We study the linear response of a two-state stochastic process, obeying the renewal condition, by means of a stochastic rate equation equivalent to a master equation with infinite memory. We show that the condition of perennial aging makes the response to coherent perturbation vanish in the long-time limit. DOI: 10.1103/PhysRevLett.95.220601 PACS numbers: 05.40.Fb, 02.50.ÿr, 82.20.Uv Many complex processes generate erratic jumps back and forth from a state ''on'' to a state ''off.'' We limit ourselves to quoting ionic channel fluctuations [1][2][3], currently triggering the search for a form of stochastic resonance valid also in the nonexponential case [4], and the intermittency of blinking nanocrystals [5,6]: It has been assessed that the intermittent fluorescence of these materials obeys the renewal theory [7]; namely, a jump from one to the other state has the effect of resetting the system's memory to zero. The nonexponential nature of the distribution of sojourn times makes this renewal process nonergodic and generates aging effects that are the object of an increasing theoretical interest [8,9]. Similar properties are found with surface-enhanced Raman spectra of single molecules [10].The authors of [11][12][13] studied the joint effect of aging and perturbation on a process of subdiffusion, without establishing, however, a direct connection with the issue of non-Poisson stochastic resonance [4]. Here we fill this gap by means of a stochastic master equation, with no time convolution. This method might be extended to an arbitrarily large number of states and to the case of arbitrarily intense perturbation.Let us assume that the distributions of on and off times are identical. We assign to the survival probability (SP) of this process, t, the inverse power law formwith > 1. This corresponds to the joint action of the time-dependent rate [14,15] qt q 0 =1 q 1 t, with q 0 ÿ 1=T and q 1 1=T, and of a resetting prescription. To illustrate this condition, let us imagine the random drawing of a number from the interval I 0; 1 at discrete times i 1; 2; . . . . The interval I is divided into two parts, I 1 and I 2 , with I 1 ranging from 0 to p i , and I 2 ranging from p i to 1. Note that p i 1 ÿ q i < 1 and q i 1, and, as a consequence, the number of times we keep drawing numbers from I 1 , without moving to I 2 , is very large.Let us evaluate the distribution of these persistence times, and let us discuss under which conditions we get the SP of Eq. (1). The SP function is the probability of remaining in I 1 after n drawings and is consequently given byUsing the condition q i 1, and evaluating the logarithm of both terms of Eq. (2), we obtainThe condition q i 1 implies that i and n of Eq. (3) are so large as to make q i virtually identical to a function of the continuous time t, q i qt q 0 =r1 q 1 t, with t t. Thus, Eq. (3) yields the SP of Eq. (1), and the corresponding waiting time distribution density, , readsWe denote as collisions the rare drawings of a number from I 2 , followed by resetting. Thus the col...