Considering a marginal probability law, averaging Eq. (25) over $\Delta tnk$ produces Display Formula
$\u2329y(t)y(t\u2032)\u232a{\Delta tnk}(2)=\u2211n=1N\u2211k=1kn\u2211k\u2032=1kn\u222b\u221e\u222b\u221ep\Delta t(\Delta tnk,\Delta tnk\u2032)\u2062\delta (t\u2212tn\u2212\Delta tnk)\delta (t\u2032\u2212tn\u2212\Delta tnk\u2032)d\Delta tnkd\Delta tnk\u2032=\u2211n=1N\u2211k=1kn\u2211k\u2032=1kn[k\xafn(tn)]\u22122pd(t,tn)pd(t\u2032,tn)=\u2211n=1N(kn2\u2212kn)[k\xafn(tn)]\u22122pd(t,tn)pd(t\u2032,tn),$(30)
where the second line is given by that $\Delta tnk$ and $\Delta tn\u2032k\u2032$ are independent of each other. The $kn2\u2212kn$ indicates the number of terms on the summations of $k$ and $k\u2032$ for $k\u2260k\u2032$, which can be switched to $Var[kn|tn]+k\xafn2(tn)\u2212k\xafn(tn)$ through the procedure of averaging Eq. (30) over $kn$. Further averaging $\u2329y(t)y(t\u2032)\u232a(2)$ for $tn$ and $N$ becomes Display Formula$Rg(2)(t,t\u2032)=\u222b\u221e[Var[kn|tn]+k\xafn2(tn)\u2212k\xafn(tn)][k\xafn(tn)]\u22122\u2062ap(tn)pd(t,tn)pd(t\u2032,tn)dtn.$(31)
If the integration for the term $k\xafn(tn)$ in $[Var[kn|tn]+k\xafn2(tn)\u2212k\xafn(tn)]$ is considered with the substitutions of Eqs. (2), (5), and (7), Display Formula$\u222b\u221e[k\xafn(tn)]\u22121pd(t,tn)pd(t\u2032,tn)ap(tn)dtn=cpc0\u222b\u221e[k\xafnI(tn)]\u22121pdI(t,tn)pdI(t\u2032,tn)[1+mp\u2009\u2062cos(\omega ptn+\psi )][1+\u2211g=1\u221emg\u2009cos(\omega gtn+\varphi g)]dtn\u2248cpc0\u222b\u221ek\xafnI(tn)\delta (t\u2212tn)\delta (t\u2032\u2212tn)\u2062[1+mp2\u2009cos(\Delta \omega p1t+\psi \u2212\varphi 1)\u2211g=1\u221emg]dtn\u2248cpc0k\xafnI[1+mp2\u2009cos(\Delta \omega p1t+\psi \u2212\varphi 1)\u2211g=1\u221emg]\delta (t\u2212t).$(32)