Let's split each period into $n$ intervals. There's a continuum $u$ of unemployed and $v$ of vacancies. During each interval, there is a total of $X/n$ job offers. That means that each unemployed gets a job offer with probability $X/(nu)$ (assuming that $n$ is eventually so small that the chance of multiple job offers to the same person will go to zero).

Single individual Number of job offers - for a single individual - in the interval is distributed $Binomial(k, n, \frac{X}{un})$. Letting $n\to\infty$, the distribution in the continuous-time analog converges to $Poisson(k, \frac{X}{u})$.

I'm interested in the probability of $x$ individuals getting at least one job interval during a whole interval. That is $(1 - Binomial(0, n, \frac{X}{nu}))^x$:

$$ (1 - (1 - \frac{X}{un})^n)^x $$

If I'm not mistaken, the continuous-time analog is

$$ (1 - e^\frac{-X}{u})^x$$

Is that correct? The combination of exponential and power function is making me quite uncomfortable.


2 Answers 2


An alternative approximating approach you could use as as check might be to say there are $X$ job offers in total and $u$ unemployed.

So the probability that an individual does not get a particular job offer is $\left(1-\dfrac{1}{u}\right)$ and so the probability the individual does not get any job offer is $\left(1-\dfrac{1}{u}\right)^X$ which is $\left(\left(1-\dfrac{1}{u}\right)^u\right)^{x/u} \approx e^{-X/u}$ much as one might expect from a Poisson distribution. You have an error in the sign of $e^{-X/u}$.

That makes the probability the individual does get a job offer about $1-e^{-X/u}$. It makes the expected number of individuals receiving job offers about $u(1-e^{-X/u})$.

The probability that $y$ individuals get job offers (rather than $x$ as you have already used $X$) is then roughly the binomial $\displaystyle {y \choose u}\left(1-e^{-X/u}\right)^y e^{-(u-y)X/u}$ and or if you wanted a Poisson approximation to this then about $\dfrac{e^{-u\left(1-e^{-X/u}\right)}u^y\left(1-e^{-X/u}\right)^y}{y!}$ and you could apply further approximations to this but it would not get much tidier

  • $\begingroup$ If $u$ is a continuum, isn't the probability of a particular individual getting the job $0$? I thought about this case first, but to get the math correctly, one would - I thought - have to have time in intervals $n$, and some measure of agents $m$, and look at the double limit when $n\to0$ and $m\to\infty$. $\endgroup$
    – FooBar
    Sep 13, 2016 at 6:11
  • $\begingroup$ Your splitting into $n$ intervals now works to get you to $(1 - e^\frac{-X}{u})$ for the probability a particular individual gets at least one job offer. Currently your problem is turning that into a probability that a given number of individuals get offers, and for that you have the binomial distribution rather than just putting it to the power of $x$ (or $y$ in my version). $\endgroup$
    – Henry
    Sep 13, 2016 at 7:29

Yes, it is correct. You can (for instance) write a Taylor expansion: \begin{align*} [1-(1-\frac{x}{un})^n]^x & = [1-e^{n ln(1-\frac{x}{un})}]^x \\ & = [1-e^{n (-\frac{x}{un} + o(\frac{1}{n}))}]^x \\ & = [1-e^{-\frac{x}{u} + o(1)}]^x \\ & \sim [1-e^{-\frac{x}{u}}]^x \text{ when } n \rightarrow +\infty \end{align*}


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