Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 / 6

The duality requested below uses the symmetric simple exclusion process, so we take that standard convention here. A configuration $\eta \in \{0,1{\}}^{\mathbb{Z}}$ records a particle at $x$ when $\eta (x)=1$. Particles attempt nearest-neighbour jumps with equal left and right rates, and any attempt into an occupied site is suppressed. Thus there is at most one particle at each site.

A convenient normalization puts independent rate-one Poisson clocks on the unoriented edges $\{x,x+1\}$. At an edge mark, interchange its two endpoint occupations. If both endpoints have the same occupation, nothing changes; if exactly one is occupied, that particle jumps to the vacant endpoint. The generator on cylinder functions isOnly finitely many terms can be nonzero for a cylinder function. A convention with total jump-attempt rate one per particle uses edge clocks of rate $1/2$, merely rescaling time.

For a pathwise construction, label all sites and swap their labels at each clock mark, carrying the initial occupations along with the labels. Each label path has jump rate two and is nonexplosive on a bounded time interval. The same is true of a backward path from any fixed space-time point. This stirring representation of symmetric exclusion therefore defines the occupations for every initial configuration, even with infinitely many particles. It also conserves the number of particles when finite.

A general exclusion process may have asymmetric jump rates, but then the plain occupation-product self-duality in the next part need not hold. The symmetry convention is therefore mathematically necessary here, rather than an irrelevant choice of time normalization.

For finite $A\subset \mathbb{Z}$, define the occupation-product functionwith empty product one. Use the stirring representation of symmetric exclusion on the time interval $[0,t]$. Trace the labels at the sites of $A$ backwards to time zero, and call their set of ancestors $B_t$. Labels remain distinct under swaps, so $|B_t|=|A|$. Pathwise,Reverse time in the Poisson clocks. Their joint law is unchanged, and each interchange is its own inverse. Thus the backward set $B_t$ has the same law as the finite-particle symmetric exclusion set $A_t$ started from $A$. Taking expectations gives the product self-duality of symmetric exclusionThe argument holds for every deterministic initial configuration $\eta$ and every finite set $A$, including the empty set, and makes no stationarity assumption.

The generator identity confirms the same mechanism. For any edge $xy$, let $A^{xy}$ interchange membership of $x,y$. ThenSumming over edges gives $L_{\eta }D=L_{A}D$. In the finite-particle generator only edges joining an occupied to a vacant site contribute; in the occupation generator the same transpositions act on the other argument. Symmetry supplies the same rate in both operations.

To see why this qualification matters, suppose right jumps have rate $a$ and left jumps rate $b\ne a$. Take $A=\{0\}$ and an initial configuration whose only particle is at $1$. The derivative at zero of the left-hand probability is $b$, since that particle must jump left into zero. The derivative of the right-hand probability for the same asymmetric finite-particle process is $a$, since its particle must jump right from zero to the initially occupied site $1$. Thus the printed same-process identity would be false in that asymmetric interpretation. Under the symmetric convention established in the preceding part, the proof is complete.

For the given exchangeable law $\mu$, putwhich is well-defined by the stated assumption. Integrate the product self-duality of symmetric exclusion against $\mu$, and use boundedness to interchange expectation and integration:The finite-particle process conserves its cardinality. Hence the right side is ${\mathbb{E}}^A{m}_{|A_t|}=m_{|A|}$. All occupation-product moments are therefore unchanged by evolution.

These moments determine the law, not merely its one-site marginals. For disjoint finite sets $B,C$, the inclusion-exclusion principle givesEvery term is unchanged, so every finite cylinder probability is unchanged. Such cylinder events generate the product sigma-algebra on $\{0,1{\}}^{\mathbb{Z}}$, and probability laws agreeing on them agree everywhere. ConsequentlyThus every exchangeable probability law is invariant for the symmetric exclusion process. This proof of exchangeable invariant laws of symmetric exclusion needs only the given cardinality dependence and duality; no representation theorem for exchangeable measures is required.