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

<p>So the probability that an individual <em>does not</em> 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 $e^{-X/u}$.</p>

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

<p>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</p>

CC BY-SA 3.0
2016-09-13T06:11:54.240Z
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 $m\to\infty$.
FooBar
CC BY-SA 3.0
2016-09-13T07:29:35.523Z
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).
Henry
CC BY-SA 3.0
2016-09-12T10:58:48.857Z
2016-09-12T10:58:48.857Z
Henry
<p>Yes, it is correct. You can (for instance) write a Taylor expansion:
\begin{align*}
[1-(1-\frac{x}{un})^n]^x &amp; = [1-e^{n ln(1-\frac{x}{un})}]^x \\
&amp; = [1-e^{n (-\frac{x}{un} + o(\frac{1}{n}))}]^x \\
&amp; = [1-e^{-\frac{x}{u} + o(1)}]^x \\
&amp; \sim [1-e^{-\frac{x}{u}}]^x \text{ when } n \rightarrow +\infty
\end{align*}</p>

CC BY-SA 3.0
2016-09-13T22:35:16.087Z
2016-09-13T22:35:16.087Z
Oliv
<p>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). </p>

<p><strong>Single individual</strong> 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 $Poisson(k, \frac{X}{u})$. </p>

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

<p>$$ (1 - (1 - \frac{X}{un})^n)^x $$</p>

<p>If I'm not mistaken, the continuous-time analog is </p>

<p>$$ (1 - e^\frac{-X}{u})^x$$</p>

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

<continuous-time><matching>
Matching problem in continuous time
FooBar
# Matching problem in continuous time

(Asked by: username_0 on 2016-09-11)

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.

**Tags:** continuous-time, matching


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## 2 answers

### Answer by username_1 on 2016-09-12. Score: 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

> **Comments:**
> **username_0**: 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 $m\to\infty$.  
> **username_1**: 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).  

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### Answer by username_2 on 2016-09-13. Score: 2

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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