Why Brownian Squares Survive Itô’s Rule and Time Squares Vanish

The rule compares sums as a time partition becomes fine. Over each interval of length

h

, time moves by

h

and Brownian motion moves by an amount with standard deviation

\sqrt h

. Summing

h^2

over all intervals tends to zero. Summing squared Brownian increments tends to the elapsed time. The shorthand

(dW_t)^2=dt

records this limiting quadratic variation.

Compare the three sums on the same partition

On a uniform partition of

[0,T]

with

n

steps, put

h=T/n

and let

Z_i

be independent standard normal variables. Then the three sums corresponding to

(dt)^2

dt\,dW_t

and

(dW_t)^2

are

\sum_{i=1}^n h^2=Th,\qquad \sum_{i=1}^n h\Delta W_i=hW_T,\qquad \sum_{i=1}^n(\Delta W_i)^2=h\sum_{i=1}^n Z_i^2.

The first two converge to zero as

h

shrinks. The third has mean

T

and variance

2Th

, so it converges in mean square to

T

. Even though each squared Brownian increment is random, its accumulated deviation from elapsed time gets small. The Itô formula uses the corresponding weighted limits when the local coefficient of each term depends on the process.

The mixed sum on this uniform grid is especially transparent. Since $h$ is the same on every interval, pull it out and telescope the Brownian increments to get $hW_T$ . It vanishes as $h$ goes to zero. The sum of squared noise increments cannot telescope this way. Each has mean $h$ , so the total keeps mean $T$ even as each summand shrinks. The variance calculation is what lets you replace that random total by its limit, and explains the shorthand used in an Itô calculation.

For an original exercise, put $X_t=3t+2W_t$ and expand the square of a single increment:

(\Delta X_i)^2=9h^2+12h\Delta W_i+4(\Delta W_i)^2.

After summing, the first term tends to zero as

9Th

, the second tends to zero as

12hW_T

, and the third tends to

4T

. Therefore

d[X]_t=4\,dt

. Applying the Itô product rule to

X_t^2

gives

d(X_t^2)=2X_t\,dX_t+4\,dt=(6X_t+4)\,dt+4X_t\,dW_t.

t=1

the resulting expectation is

9+4=13

. This agrees with computing the second moment directly from

X_1\sim N(3,4)

, where the second argument denotes variance.

The test for any proposed correction is whether it survives aggregation. A single large Brownian step may give a squared increment several times larger than $h$ . The rule still holds in the fine-grid limit.

On a uniform grid with $T=1$ and $n=100$ , the Brownian squared sum has standard deviation $\sqrt{2/100}\approx0.141$ . With $10{,}000$ steps the standard deviation is approximately $0.014$ . This gives a concrete scale for the error when the differential rule is approximated on a computer.

Why Brownian Squares Survive Itô’s Rule and Time Squares Vanish

Compare the three sums on the same partition

Explore a Brownian path at different resolutions

References and next steps