Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 27 6 b Solution Created 2026-10-03 Updated 2026-10-14
For a fixed environment, use two independent copies of the walk to express . The product is finite and nonnegative, so we can interchange the two expectations. At each time , if the walks occupy different sites, the two environment signs are independent and their factors have product expectation . If they coincide, both factors use the same sign andDifferent time layers are independent even when a site is visited again. Therefore, with the simultaneous collision count ,This is the collision-count second moment for independent path weights. It involves equal-time intersections, not arbitrary intersections of the two spatial ranges.
The difference walk has independent bounded increments symmetric about zero and a full-dimensional step distribution. It may hold at zero for one step; such a hold counts as a positive-time return. The allowed transience fact givesMoreover since the two first steps can be equal. Restarting the independent increments after every return to zero shows that the total positive-time return count has the zero-based geometric distributionIndeed the probability of at least returns is , and subtracting successive tails gives this law. Consequently, whenever ,A sufficient small-noise condition is .
The L2 martingale convergence theorem now gives in as well as almost surely. One can see the mean preservation directly: for , the martingale identity gives . The second moments are increasing and bounded, so the sequence is Cauchy in and its limit is the already identified almost-sure limit. Thus