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Let the energy of a flow be $\mathcal{E}(\theta )={\sum }_e\theta (e{)}^2$, with each unoriented edge counted once. Expanding as in part (i), dropping orientation signs, and summing over common edges givesuniformly in $r$.

Part (i), $\mathbb{E}{X}_r=1$, and the Paley-Zygmund inequality give a constant $a>0$ such that $\mathbb{P}(X_r>1/2)\ge a$ for every $r$. The preceding uniform expectation bound and Markov inequality allow a constant $L$ such thatOn this event, ${\theta }_{\omega }/X_r$ is a unit open flow from $o$ to $z_r$ with energy at most $4L$.

The events that there is such a bounded-energy unit flow from $o$ out of $B(o,r)$ are decreasing in $r$. Their intersection still has probability at least $a/2$. A diagonal compactness argument produces on this intersection a unit flow from $o$ to infinity, supported on its open cluster, with finite energy. The finite-energy flow criterion for transience makes that open cluster transient.

Thus a transient open cluster exists with positive probability. This existence event is a tail event: changing finitely many edges cannot destroy transience in every infinite component, because transience is invariant under finite graph modifications. The Kolmogorov zero-one law upgrades its probability to one.

Solved by gpt-5.6-sol high.

Because $V$ is measurable for the cylinder sigma-algebra, membership in $V$ depends on only countably many coordinates $T_0=\{t_1,t_2,\dots \}$. Its image in ${\mathbb{R}}^{T_0}$ is a measurable linear subspace $V_0$, and

By successively applying the finite-dimensional Gaussian regression formula, realize this Gaussian sequence as a lower-triangular linear transform of independent standard normal variables:where every coordinate of the sum contains only finitely many terms. If some deterministic column $a_k$ does not belong to $V_0$, then, after conditioning on every $g_j$ except $g_k$, at most one value of $g_k$ can put the sum in $V_0$. The continuous normal distribution gives probability zero. If every $a_k$ belongs to $V_0$, changing finitely many $g_k$ does not change the membership event. It is then a tail event, and the Kolmogorov zero-one law gives probability zero or one. This proves the Gaussian zero-one law for measurable linear subspaces.

Now let $X(t)=\sqrt{t},g_t$ for independent standard normal variables $g_t$. DefineBoth are cylinder-measurable infinite-dimensional linear subspaces. For every $\epsilon >0$,so the Borel-Cantelli lemmas imply $g_t/\sqrt{t}\to 0$ almost surely and hence $\mathbb{P}(X\in V)=1$. On the other hand, $g_t^2\ge 1$ infinitely often almost surely, again by Borel-Cantelli, soalmost surely. Therefore $\mathbb{P}(X\in W)=0$.

Solved by gpt-5.6-sol high.