Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 27 / 6

Write $Z_t=⟨{\kappa }_t⟩$, where the angle brackets average only over walk paths, and use $\mathbb{E}$ only for the environment. Every nearest-neighbor path of length $t$ has probability $(2d{)}^{-t}$, so the finite path sum is exactly the walk expectation. It depends only on environment layers through time $t$, and hence is ${\mathcal{F}}_t$-measurable. Its factors lie in $[1-\epsilon ,1+\epsilon ]$, so $Z_t$ is positive and integrable at each fixed time.

For any fixed path through time $t+1$, the environment variable $h(t+1,{\xi }_{t+1})$ is independent of ${\mathcal{F}}_t$ and has mean zero. Conditional expectation therefore removes its last factor. Summing over the $2d$ possible last steps of each length-$t$ path cancels the extra factor $2d$ in the path probability. ThusThis is the positive random-environment path-weight martingale. It is nonnegative and bounded in $L^1$, so the martingale convergence theorem gives a finite limit $\zeta \ge 0$ almost surely. At this stage one can conclude $\mathbb{E}\zeta \le 1$ by Fatou's lemma, but not yet equality; the second-moment bound in part (b) will prevent loss of mean.

For a fixed environment, use two independent copies of the walk to express $Z_t^2=⟨{\kappa }_t^{(1)}{\kappa }_t^{(2)}⟩$. The product is finite and nonnegative, so we can interchange the two expectations. At each time $j$, if the walks occupy different sites, the two environment signs are independent and their factors have product expectation $1$. 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 $I_t={\sum }_{j=1}^t{1}_{\{{\xi }_j^{(1)}={\xi }_j^{(2)}\}}$,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 $D_j={\xi }_j^{(1)}-{\xi }_j^{(2)}$ 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 $\rho >0$ 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 $I_{\infty }$ has the zero-based geometric distributionIndeed the probability of at least $k$ returns is ${\rho }^k$, and subtracting successive tails gives this law. Consequently, whenever $\rho (1+{\epsilon }^2)<1$,A sufficient small-noise condition is $0<\epsilon <\min (1,\sqrt{(1-\rho )/\rho })$.

The L2 martingale convergence theorem now gives $Z_t\to \zeta$ in $L^2$ as well as almost surely. One can see the mean preservation directly: for $t\ge s$, the martingale identity gives $\mathbb{E}(Z_t-Z_s{)}^2=\mathbb{E}{Z}_t^2-\mathbb{E}{Z}_s^2$. The second moments are increasing and bounded, so the sequence is Cauchy in $L^2$ and its limit is the already identified almost-sure limit. Thus

For each fixed $m\ge 0$ and $T\ge m$, remove the first $m$ weight factors but retain the same walk law:This quantity is measurable with respect to ${\mathcal{G}}_{m+1}=\sigma (h(j,x):j\ge m+1,x\in {\mathbb{Z}}^d)$, since the averaging over all walk positions is deterministic and involves no earlier environment variables. The bounds on each deleted factor give, pointwise for every environment,Both constants are finite and strictly positive. Taking lower limits gives the exact identity of eventsThere is no exceptional-set issue in this comparison. Choose the version of $\zeta$ to equal ${\mathrm{liminf}}_T{Z}_T$ when this is finite and to equal $1$ otherwise; it agrees with the almost-sure martingale limit, and its zero event is precisely the left-hand event above. Since the identity holds for every $m$, $\{\zeta =0\}$ belongs to the printed tail sigma-algebra ${\bigcap }_{n\ge 1}{\mathcal{G}}_n$. This proves the tail event for a vanishing positive path-weight limit without pretending that the whole value of $\zeta$ is tail-measurable.

The independent time layers of the environment are independent random vectors, so the Kolmogorov zero-one law applies to their tail sigma-algebra. It yields $\mathbb{P}(\zeta =0)\in \{0,1\}$. Under the small-noise condition from part (b), $\mathbb{E}\zeta =1$ and $\zeta \ge 0$, so that probability cannot be $1$. HenceThe tail-event argument itself works for every $0<\epsilon <1$; the deduction of positivity here uses the mean-one conclusion supplied by part (b).