Quant Academy Math · Probability · Quant Finance
Phase 8 - Lesson 8.4

Quadratic Variation and Path Properties of Brownian Motion

The single fact that makes stochastic calculus different: smooth paths have zero quadratic variation, but Brownian motion accumulates \([B]_t=t\).

⏱ 55 min● Advanced🔗 Prereqs: 8.3
↖ Phase 8 hub
Builds on: 8.3 gave BM’s axioms and its jagged, non-differentiable paths.
Leads to: \([B]_t=t\), i.e. \((dB)^2=dt\), is the engine of Ito’s lemma in Phase 9.

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Key Vocabulary

Quadratic variation
\([f]_t=\lim_{\|\Pi\|\to0}\sum_k (f(t_{k})-f(t_{k-1}))^2\) over partitions \(\Pi\) of \([0,t]\).
Total (first-order) variation
\(\sum_k|f(t_k)-f(t_{k-1})|\); finite for smooth paths, infinite for BM.
Partition mesh
\(\|\Pi\|=\max_k(t_k-t_{k-1})\); the size of the largest sub-interval.
Realized variance
The empirical quadratic variation \(\sum(\Delta B)^2\) of a price path; converges to \(\int\sigma^2\,dt\).
Ito multiplication rule
\((dB_t)^2=dt,\ (dt)^2=0,\ dt\,dB_t=0\); the algebra of second-order terms.
Predictable variation
For \(B\), the compensator making \(B_t^2-t\) a martingale is exactly \([B]_t=t\).

Intuition & Motivation

Intuition
Ordinary calculus quietly assumes that squared increments are negligible: over a tiny step \(dt\), a smooth function changes by \(f'(t)\,dt\), so \((df)^2\sim(dt)^2\) vanishes faster than \(dt\) and drops out. This is why the second-order term in a Taylor expansion never survives for smooth paths - their quadratic variation is zero. Brownian motion breaks this. Its increment over \(dt\) is of size \(\sqrt{dt}\), so the square is of size \(dt\) - the same order as \(dt\) itself, and it does not vanish. Summed up, the squared increments converge to \(t\): \([B]_t=t\). That surviving second-order term, written \((dB)^2=dt\), is the entire reason Ito’s formula has an extra half-derivative term that classical calculus lacks. This lesson is the hinge of the whole subject.

Quadratic variation defined

Definition - Quadratic variation along a partition
For a partition \(0=t_0\lt t_1\lt \cdots\lt t_N=t\) with mesh \(\|\Pi\|\), the quadratic variation of \(f\) on \([0,t]\) is
\[[f]_t=\lim_{\|\Pi\|\to0}\ \sum_{k=1}^{N}\big(f(t_k)-f(t_{k-1})\big)^2,\] (8.6)
when the limit exists (for BM, in the mean-square / probability sense).

Smooth paths: quadratic variation is zero

Theorem - Continuously differentiable paths have zero QV
If \(f\) is \(C^1\) (indeed, Lipschitz) on \([0,t]\), then \([f]_t=0\).
Proof
By the mean value theorem, \(f(t_k)-f(t_{k-1})=f'(\xi_k)(t_k-t_{k-1})\) for some \(\xi_k\). With \(M=\max|f'|\), \(\sum_k (f(t_k)-f(t_{k-1}))^2\le M^2\sum_k(t_k-t_{k-1})^2\le M^2\,\|\Pi\|\sum_k(t_k-t_{k-1})=M^2\,\|\Pi\|\,t\to0\) as the mesh \(\|\Pi\|\to0\). The squared increments are \(O(\|\Pi\|^2)\) each and there are \(O(1/\|\Pi\|)\) of them, so the sum vanishes.
Key Idea
For smooth paths, \((\Delta f)^2\) is second-order small, so the quadratic variation is 0 and classical calculus keeps only first-order terms. Brownian motion is exactly where this fails.

