Chapter 03

Hard

Quadratic Variation and Why Calculus Breaks

00 · Symbol Glossary

$\Pi_n$partition of $[0,t]$

A finite set of points $0=t_0<t_1<\cdots<t_n=t$ dividing $[0,t]$ into subintervals, used to build a Riemann-sum-style approximation. Mesh $|\Pi_n| = \max_i (t_i-t_{i-1})$ is the width of the widest subinterval.

$V_t$total variation

$V_t = \sup_{\Pi} \sum_i |f(t_i)-f(t_{i-1})|$ , the supremum over all partitions of the sum of absolute increments — the total distance a path travels, not its net displacement.

$[W,W]_t$quadratic variation of $W$ on $[0,t]$

The limit in probability of $\sum_i (W_{t_i}-W_{t_{i-1}})^2$ over partitions with mesh $\to 0$ . This chapter's central computation: $[W,W]_t = t$ .

$(\Delta W_i)^2$squared increment

Shorthand for $(W_{t_i}-W_{t_{i-1}})^2$ , the term summed to build quadratic variation. Each term has mean $t_i-t_{i-1}$ and is itself random — quadratic variation is a limit of a random sum, and its value $t$ is the fact that the randomness washes out in that particular limit.

01 · Total Variation vs. Quadratic Variation

A smooth deterministic path and a Brownian path can both be continuous, yet they measure "how much a path moves" in completely different ways once the increments are squared instead of just summed.

Definition — Total Variation

For a function $f$ on $[0,t]$ and a partition $\Pi_n=\{0=t_0<\cdots<t_n=t\}$ , the total variation is

V_t(f) = \sup_{\Pi_n} \sum_{i=1}^{n} \big| f(t_i)-f(t_{i-1}) \big|

$f$ has bounded variation if $V_t(f)<\infty$ . Every continuously differentiable function on $[0,t]$ has bounded variation — $V_t(f)=\int_0^t |f'(s)|\,ds$ .

Plain language: total variation adds up the absolute size of every wiggle, so it grows large for a path that oscillates a lot, even if the oscillations cancel and the path ends up near where it started.

Definition — Quadratic Variation

The quadratic variation of $f$ on $[0,t]$ is the limit, if it exists, of

[f,f]_t = \lim_{|\Pi_n|\to 0} \sum_{i=1}^{n} \big(f(t_i)-f(t_{i-1})\big)^2

as the mesh $|\Pi_n|\to 0$ .

For a continuously differentiable $f$ , this limit is always $0$ : each squared increment is $O((\Delta t_i)^2)$ , and summing $n\sim t/|\Pi_n|$ such terms gives a sum of order $|\Pi_n|\cdot t \to 0$ . Smooth functions have zero quadratic variation and (generically) positive, finite total variation. Brownian motion reverses both of these facts.

02 · The Central Result: $[W,W]_t = t$

Theorem — Quadratic Variation of Brownian Motion

For a standard Wiener process $W$ and any sequence of partitions of $[0,t]$ with mesh $\to 0$ ,

[W,W]_t = \lim_{|\Pi_n|\to 0} \sum_{i=1}^{n} (W_{t_i}-W_{t_{i-1}})^2 = t \quad \text{(limit in probability, or a.s. along a suitable subsequence)}

The computation behind the theorem is a mean/variance argument, not a deep analytic one. Write $\Delta_i = W_{t_i}-W_{t_{i-1}}$ and $\Delta t_i = t_i - t_{i-1}$ .

Step-by-step — Why the random sum converges to the deterministic value $t$

Take the mean of the sum:

\mathbb{E}\!\left[\sum_i \Delta_i^2\right] = \sum_i \mathbb{E}[\Delta_i^2] = \sum_i \mathrm{Var}(\Delta_i) = \sum_i \Delta t_i = t

, using

\Delta_i \sim N(0,\Delta t_i)

from axiom (iii) of Chapter 02.

Take the variance of the sum: by independence of increments (axiom (ii)),

\mathrm{Var}\!\left(\sum_i \Delta_i^2\right) = \sum_i \mathrm{Var}(\Delta_i^2)

. For

Z\sim N(0,h)

\mathrm{Var}(Z^2)=2h^2

, so this equals

\sum_i 2(\Delta t_i)^2

Bound the variance by the mesh:

\sum_i 2(\Delta t_i)^2 \leq 2|\Pi_n|\sum_i \Delta t_i = 2|\Pi_n|\,t \to 0

|\Pi_n|\to 0

, since each

\Delta t_i \le |\Pi_n|

Conclude by Chebyshev: the sum has mean exactly

t

(step 1) and variance shrinking to

0

(step 3), so the sum converges to the constant

t

in probability — a random quantity with vanishing variance converges to its own mean.

