Stochastic Calculus

Lesson 4 of 11

The Itô Integral

Left-endpoint and right-endpoint sums for ∫W dW converge to different answers, ½W_T² − ½T and ½W_T² + ½T, and finance takes the left one because a position has to be set before the price moves.

Two sums, two answers

In ordinary calculus, ∫0Tx dx=12T2\int_0^T x\,dx = \tfrac12 T^2, and it makes no difference where in each strip you evaluate xx. Left endpoints, right endpoints and midpoints all converge to the same area. So the natural guess for Brownian motion is ∫0TW dW=12WT2\int_0^T W\,dW = \tfrac12 W_T^2.

To test this, chop [0,T][0, T] into nn steps ti=iT/nt_i = iT/n, write ΔWi=Wti+1−Wti\Delta W_i = W_{t_{i+1}} - W_{t_i}, and build two sums:

Ln=∑i=0n−1Wti ΔWi,Rn=∑i=0n−1Wti+1 ΔWi.L_n = \sum_{i=0}^{n-1} W_{t_i}\,\Delta W_i, \qquad R_n = \sum_{i=0}^{n-1} W_{t_{i+1}}\,\Delta W_i.

Simulate 20,000 paths on [0,1][0,1] with n=1000n = 1000 steps. The left sums average about 00 and the right sums average about 11. On a single path that ends at W1=1.2W_1 = 1.2, the left sum lands near 0.220.22 and the right sum near 1.221.22. The guess 12W12=0.72\tfrac12 W_1^2 = 0.72 sits halfway between them.

A finer grid does not close the gap. The two sums differ by

Rn−Ln=∑i(ΔWi)2,R_n - L_n = \sum_i (\Delta W_i)^2,

which is the quadratic variation from the earlier lesson. For a smooth path that sum shrinks to zero, which is why the choice of point never mattered in ordinary calculus. For Brownian motion it converges to TT, so the two answers stay a distance TT apart however small the steps get.