Springer Finance

102 3 Brownian Motion

Most functions have continuous derivatives, and hence their quadratic vari­

ations are zero. For this reason, one never considers quadratic variation in

ordinary calculus. The paths of Brownian motion, on the other hand, can­

not be differentiated with respect to the time variable. For functions that do

not have derivatives, the Mean Value Theorem can fail and Remark 3.4.2 no

longer applies. Consider, for example, the absolute value function f(t) = ltl in

Figure 3.4.3. The chord connecting (t1,J(ti)) and (t 2 , j(t 2 )) has slope!, but

nowhere between h and t 2 does the derivative of f(t) = ltl equal !· Indeed,

this derivative is always -1 for t < 0, is always 1 for t > 0, and is undefined

at t = 0, where the the graph of the function f(t) = ltl has a "p oint." Figure

3.2.2 suggests correctly that the paths of Brownian motion are very "pointy."

Indeed, for a Brownian motion path W(t), there is no value of t for which

ft W ( t) is defined.

Fig. 3.4.3. Absolute value function.

Theorem 3.4.3. Let W be a Brownian motion. Then W, W = T fo r all
T 2: 0 almost surely.

We recall that the terminology almost surely means that there can be some
paths of the Brownian motion for which the assertion [W, W] (T) = T is not
true. However, the set of all such paths has zero probability. The set of paths
for which the assertion of the theorem is true has probability one.

PROOF OF THEOREM 3.4.3: Let II= {to, t1, ... , tn} be a partition of [0, T).
Define the sampled quadratic variation corresponding to this partition to be

n- 1

Qrr = L (W(ti+I)-W(ti))

2

• j=O

We must show that this sampled quadratic variation, which is a random vari­
able (i.e., it depends on the path of the Brownian motion along which it is