this curve
and repeatedly returns. A symmetric statement holds near
t = 0 (by self-similarity), which is the precise sense in which the
path is infinitely jagged at the smallest scales.