odd, this results in $|\xi_i-\xi_j|\le 2M$ for all $i$ and $j$, while in the case when $M$ is even this is true only whenever $i-j$ is even. The observation that $\sum_j \xi_j(t)=t$ thus proves (\[eq\_xieo\]) for odd $M$, since $$\left|t-M\xi_i(t)\right|=\left|\sum_{j=1}^{M}
\left[\xi_j(t)-\xi_i(t)\right]\right|\le (M-1)\times 2M.$$ Now, when $M=2L$ is even, denote H(t)=. Suppose $i\in\M$ is even. Then |t-M\_i(t)+MH(t)|&=&|2\_[j=1]{}\^L \_[2j]{}(t)-2L\_i(t) |\
& =& |2\_[j=1]{}\^L | 4M(L-1)&lt;2M\^2. A similar argument holds for odd $i$. Hence we have established (\[eq\_xieo\]) for even $M$ as well as for odd $M$.

To finish the proof, denote by $\tau_m$, $m\ge 0$, the consecutive renewal times of the Markov chain $v(t)$ after $t_{j_0}$, that is \_0&=&{tt\_[j\_0]{}: v\_1(t)=v\_2(t)=…=v\_[M]{}(t)=0},\
\_m&=&{t\_[m-1]{}: v\_1(t)=v\_2(t)=…=v\_[M]{}(t)=0}, m1. By Proposition \[prop\_a\_conv\], these stopping times are well-defined; moreover, $\tau_{m+1}-\tau_m$ are i.i.d. and have exponential tails. Let $\zeta_{m+1}=H(\tau_{m+1})-H(\tau_{m})$. Then the $\zeta_m$ are also i.i.d.; moreover, their distribution is symmetric around $0$, and $|\zeta_{m+1}|\le \tau_{m+1}-\tau_m$ hence the $\zeta_m$ also have exponential tails. The rest follows from the standard Donsker–Varadhan invariance principle; see e.g. [@DKL], pp. 590–592, for a proof in a very similar set-up.

Symmetric interaction
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In the symmetric case, the potential of site $k$ at time $t$ is \[eq:sympot\] u\_k(t)=\_[k-1]{}(t)+\_k(t)+\_[k+1]{}(t), k, and the transition probabilities of the Markov 