Exercise 13.3.10 We can prove Eq. (13.17) without using Eq. (13.14). Let fn(X) ≡

2 n−1 k=0



X



(k+1) t 2n



− X



kt 2n



.

(1) Prove that |X( (k+1) t/2n )− X(kt/2n)| has mean 2−n/2

√

2/π and variance 2−n(1−2/π). ( fn(X) thus has mean 2n/2

√

2/π and variance 1−2/π.) (2) Show that fn(X)→∞ with probability one. (Hint: Prob[ |X− E[ X]| ≥k] ≤ Var[ X]/k2 by Chebyshev’s inequality.)