Computing an explicit formula for the square of a martingale, a question concerning a paper from Kipnis and Cocozza

Question

In the article from 1997 Cocozza, C. and Kipnis, C. 1977 Existence de processus Markoviens pour des systemes infinis de particules. Ann. lnst. H. Poincare, Sect. B, 13, 239-257. , one reads

and in page 244, one reads:

I couldn't derive the expression for $(\tilde{\eta^x_t})^2 - \int_0^t c(x,\eta_s)\, ds$ from lemma I.1

From Lemma I.1 b)taking $x=y$ one obtains $$ (\tilde{\eta^x_t})^2 \equiv ({\eta^x_t})^2 - 2\int_0^t c(x,\eta_s)(1 - 2\eta^x_s)\eta^x_s \, ds $$

So

$$ (\tilde{\eta^x_t})^2 - \int_0^t c(x,\eta_s)\, ds\equiv ({\eta^x_t})^2 - 2\int_0^t c(x,\eta_s)(1 - 2\eta^x_s)\eta^x_s \, ds - \int_0^t c(x,\eta_s)\, ds $$

But how do one arrive at $$ (\tilde{\eta^x_t})^2 - \int_0^t c(x,\eta_s)\, ds\equiv ({\eta^x_t})^2 - 2\int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds - \int_0^t c(x,\eta_s)\, ds $$

Is it true that $$ \int_0^t c(x,\eta_s)(1 - 2\eta^x_s)\eta^x_s \, ds = \int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds ? $$ If we divide in cases we see that the left side is equal to $$ - \int_0^t c(x,\eta_s)1_{\eta^x_s = 1} \, ds$$ while the right side is equal to $$ - \int_0^t c(x,\eta_s)1_{\eta^x_s = 1} \, ds + \int_0^t c(x,\eta_s)1_{\eta^x_s = 0} \, ds$$ Is this a typo?

Assuming it is a typo:

We continue our computations from the following equivalence $$ (\tilde{\eta^x_t})^2 - \int_0^t c(x,\eta_s)\, ds\equiv ({\eta^x_t})^2 - 2\int_0^t c(x,\eta_s)(1 - 2\eta^x_s)\eta^x_s \, ds - \int_0^t c(x,\eta_s)\, ds \\ \equiv ({\eta^x_t})^2 + 2\int_0^t c(x,\eta_s)1_{\eta^x_s = 1} \, ds - \int_0^t c(x,\eta_s)\, ds\\ \equiv ({\eta^x_t})^2 + \int_0^t c(x,\eta_s)1_{\eta^x_s = 1} \, ds - \int_0^t c(x,\eta_s) 1_{\eta^x_s = 0}\, ds\\ \equiv ({\eta^x_t})^2 - \int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds $$ Question is now, how do we see that

$$ ({\eta^x_t})^2 - \int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds \equiv \tilde{\eta^x_t}?$$

Answer 1

Note that $\eta^x_t \in\{0,1\}$ this implies that $(\eta^x_t)^2 = \eta^x_t$ and therefore

$$ ({\eta^x_t})^2 - \int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds = {\eta^x_t} - \int_0^t c(x,\eta_s)(1 - 2\eta^x_s) \, ds = \tilde{\eta^x_t} $$

This is indeed a typo