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In Liggett's book, interacting particle systems, on pages 425 426 one reads

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and just before we see

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The proof of $E[\eta_t(u)]\leq (e^{Bt}\eta)(u)$ is left to the reader.

Here is my attempt:

$$E[\eta_t(u)] = \sum_{k=0}^\infty E[\eta^k_t(u)] P(\tau_k \leq t < \tau_{k+1})\\ = \sum_k \frac{e^{-t}t^k}{k!} E[\eta^k_t(u)] $$

where $eta^k_t(x)$ is given by (1.9) above.

Now it remains to bound $E[\eta^k_t(u)]$ for this purpose we compute

$$E[A^{(1)}\zeta(u)] = \frac{1}{n}\sum_x E A_x\zeta(u) = \\ = \frac{1}{n}\sum_x \sum_v E A_x(u,v)\zeta(v)\\ = \frac{1}{n}\sum_x \sum_{v\neq u} E A_x(u,v)\zeta(v) + E (A_x(u,u)-1)\zeta(u) + \zeta(u)\\ \leq (C+I)\zeta(u) $$

where $$C(u,v) = \begin{cases} \frac{1}{n}\sum_x \sum_{v\neq u} E A_x(u,v) & v \neq u\\ E |A_x(u,u)-1|& v = u \end{cases}$$

Assume for the moment that $A$ and $C$ commute:

then

$$E[\eta_t(u)] = \sum_k \frac{e^{-t}t^k}{k!} E[\eta^k_t(u)]\\ \leq \sum_k \frac{e^{-t}t^k}{k!}(C+I)^k e^{tA}\eta(u)\\ = e^{t(A + C + I)}\eta (u) $$

Now note that $A+C \leq B$ So we conclude that $$E[\eta_t(u)] \leq e^t e^{t(B)}\eta (u) $$

This is not quite the result we would wish.

I must be missing something on the way.

Any ideas?

Conrado Costa
  • 6,071

1 Answers1

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Consider the equation obtained via the Dynkin martingale:

$$\eta_t(u) = \eta_0(u) + \int_0^t \Omega\eta_s(u)\, ds + M^x_t$$

consider the equation for the expected value $E[\eta_t(u)] = \varphi(t)$

$$\varphi(t)= \varphi(0) + \int_0^t \Omega \varphi(s)\,ds $$

Now we compute

$$\Omega \eta(u) = \sum_x \Bbb{E}[A_x \eta(u) - \eta(u)] + \sum_{v} a(u,v)\eta(v) \\ = \sum_{x,v} \Bbb{E}[A_x(u,v) \eta(v) - \eta(u)] + \sum_{v} a(u,v)\eta(v) \\ \leq B \eta(u) $$

From where we deduce that

$$\varphi(t)\leq \varphi(0) + \int_0^t B \varphi(s)\,ds $$

and therefore $\varphi(t) \leq e^{Bt} \varphi(0) = e^{Bt}\eta(u)$

Conrado Costa
  • 6,071