I was studying chapter 1 of the book "Equilibrium Unemployment Theory" and I got confused about the way Pissarides has defined the probability of a firm not finding a worker in a short time interval $δt$ as $1-q(θ)δt$ (page 7 of the book).

The first reason I got confused is that, to me that cannot represent the probability because if we take $δt$ large enough, the result can be negative.

Second, if we take $q(θ)δt$ as the rate (NOT the probability) of a firm finding a match in a time interval $δt$, then the rate of the firm not finding a match in the same time interval shouldn't be $[1-q(θ)]δt$ instead of $1-q(θ)δt$?

Any help would be much appreciated.


2 Answers 2


$q(\theta)$ is defined as the job-filling rate. Note that market tightness $\theta$ is not necessarily constant over time (Pissarides makes a dynamic analysis at some point). It may help to denote it $\theta_t$. As an approximation, $q(\theta_t)\delta t$ is a probability for a firm to meet a worker between $t$ and $t+\delta t$ for $\delta t$ small enough.

1) If you choose a large $\delta t$, this approximation is not valid anymore. In other words, assuming a large $\delta t$ prevent you from interpreting $q(\theta_t)\delta t$ as a probability. To define an equilibrium, Pissarides will anyway take the limit $\delta t\to 0$.

2) $q(\theta_t)$ is a rate (that is not constrained to be lower than 1), whereas $q(\theta_t)\delta t$ is a probability. Thus, the probability of a firm not meeting a worker between time $t$ and $t+\delta t$ is $1-q(\theta_t)\delta_t$.

$1-q(\theta_t)$ has no clear interpretation (it can even be negative). I guess (but I would like confirmation) that the rate of a firm not finding a worker cannot be defined in this case (or it would be $+\infty$).


Looking at eq. (1.1) in the previous page of the book $m\cdot L$ is the number of matches, "per one unit of time" as the author writes. This is a discrete-time setup. Then $m \equiv \frac {mL}{L}$ is a straightforward proportion "number of matches divided by number of workers seeking to work, per one unit of time". It can very well represent a probability per one unit of discrete time. Formally,

$$Prob (\text{match per unit of time}) = m $$

What the author wants is to consider continuous time. Continuous time is a difficult concept. Still, think about breaking the time-interval (it may be a year or a day) that has been "standardized" to have length $1$, in "infinite" small intervals (say milli-seconds), each having essentially zero length but all together summing up to unity (I said it is a difficult concept). We represent these infinitesimal intervals by $\delta t$ or $dt$. So this symbol is never allowed to be "large enough", it is explicitly used to denote only something infinitesimal, of virtually no length. Then the probability in this almost non-existent interval is the corresponding fraction of the probability for a whole unit of time, so naturally,

$$Prob (\text{match per infinitesimal length of time}) = m \cdot dt$$

Summing up these minuscule probabilities over the domain, i.e. integrating in $[0,1]$ we verify this

$$\int_0^1 m dt = m\int_0^1 dt = m\cdot (1-0) = m$$

  • $\begingroup$ Thanks for your answer. But if this is the case, shouldn't it work the other way around too? I mean we can similarly say that Prob(no match per unit of time)=1-m and Prob(no match per infinitesimal length of time)=(1-m)⋅dt. Right? But these two (m.dt and (1-m).dt ) wouldn't sum up to 1. $\endgroup$
    – Saba
    Commented Feb 27, 2018 at 9:54
  • $\begingroup$ @Saba $mdt + (1-m)dt = mdt +dt - mdt = dt$ per infinitesimal unit of time. I.e. this is a probability flow per some unit, don't think that "it has to sum up to one" on its own. The total "probability mass" in the infinitesimal interval is $dt$, not $1$. Also, integrating $dt$ in $[0,1]$ gives you $1$, so it remains consistent with the discrete extension. $\endgroup$ Commented Feb 27, 2018 at 10:08
  • $\begingroup$ Thanks again. So you're saying that Prob(no match per infinitesimal length of time)=(1-m)⋅dt is correct right? Then why the author has defined it as 1-m.dt in the book? Actually that's the main source of my confusion. $\endgroup$
    – Saba
    Commented Feb 27, 2018 at 10:27

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