Lesson 2.1: The Failure of Path Length (Proving "Infinite Wiggliness")

Welcome to Module 2! In our previous module, we established two fundamental concepts: a 'Normal' Path (smooth and predictable) and a 'Random' Path (infinitely wiggly, scaling with the square root of time).

In this lesson, we are going to use the "weird" Δt\sqrt{\Delta t} rule to prove that the path of a Brownian Motion, WtW_t, is infinitely long. This will demonstrate, with math, why "normal" calculus fails and why we need to invent a new set of rules.

Part 1: How We Measure Path Length in the "Normal World"

Let's start with a simple, predictable path. Imagine a car traveling at a constant speed of 60 mph for 1 hour. The total distance traveled is simply 60 miles.

Let's prove this using "calculus-style" summation. We break the 1-hour journey into nn tiny time steps, each with size Δt\Delta t.

Total Length=Δf\text{Total Length} = \sum \Delta f

In the "normal" world, the tiny distance Δf\Delta f is proportional to the time step Δt\Delta t: Δf=60Δt\Delta f = 60 \cdot \Delta t. So our sum is:

Total Length=(60Δt)=60Δt=60T=60 miles\text{Total Length} = \sum (60 \cdot \Delta t) = 60 \sum \Delta t = 60 \cdot T = 60 \text{ miles}

Key Takeaway: For a normal path, the sum is finite because the change Δf\Delta f is proportional to Δt\Delta t.

Part 2: Trying to Measure Path Length in the "Random World"

Now, let's try to find the "path length" of our Brownian Motion, WtW_t, from time 0 to TT. This is the "Coastline of Britain" problem: how long is a fractal?

We use the same summation logic. The "path length" is the sum of all the tiny absolute changes in position. Our goal is to calculate the value of:

Total Length=limni=1nΔWi\text{Total Length} = \lim_{n \to \infty} \sum_{i=1}^n |\Delta W_i|

Part 3: The Step-by-Step Derivation of "Infinity"

Derivation Steps

Step 1: What is the "typical size" of one step, |ΔWᵢ|?

In Lesson 1.3, we proved the "weird scaling" property. The Variance of ΔWi\Delta W_i is Δt\Delta t. The Standard Deviation (the "typical size") is Variance\sqrt{\text{Variance}}. Therefore, the "typical size" of a single random step ΔWi|\Delta W_i| is Δt\sqrt{\Delta t}.

Step 2: Replace the random step with its "typical size."

We are now trying to calculate:

Total Lengthi=1nΔt\text{Total Length} \approx \sum_{i=1}^n \sqrt{\Delta t}

Step 3: Solve the sum.

We are summing the same number, Δt\sqrt{\Delta t}, a total of nn times.

Total LengthnΔt\text{Total Length} \approx n \cdot \sqrt{\Delta t}

Step 4: Express n in terms of Δt.

Our total time is TT. We have nn steps. The size of each step is Δt\Delta t. This means Δt=T/n\Delta t = T / n, so n=T/Δtn = T / \Delta t.

Step 5: Substitute n back into our 'Total Length' equation.

Total Length(TΔt)Δt\text{Total Length} \approx \left( \frac{T}{\Delta t} \right) \cdot \sqrt{\Delta t}

Step 6: Simplify the algebra.

ΔtΔt=Δt(Δt)(Δt)=1Δt\frac{\sqrt{\Delta t}}{\Delta t} = \frac{\sqrt{\Delta t}}{(\sqrt{\Delta t}) \cdot (\sqrt{\Delta t})} = \frac{1}{\sqrt{\Delta t}}

So, our final equation is:

Total LengthTΔt\text{Total Length} \approx \frac{T}{\sqrt{\Delta t}}

Step 7: The "Aha!" Moment - Take the Limit.

Our original goal was to find the true path length, which is the limit as our "ruler" (the step size Δt\Delta t) gets infinitely small.

Total Length=limΔt0TΔt\text{Total Length} = \lim_{\Delta t \to 0} \frac{T}{\sqrt{\Delta t}} \to \infty

We have just proven that the "path length" of a Brownian Motion is infinite.

Why This Matters
  • It Proves Lesson 1.1: We just proved, mathematically, that a random path is "infinitely wiggly." Its total path length is literally infinite, just like a fractal.
  • It Proves "No Derivative": The "slope" ΔWΔt\frac{\Delta W}{\Delta t} would be "typically" ΔtΔt=1Δt\frac{\sqrt{\Delta t}}{\Delta t} = \frac{1}{\sqrt{\Delta t}}, which goes to infinity.
  • It Forces Us to Find a New Way: Summing the first power (ΔW1\sum |\Delta W|^1) is a dead end. We need a new way to measure the "total variation" of a random path.

Up Next: Quadratic Variation