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

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.