Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
http://www.math.unl.edu
Voice: 402-472-3731
Fax: 402-472-8466

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Rating

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Question of the Day

What is an example of a function that “varies a lot”? What is an example of a function that does not “vary a lot”? How would you measure the “variation” of a function?

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Key Concepts

Vocabulary

Mathematical Ideas

Variation

The idea is that we measure the total (hence the absolute value) up-and-down movement of a function. This definition is similar to other partition based definitions such as the Riemann integral and the arclength of the graph of the function. A monotone increasing or decreasing function has bounded variation. A function with a continuous derivative has bounded variation. Some functions, for instance Brownian Motion, do not have bounded variation.

Again, the idea is that we measure the total (hence the positive terms created by squaring) up-and-down movement of a function. However, the squaring will make small ups-and-downs smaller, so that perhaps a function without bounded variation may have quadratic variation. In fact, this is the case for the Wiener Process.

Definition. The total quadratic variation of $Q$ of a function $f$ on an interval $[a, b]$ is

Q = sup_{P} \sum_{i = 0}^{n} {(f (t_{i + 1}) - f (t_{i}))}^{2}

where the supremum is taken over all partitions $P$ with $a = t_{0} < t_{1} < \dots < t_{n} < t_{n + 1} = b$ , with mesh size going to zero as the number of partition points $n$ goes to infinity.

Quadratic Variation of the Wiener Process

We can guess that the Wiener Process might have quadratic variation by considering the quadratic variation of our coin-flipping fortune record first. Consider the function piecewise linear function

Ŵ (t)

defined by the sequence of sums

T_{n} = Y_{1} + \dots + Y_{n}

from the Bernoulli random variables

Y_{i} = + 1

with probability

p = 1 ∕ 2

and

Y_{i} = - 1

with probability

q = 1 - p = 1 ∕ 2

. With some analysis, it is possible to show that we need only consider the quadratic variation at points

1, 2, 3, \dots, n

Then each term

{(Ŵ (i + 1) - Ŵ (i))}^{2} = Y_{i + 1}^{2} = 1

. Therefore, the quadratic variation is the total number of steps,

Q = n

. Now remember the Wiener Process is approximated by

W_{n} (t) = (1 ∕ \sqrt{n}) Ŵ (n t)

. Each step is size

1 ∕ \sqrt{n}

, then the quadratic variation of the step is

1 ∕ n

and there are

n

steps on

[0, 1]

. The total quadratic variation of

W_{n} (t) = (1 ∕ \sqrt{n}) Ŵ (n t)

[0, 1]

1

We will not completely rigorously prove that the total quadratic variation of the Wiener Process is

t

, as claimed, but we will prove a theorem close to the general definition of quadratic variation.

Proof. Introduce some briefer notation for the proof, let:

Δ_{n k} = W (\frac{k}{2^{n}} t) - W (\frac{k - 1}{2^{n}} t) k = 1, \dots, 2^{n}

and

W_{n k} = Δ_{n k}^{2} - t ∕ 2^{n} k = 1, \dots, 2^{n} .

We want to show that $\sum_{k = 1}^{2^{n}} Δ_{n k}^{2} \to t$ or equivalently: $\sum_{k = 1}^{2^{n}} W_{n k} \to 0$ . For each $n$ , the random variables $W_{n k}, k = 1, \dots, 2^{n}$ are independent and identically distributed by properties 1 and 2 of the definition of standard Brownian motion. Furthermore,

E [W_{n k}] = E [Δ_{n k}^{2}] - t ∕ 2^{n} = 0

by property 1 of the definition of standard Brownian motion.

A routine (but omitted) computation of the fourth moment of the normal distribution shows that

E [W_{n k}^{2}] = 2 t^{2} ∕ 4^{n} .

Finally, by property 2 of the definition of standard Brownian motion

E [W_{n k} W_{n j}] = 0, k \neq j .

Now, expanding the square of the sum, and applying all of these computations

E [{\{\sum_{k = 1}^{2^{n}} W_{n k}\}}^{2}] = \sum_{k = 1}^{2^{n}} E [W_{n k}^{2}] = 2^{n + 1} t^{2} ∕ 4^{n} = 2 t^{2} ∕ 2^{n} .

Now apply Chebyshev’s Inequality to see:

ℙ [|\sum_{k = 1}^{2^{n}} W_{n k}| > ϵ] \leq \frac{2 t^{2}}{ϵ^{2}} {(\frac{1}{2})}^{n} .