Brownian motion: quadratic variation equals t

Theorem - Quadratic variation of Brownian motion
For standard Brownian motion, \([B]_t=t\); precisely, \(\sum_{k}(B_{t_k}-B_{t_{k-1}})^2\to t\) in \(L^2\) (and in probability) as \(\|\Pi\|\to0\).
Proof
Let \(\Delta_k=B_{t_k}-B_{t_{k-1}}\sim\Normal(0,\delta_k)\) with \(\delta_k=t_k-t_{k-1}\), independent across \(k\). Then \(\E[\sum_k\Delta_k^2]=\sum_k\delta_k=t\). For the variance, using \(\Var(\Delta_k^2)=2\delta_k^2\) for a Gaussian, \(\Var(\sum_k\Delta_k^2)=\sum_k 2\delta_k^2\le 2\|\Pi\|\sum_k\delta_k=2\|\Pi\|\,t\to0.\) So the sum has mean \(t\) and variance \(\to0\): it converges to the constant \(t\) in \(L^2\).

The mean is \(t\) for every partition; what the mesh-zero limit adds is that the fluctuation around \(t\) disappears, so the random sum collapses to a deterministic number. This is why \([B]_t=t\) is not a statement about one path’s luck but a law.

Two immediate corollaries: (i) BM has infinite total variation on every interval (else its QV would be 0), so you cannot integrate against \(dB\) path-by-path as a Stieltjes integral - you need the Ito integral. (ii) \(B_t^2-t\) is a martingale, with \(t\) the compensator equal to \([B]_t\).

The heuristic that runs Ito calculus

\[(dB_t)^2=dt,\qquad (dt)^2=0,\qquad dt\,dB_t=0.\] (8.7)

Read (8.7) as the differential shorthand for \([B]_t=t\). When you Taylor-expand \(f(B_t)\), the second-order term \(\tfrac12 f''(B_t)(dB_t)^2\) does not vanish - it equals \(\tfrac12 f''(B_t)\,dt\). That surviving term is the Ito correction, the difference between \(df=f'\,dB\) (wrong) and Ito’s formula (right).

Worked Example - Contrast: a line vs a Brownian path over [0,1]
1
Smooth path \(f(t)=3t\). Partition \([0,1]\) into \(N\) equal pieces. Each increment is \(3/N\), squared is \(9/N^2\); summing \(N\) of them gives \(9/N\to0\). So \([f]_1=0\).
2
Its total variation is \(\sum|Δf|=N\cdot(3/N)=3\), finite - the smooth path is well-behaved for ordinary integration.
3
Brownian path \(B\). Each increment \(\Delta_k\sim\Normal(0,1/N)\); \(\E[\Delta_k^2]=1/N\); summing \(N\) gives mean \(N\cdot(1/N)=1\), and the variance \(2N\cdot(1/N)^2=2/N\to0\).
4
So \([B]_1=1\) with vanishing fluctuation, while its total variation \(\sum|\Delta_k|\approx N\cdot\sqrt{1/N}=\sqrt N\to\infty\).
5
The line has finite length and zero QV; the Brownian path has infinite length and QV equal to the elapsed time \(t=1\). That gap is the entire reason for a new calculus.

Interactive: estimate quadratic variation numerically

Common Mistakes to Avoid
  • Thinking \([B]_t\) is random in the limit; the fluctuation vanishes, so \([B]_t=t\) is deterministic.
  • Dropping the \((dB)^2\) term as you would for smooth calculus - for BM it equals \(dt\) and must be kept.
  • Confusing quadratic variation (\(=t\), finite) with total variation (\(=\infty\)) for BM; they are different limits.
  • Believing a finer partition makes the QV grow without bound; it converges to \(t\), it does not diverge.
Quant Practitioner Tips
  • Memorize \((dB)^2=dt\); it is the single rule you will apply thousands of times in Ito calculus.
  • Realized variance from high-frequency returns is a direct estimate of \(\int_0^T\sigma_t^2\,dt\) - quadratic variation is measurable in markets.
  • If a manipulation of \(B\) seems to need its derivative, replace the second-order term using \((dB)^2=dt\) instead.