Plain language: the sum of squared increments is random for any fixed partition, but its randomness averages away as the partition gets finer, leaving the deterministic answer $t$ . This is the opposite of total variation, which for Brownian motion is infinite almost surely — the path moves "a lot" in the total-variation sense while its squared movement converges to something perfectly deterministic.

Infinite total variation — stated, not proven

$V_t(W) = \infty$ almost surely for Brownian motion. This chapter does not reproduce that proof (it uses a similar partition argument combined with the law of the iterated logarithm); the deliberate omission does not affect anything downstream, since quadratic variation, not total variation, is the quantity stochastic calculus is built on.

03 · Why the Riemann–Stieltjes Integral Fails

Classical calculus defines $\int_0^t f\,dg$ as a Riemann–Stieltjes integral: a limit of sums $\sum_i f(\xi_i)\big(g(t_i)-g(t_{i-1})\big)$ over shrinking partitions, for a sample point $\xi_i$ in each subinterval. That construction requires the integrator $g$ to have bounded variation — otherwise the sums do not converge to a value independent of how $\xi_i$ is chosen inside $[t_{i-1},t_i]$ .

❌ Defining $\int_0^t W_s\, dW_s$ pathwise as a Riemann–Stieltjes integral

Approximate $\int_0^t W_s\,dW_s$ by $\sum_i W_{\xi_i}(W_{t_i}-W_{t_{i-1}})$ for two natural choices of sample point: left endpoint $\xi_i = t_{i-1}$ , and midpoint $\xi_i = (t_{i-1}+t_i)/2$ .

Why it breaks: because $W_t(\omega)$ has infinite total variation for almost every path $\omega$ (Section 02's NoteBlock), the Riemann–Stieltjes sums do not converge to a common value as $\xi_i$ ranges over $[t_{i-1},t_i]$ . The left-endpoint sum converges (in probability) to $\tfrac12 W_t^2 - \tfrac12 t$ ; the midpoint sum converges to $\tfrac12 W_t^2$ . The two limits differ by exactly $\tfrac12 t$ , and no choice of $\xi_i$ is more "correct" than another under the classical definition.

Consequence: $\int_0^t W_s\,dW_s$ has no meaning as an ordinary Riemann–Stieltjes integral — the answer depends on an arbitrary convention. Chapter 05 resolves this by defining the Itô integral with the left-endpoint convention fixed as part of the definition, which is what makes the Itô integral a specific object rather than an ill-posed one.

04 · Second-Order Terms Do Not Vanish

In ordinary calculus, a Taylor expansion of $f(x+dx)$ keeps the first-order term $f'(x)\,dx$ and discards $(dx)^2$ as negligible — legitimate because for a smooth deterministic path, $(dx)^2 = O(dt^2)$ , which vanishes faster than $dt$ itself when integrated.

Example — Why $(dx)^2$ is negligible for smooth paths

If $x_t$ is differentiable with $dx = x'(t)\,dt$ , then $(dx)^2 = (x'(t))^2\,dt^2$ . Summing over $n=t/\Delta t$ subintervals gives a total contribution of order $n\cdot\Delta t^2 = t\cdot\Delta t \to 0$ . Discarding second-order terms in ordinary calculus is not an approximation — it is exact in the limit.

For Brownian motion, the same bookkeeping gives a different answer. $(dW)^2$ is not $O(dt^2)$ ; by Section 02, when summed over a partition, $\sum_i (\Delta W_i)^2 \to t = \int_0^t ds$ . Heuristically, $(dW_t)^2$ behaves like $dt$ itself, not like a negligible higher-order term.

dW_t \cdot dW_t = dt \qquad \text{(informal shorthand for } [W,W]_t = t\text{)}

Plain language: the "square of a small random step" is the same order of magnitude as "a small step in time" — not smaller. Any Taylor expansion of $f(W_t)$ that drops the $(dW_t)^2$ term the way ordinary calculus drops $(dx)^2$ is discarding a term that does not actually vanish, which understates the drift of $f(W_t)$ by exactly the term $\tfrac12 f''(W_t)\,dt$ . Correcting for this missing term is precisely Itô's Lemma, covered in Chapter 06.

