The Poisson-Inhibition Model and the Poisson-Erlang Model

The Poisson-Inhibition model, as well as previously developed inhibition models, are developed to account for the sequence of RTs which can be observed when a single test is administrated to a single subject. The model is designed to explain the statistical properties of a series of reaction times, $T_1, T_2, \ldots, T_j, \ldots, T_n$, representing the amounts of time a person uses for each one of a sequence of n responses executed consecutively. It is assumed that each response requires the same amount of processing time: A. The actual time T_dj spent (the reaction time), exceeds A because of distractions interrupting work on the task. So, T_j = A + D_j, where $D_{j} = D_{1j} + D_{2j} + \ldots + D_{Nj}$ is the sum of the distraction intervals for particular response j. Within reaction time j, the duration of the distractions D_ijare random variables and so is their number N. In the previously mentioned Poisson-Erlang model, PE model for short, N has a Poisson distribution and the D_ij are exponentially distributed with a constant transition rate $\delta$, so D_j has an Erlang distribution. In the Poisson-Inhibition model, PI model for short, the hypothetical construct inhibition, denoted by Y(t), is introduced. Inhibition Y(t) increases linearly with rate a₁ during work intervals:

$$Y'(t) = a_1 \mbox{ during work intervals}$$ (1)

and decreases linearly with rate a₀ during distraction intervals:

$$Y'(t) = -a_0 \mbox{during rest intervals}$$ (2)

The transition rates $\lambda_1$, from work to rest (distraction), and $\lambda_0$, from rest to work are no longer constant. From now on they depend on inhibition: $\lambda_1 = l_1(Y)$, $\lambda_0 = l_0(Y)$, where l₁ is a non-decreasing function and l₀ is a non-increasing function. Specification of the functions l₁ and l₀ leads to various "special inhibition models". In the PI model l₁ and l₀ are as follows:

$$l_1(y) = c_1, \mbox{ ($c_1$\space a positive constant)}$$ (3)

The transition rate from work to rest is constant. Since a task requires, for its completion, an amount of working time A, and during this time, interruptions occur with rate c₁, it follows, that the number of distractions is Poisson distributed with mean c₁ A. This was the reason for the "Poisson" in the name of the model.

$$l_0(y) = \frac{c_0}{y}, \mbox{ for $y>0$ . ($c_0$\space a positive constant)}$$ (4)

The transition rate from rest to work is given by $\lambda_0(t) = l_0(Y(t)) = c_0/Y(t)$. Note, that as Y(t) goes to zero (during a distraction), the transition rate $\lambda_0(t)$ goes to infinity and this forces a transition from rest to work before the inhibition could reach zero.

In the case of the PE model the inhibition has no influence on the development of the process. However, one might as well consider it rising with rate a₁ during work and falling with rate a₀ during distractions. It seems natural to assume that as the process goes on, the inhibition will neither increase to $+\infty$ or decrease to $-\infty$, but will fluctuate around a more or less stable mean. Thus the mean increase of inhibition during a work interval: a₁/c₁should be equal to the mean decrease during a distraction interval: a₀/c₀. According to the PI model (Smit and van der Ven, 1995) the parameters A, a₀, a₁, c₀ and c₁ satisfy the equations () through (). According to Smit (personal communication) one might write a = a₀, c = c₀, k = a₁/a₀ = c₁/c₀. This is a nice way to reduce the number of parameters by one: A, a, c and k instead of A, a₀, a₁, c₀, c₁. No generality is lost by this reduction. For example, suppose one applies a linear scale transformation to inhibition. Say one measures inhibition in J (Joule) instead of in KJ (Kilo Joules). This would change a₀ and a₁ by a factor 1000 (a_i becomes 1000a_i) and also the numerical values of Y (inhibition) would be multiplied by 1000 and hence, since the transition rate $\lambda_0 = l_0(y) = c_0/y = \frac{1000c_0}{1000y}$has to stay the same, c₀ should also be changed to 1000 c₀. Of course c₁ is not affected. So, by a scale transformation of the inhibition, one can change the quotient $\frac{c_1}{c_0}$ to any value one may desire. Now the following result can be shown to hold. The PI model with parameters A, a₀ =a, a₁ = ka, c₀ = c and c₁ = kc tends to the PE model with parameters A, c₀ = c and c₁ = kc (i.e. the mean number of distractions is given by c₁A = kcA and the mean individual distraction time by $\frac{1}{c}$) when $a \rightarrow 0$.

According to the Poisson-Inhibition model the long-term trend can be described by an exponential curve of the following form:

$$y_j = \alpha + \beta \gamma^{j-1} \mbox{ for } j = 1, 2, \ldots, n,$$ (5)

where n is equal to the number of reaction times. $\alpha$, $\beta$ and $\gamma$ each are functions of the parameters A, a₀, a₁, c₀ and c₁. In order to evaluate the stationary mean and variance one may compute the mean and the variance of the reaction times in the last part of the task, assuming that the long-term trend can be neglected in that part of the test.