Knowledge Check

Q1 Medium
The quadratic variation of a continuously differentiable path on \([0,t]\) is:
\(t\)
0
infinite
equal to its total variation
Q2 Medium
For standard Brownian motion, \(\sum_k(B_{t_k}-B_{t_{k-1}})^2\) as the mesh \(\to0\) converges to:
0
\(t\)
\(t^2\)
a random Gaussian
Q3 Hard
The identity \((dB_t)^2=dt\) is important because it:
Makes Brownian paths differentiable
Produces the extra second-order term in Ito’s formula
Shows BM has finite total variation
Implies \(\Var(B_t)=t^2\)

Practical Exercise

Consider \(X_t=\sigma B_t\) with constant \(\sigma\gt 0\). (a) Show \([X]_t=\sigma^2 t\). (b) Explain why the smooth drift term in \(Y_t=\mu t+\sigma B_t\) contributes nothing to \([Y]_t\), so \([Y]_t=\sigma^2 t\).

▶ Show full solution

(a) Increments \(\Delta X_k=\sigma\Delta B_k\), so \(\sum(\Delta X_k)^2=\sigma^2\sum(\Delta B_k)^2\to\sigma^2\cdot t=\sigma^2 t\) using \([B]_t=t\).

(b) Write \(\Delta Y_k=\mu\,\delta_k+\sigma\Delta B_k\). Then \((\Delta Y_k)^2=\mu^2\delta_k^2+2\mu\sigma\,\delta_k\Delta B_k+\sigma^2(\Delta B_k)^2\). Summing: the \(\mu^2\delta_k^2\) term is \(\le \mu^2\|\Pi\|t\to0\) (smooth part, zero QV); the cross term \(\to0\) as well (order \(\|\Pi\|^{1/2}\)); only \(\sigma^2\sum(\Delta B_k)^2\to\sigma^2 t\) survives. Hence \([Y]_t=\sigma^2 t\): drift is invisible to quadratic variation, only diffusion matters.

After the reveal, answer for yourself: Why does this mean the drift \(\mu\) cannot be estimated from the quadratic variation of a single path?

Lesson Summary

Quadratic variation measures accumulated squared increments. Smooth (\(C^1\)) paths have \([f]_t=0\) because their squared increments are second-order small, so classical calculus keeps only first-order terms. Brownian motion instead has \([B]_t=t\): its \(\sqrt{dt}\)-sized increments make \((dB)^2\) order \(dt\), which survives as the Ito correction \((dB)^2=dt\) - the defining break from ordinary calculus.

Formula Sheet Additions

Brownian QV
\[[B]_t=t,\qquad (dB_t)^2=dt\]
The surviving second-order term; the engine of Ito’s lemma.
Scaled QV
\[X_t=\sigma B_t\Rightarrow [X]_t=\sigma^2 t\]
Realized variance measures accumulated \(\sigma^2\).
Error Log Checklist
  • Did I keep the \((dB)^2=dt\) term instead of discarding it?
  • Did I distinguish quadratic variation (\(=t\)) from total variation (\(=\infty\))?
  • Did I remember drift contributes 0 to QV?

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: Why is the quadratic variation of a smooth path zero?
A: Each squared increment is \(O(\|\Pi\|^2)\) and there are \(O(1/\|\Pi\|)\) of them, so the sum is \(O(\|\Pi\|)\to0\).
Q: State \([B]_t\) and the differential rule it yields.
A: \([B]_t=t\) (in \(L^2\)); equivalently \((dB_t)^2=dt\).
Q: Why does BM need a new calculus rather than Stieltjes integration?
A: Its total variation is infinite, so path-by-path Stieltjes integration against \(dB\) fails; the finite quadratic variation \(t\) forces the Ito construction.

Flashcards

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Smooth vs Brownian QV
\(C^1\) path: \([f]_t=0\). Brownian motion: \([B]_t=t\).
Ito rule
\((dB)^2=dt\), \((dt)^2=0\), \(dt\,dB=0\).

Completion Checklist

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Source References

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