05 · Exercises

EXERCISE 3.1

Use $\mathbb{E}[\Delta_i^2]=\mathrm{Var}(\Delta_i)$ and sum over a partition with $n$ equal subintervals of length $t/n$ .

With $n$ equal subintervals, $\Delta t_i = t/n$ for each $i$ , so $\mathbb{E}[\sum_i \Delta_i^2] = \sum_{i=1}^n \Delta t_i = n\cdot(t/n) = t$ regardless of $n$ . The expected value of the sum is exactly $t$ for every partition, not just in the limit — the limit theorem in Section 02 is about the sum's variance collapsing to $0$ , not about its mean changing.

For a partition of $[0,t]$ into $n$ equal subintervals, compute $\mathbb{E}\!\left[\sum_i (\Delta W_i)^2\right]$ exactly (not just in the limit) and explain why the answer does not depend on $n$ .

EXERCISE 3.2

Apply the variance bound from Step 3 of Section 02 with mesh $|\Pi_n| = t/n$ .

$\mathrm{Var}\!\left(\sum_i \Delta_i^2\right) \leq 2|\Pi_n|\,t = 2(t/n)\,t = 2t^2/n$ . This tends to $0$ as $n\to\infty$ at rate $1/n$ , confirming convergence in probability (and, with more work using a Borel–Cantelli argument along a subsequence like $n=2^k$ , almost-sure convergence). The rate $1/n$ also shows how fast a numerical simulation's estimate of quadratic variation improves as the partition is refined.

For $n$ equal subintervals of $[0,t]$ , bound $\mathrm{Var}\!\left(\sum_i (\Delta W_i)^2\right)$ in terms of $n$ and $t$ , and state the rate at which this variance shrinks as $n\to\infty$ .

EXERCISE 3.3

Expand $W_{t_i}(W_{t_i}-W_{t_{i-1}})$ and $W_{t_{i-1}}(W_{t_i}-W_{t_{i-1}})$ using $a b = \tfrac12(a^2+b^2-(a-b)^2)$ style identities, then sum by telescoping.

Left endpoint: $\sum_i W_{t_{i-1}}\Delta_i = \sum_i W_{t_{i-1}}(W_{t_i}-W_{t_{i-1}})$ . Using $ab=\tfrac12\big[(a+b)^2-a^2-b^2\big]$ with $a=W_{t_{i-1}}$ , $b=\Delta_i$ : $W_{t_{i-1}}\Delta_i = \tfrac12\big[W_{t_i}^2 - W_{t_{i-1}}^2 - \Delta_i^2\big]$ . Summing telescopes the first two terms to $\tfrac12 W_t^2$ , leaving $\tfrac12\sum_i \Delta_i^2 \to \tfrac12 t$ . So the left-endpoint sum $\to \tfrac12 W_t^2 - \tfrac12 t$ . Midpoint sums instead symmetrize the increment and the quadratic-variation correction cancels, converging to $\tfrac12 W_t^2$ . The two differ by $\tfrac12 t$ , matching Section 03's FailBlock.

Show algebraically that the left-endpoint Riemann–Stieltjes sum for $\int_0^t W_s\,dW_s$ converges to $\tfrac12 W_t^2 - \tfrac12 t$ , using the telescoping identity for $\sum_i W_{t_{i-1}}(W_{t_i}-W_{t_{i-1}})$ .

06 · Chapter Summary

Concept	Meaning
Total variation $V_t$	sum of absolute increments; infinite a.s. for $W$
Quadratic variation $[W,W]_t$	limit of squared increments; equals $t$ exactly
Smooth functions	zero quadratic variation, generically finite total variation
Brownian motion	infinite total variation, finite deterministic quadratic variation
Riemann–Stieltjes fails	requires bounded variation; $\int W_s\,dW_s$ depends on sample-point convention without one
$(dW_t)^2 = dt$	informal shorthand — second-order terms do not vanish in Itô calculus

Next: Chapter 04 — Filtrations, Adapted Processes, and Martingales, which supplies the information structure ( $\mathcal{F}_t$ , adaptedness) that the Itô integral of Chapter 05 needs before it can fix the sample-point convention this chapter left unresolved.