A trader argues that since Brownian increments have zero mean, E[f(W_T)] = f(0) for any smooth f. Give the correct identity, compute the error for f(x) = x^2 and for f(x) = e^x at T = 1, and name the term responsible.

A trader argues that since Brownian increments have zero mean, E[f(W_T)] = f(0) for any smooth f. Give the correct identity, compute the error for f(x) = x^2 and for f(x) = e^x at T = 1, and name the term responsible.

Approach: Apply Ito to f(W_t), take expectations to remove the stochastic integral, and read the remaining integral of the second derivative.

E[f(W_T)] = f(0) + (1/2) ∫_0^T E[f''(W_s)] ds, so the error is the Ito correction: it is 1 for f(x) = x^2 and e^{1/2} - 1 = 0.6487 for f(x) = e^x at T = 1. Ito gives f(W_T) = f(0) + ∫_0^T f'(W_s) dW_s + (1/2) ∫_0^T f''(W_s) ds, and the stochastic integral has mean zero when f' has suitable growth. For the square, f'' = 2 and the correction is T = 1, matching E[W_1^2] = 1. For the exponential, E[f''(W_s)] = E[e^{W_s}] = e^{s/2}, and ∫_0^1 (1/2) e^{s/2} ds = e^{1/2} - 1. The second order Taylor term survives because the quadratic variation of Brownian motion is of order dt rather than of order (dt)^2, so squared increments contribute at first order. The sign of the error is the sign of convexity, which is the same statement as Jensen's inequality applied to the terminal distribution.

Follow-up: For which f does the identity fail because the stochastic integral is only a local martingale, and give a concrete f on [0,1] where E[f(W_1)] differs from the formula?

Key concepts: Ito correction, second order Taylor term, convexity, quadratic variation.