Chebyshev’s inequality and the Borel-Cantelli lemma are seemingly disparate results from probability theory but they combine beautifully in demonstrating a curious property of Brownian motion: that it has finite quadratic variation even though it has unbounded linear variation. Not only do the proofs of Chebyshev’s inequality and the Borel-Cantelli lemma have some interesting features themselves, but their application to the variation of Brownian motion also provides a nice example of how to prove explicitly that a sequence of random variables converges *almost surely* by showing that the probability of the divergence set is zero. In this note I want to bring out this last aspect in particular.

There are two equivalent forms of Chebyshev’s inequality. Letting denote an integrable random variable with mean and finite non-zero variance , one form of Chebyshev’s inequality says

where is a real number.

Equivalently, if we let , then we can write Chebyshev’s inequality as

**Proof of Chebyshev’s inequality**

The following is a proof of the second form. Since , we have

Let so that .

Let

If then and

(by definition of ). However, if then so it is still true that

Thus, this inequality always holds. Taking expectations we get

Division of both sides by gives Chebyshev’s inequality. QED

The Borel-Cantelli lemma actually consists of a pair of results. Let be a sequence of events, i.e., a sequence of subsets of . Then

is the set of all elements that are in an infinity of the subsets . The Borel-Cantelli lemma says the following:

(a) If then

(b) If is independent and then

**Proof of the Borel-Cantelli lemma**

(a) For any integer , we can write

(the last inequality by subadditivity). Since converges, the tail sums as . Therefore (by a ‘sandwich’ argument) .

(b) We have

(by De Morgan’s laws). Therefore

By the product rule for independent events,

Now, for all we have . (To see this, let . Then and . Therefore is an increasing function, so for ). It follows that

and therefore also

But as . Therefore and , so . QED

To apply these results to the variation of Brownian motion, let

be a partition of the interval . Define the *mesh* of the partition by

Let us restrict ourselves only to sequences of partitions for which , and in particular for which . For every let

and let

where is a real-valued function defined on . If we will say that has finite variation on . In the case , if we will say that is of bounded linear variation on . In the case , if we will say that is of bounded quadratic variation on

If we replace in the above by a Brownian motion , then we find a curious result: is of unbounded linear variation almost surely, but it has a finite quadratic variation equal to almost surely. In other words,

(a.s.)

(a.s.)

We will use Chebyshev’s inequality and the Borel-Cantelli lemma to prove these.

**Theorem 1**

If is a sequence of partitions of with (and therefore ), then

(a.s.)

as . In other words, Brownian motion has bounded quadratic variation on equal to , almost surely.

**Proof:**

Observe that, by ‘telescoping’,

Let

Note that because . Therefore

(cross product terms vanish since Brownian motion has independent increments)

(since the fourth moment of a zero-mean normal random variable is and here )

(Since , we see already at this stage that in ).

By the second form of Chebyshev’s inequality we can now write for any that

In particular, we can write for any integer that

By the first result in the Borel-Cantelli lemma we can then say

where

Now,

so if then we must also have

But

is the divergence set when considering the almost sure convergence of to zero. Since the probability of the divergence set is zero, we conclude that a.s., and therefore (a.s.). QED

**Theorem 2**

If is a sequence of partitions of with , then

(a.s.)

as . In other words, Brownian motion has unbounded linear variation on , almost surely.

**Proof:**

By contradiction. Suppose Brownian motion has bounded linear variation, i.e., . Then we can write

Since is uniformly continuous on ( is continuous on and any continuous function is uniformly continuous when restricted to a compact set) we can write

as

which if implies

(a.s.)

This contradicts Theorem 1. Therefore cannot have bounded linear variation. QED