Now since $\sum {(1 ∕ 2)}^{n}$ is a convergent series, the Borel-Cantelli lemma implies that the event

|\sum_{k = 1}^{2^{n}} W_{n k}| > ϵ

can occur for only finitely many $n$ . That is, for any $ϵ > 0$ , there is an $N$ , such that for $n > N$

|\sum_{k = 1}^{2^{n}} W_{n k}| < ϵ .

Therefore we must have that $lim_{n \to \infty} \sum_{k = 1}^{2^{n}} W_{n k} = 0$ , and we have established what we wished to show. □

Remark. Here’s a less rigorous and somewhat different explanation of why the squared variation of Brownian motion may be guessed to be $t$ , see [1]. Consider

\sum_{k = 1}^{n} {(W (\frac{k t}{n}) - W (\frac{(k - 1) t}{n}))}^{2} .

Now let

Z_{n k} = \frac{(W (\frac{k t}{n}) - W (\frac{(k - 1) t}{n}))}{\sqrt{t ∕ n}}

Then for each $n$ , the sequence $Z_{n k}$ is a sequence of independent, identically distributed standard normal $N (0, 1)$ random variables. Now we can write the quadratic variation as:

\sum_{k = 1}^{n} Δ_{n k}^{2} = \sum_{k = 1}^{n} \frac{t}{n} Z_{n k}^{2} = t (\frac{1}{n} \sum_{k = 1}^{n} Z_{n k}^{2})

But notice that the expectation $E (Z_{n k}^{2})$ of each term is the same as calculating the variance of a standard normal $N (0, 1)$ which is of course $1$ . Then the last term in parentheses above converges by the weak law of large numbers to $1$ ! Therefore the quadratic variation of Brownian motion converges to $t$ . This little proof is in itself not sufficient to prove the theorem above because it relies on the weak law of large of numbers. Hence the theorem establishes convergence in distribution only while for the theorem above we want convergence almost surely.

Remark. Starting from

lim_{n \to \infty} \sum_{n = 1}^{2^{n}} {[W (\frac{k}{2^{n}} t) - W (\frac{k - 1}{2^{n}} t)]}^{2} = t

and without thinking too carefully about what it might mean, we can imagine an elementary calculus limit to the left side and write the formula:

\int_{0}^{t} {[d W (τ)]}^{2} = t = \int_{0}^{t} d τ .

In fact, with more advanced mathematics this can be made sensible ad mathematically sound. Now from this relation, we could write the integral equality in differential form:

d W {(τ)}^{2} = d τ .

The important thing to remember here is that the formula suggests that Brownian motion has differentials that cannot be ignored in second (or squared, or quadratic) order. Brownian motion “wiggles” so much that even the total of the squared differences add up! In retrospect, this is not so surprising given the law of the iterated logarithm. We know that in any neighborhood $[t, t + d t]$ to the right of $t$ , Brownian motion must come close to $\sqrt{2 t log log t}$ . That is, intuitively, $W (t + d t) - W (t)$ must be about $\sqrt{2 d t}$ in magnitude, so we would guess $d W^{2} \approx 2 d t$ The theorem makes it precise.

Proof.

\sum_{n = 1}^{2^{n}} |W (\frac{k}{2^{n}} t) - W (\frac{k - 1}{2^{n}} t)| \geq \frac{\sum_{n = 1}^{2^{n}} {|W (\frac{k}{2^{n}} t) - W (\frac{k - 1}{2^{n}} t)|}^{2}}{max_{j = 1, \dots, 2^{n}} |W (\frac{k}{2^{n}} t) - W (\frac{k - 1}{2^{n}} t)|}

The numerator on the right converges to $t$ , while the denominator goes to $0$ because Brownian paths are continuous, therefore uniformly continuous on bounded intervals. Therefore the faction on the right goes to infinity. □

Sources

The theorem in this section is drawn from A First Course in Stochastic Processes by S. Karlin, and H. Taylor, Academic Press, 1975. The heuristic proof using the weak law was taken from Financial Calculus: An introduction to derivative pricing by M Baxter, and A. Rennie, Cambridge University Press, 1996, page 59. The mnemonic statement of the quadratic variation in differential form is derived from Steele’s text.

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Problems to Work for Understanding

Reading Suggestion:

References

[1] M. Baxter and A. Rennie. Financial Calculus: An introduction to derivative pricing. Cambridge University Press, 1996. HG 6024 A2W554.

[2] S. Karlin and H. Taylor. A Second Course in Stochastic Processes. Academic Press, 1981.

[3] J. Michael Steele. Stochastic Calculus and Financial Applications. Springer-Verlag, 2001. QA 274.2 S 74.

Outside Readings and Links:

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