1 files changed, 5 insertions, 27 deletions
diff --git a/content/know/concept/wiener-process/index.pdc b/content/know/concept/wiener-process/index.pdc
index 3602b44..f8610a2 100644
--- a/content/know/concept/wiener-process/index.pdc
+++ b/content/know/concept/wiener-process/index.pdc
@@ -13,14 +13,13 @@ markup: pandoc
 
 # Wiener process
 
-The **Wiener process** is a stochastic process that provides
-a pure mathematical definition of the physical phenomenon of **Brownian motion**,
+The **Wiener process** is a [stochastic process](/know/concept/stochastic-process/)
+that provides a pure mathematical definition
+of the physical phenomenon of **Brownian motion**,
 and hence is also called *Brownian motion*.
 
 A Wiener process $B_t$ is defined as any
-time-indexed [random variable](/know/concept/random-variable/)
-$\{B_t: t \ge 0\}$ (i.e. stochastic process)
-that has the following properties:
+stochastic process $\{B_t: t \ge 0\}$ that satisfies:
 
 1.  Initial condition $B_0 = 0$.
 2.  Each **increment** of $B_t$ is independent of the past:
@@ -49,28 +48,7 @@ Another consequence is invariance under "time inversion",
 by defining $\sqrt{\alpha} = t$, such that $W_t = t B_{1/t}$.
 
 Despite being continuous by definition,
-the **total variation** $V(B)$ of $B_t$ is infinite
-(informally, the curve is infinitely long).
-For $t_i \in [0, 1]$ in $n$ steps of maximum size $\Delta t$:
-
-$$\begin{aligned}
-    V_t
-    = \lim_{\Delta t \to 0} \sup \sum_{i = 1}^n \big|B_{t_i} - B_{t_{i-1}}\big|
-    = \infty
-\end{aligned}$$
-
-However, curiously, the **quadratic variation**, written as $[B]_t$,
-turns out to be deterministically finite and equal to $t$,
-while a differentiable function $f$ would have $[f]_t = 0$:
-
-$$\begin{aligned}
-    \:[B]_t
-    = \lim_{\Delta t \to 0} \sum_{i = 1}^n \big|B_{t_i} - B_{t_{i - 1}}\big|^2
-    = t
-\end{aligned}$$
-
-Therefore, despite being continuous by definition,
-the Wiener process is not differentiable,
+the Wiener process is not differentiable in general,
 not even in the mean square, because:
 
 $$\begin{aligned}