<p><strong>Setup</strong> Say you have two types of workers, high and low. The share of low-types among the unemployed population is $P$. I want to find the job-finding rate for these types. </p>

<p><strong>Matching</strong> Matching is through an urn-ball model: Each unemployed has one single ball (application) that he tosses at random into an urn (vacancy). Each vacancy will then have $ x \in [0, \infty)$ many applications, mixed from both types. The vacancy will pick a high type (at random) whenever there is one, otherwise it will pick randomly among the whole (low-type) pool of its applicants. Denote by $g(x)$ the probability mass function of a vacancy having $x$ many  applications.</p>

<p><strong>Job-finding rates</strong> The average pool job-finding rate of each pool is equal to the likelihood of a specific unemployed of that pool of finding a job. </p>

<p>Consider the matching-rate of the low-types: Denote $P(M)$ probability of a specific low-type matching, $P(x_h = 0)$ the probability of a vacancy unconditionally having zero high-types. Then, the low-type's job-finding rate is</p>

<p>$$ P(M) P(x_h == 0| M) \frac{1}{E[x | x_h == 0 \wedge M]}$$</p>

<p>A specific low-type will only find a job if </p>

<ul>
<li>he matches</li>
<li>everyone where he matched was of the low-type (no high-type matched)</li>
<li>and then, with the inverse rate of people that on average matched wherever he matched (if there's 10, he has a 1/10 chance of getting the job)</li>
</ul>

<p>That's at least what I thought. Denote $P(M) = 1$, people toss a ball that always lands in one urn. Say $g(x) = 1/4$ for $x\in [0, 3]$: The tossing is in a way that applications are equally distributed over vacancies with boundaries 0, 3.</p>

<p>I think that</p>

<p>$$ E[x | x_h == 0 \wedge M] = \sum_x P(x | x_h == 0 \wedge M) x \\
= \frac{1}{P(x_h == 0 \wedge M)}\sum_x P(x \wedge x_h == 0 \wedge M) x$$</p>

<p>Where we can rewrite </p>

<p>$$ P(x_h == 0 \wedge M) = \sum_x P(x_h == 0 \wedge M \wedge x)\\
= \sum_x P(x_h == 0 | \wedge M \wedge x)$$</p>

<p>As I said, $P(M) = 1$, and $P(x_h == 0 | \wedge M \wedge x)$ is the binomial chance of drawing $0$ high-types out of $x-1$ applications with probability $1-P$. All sums start at $x=1$.</p>

<p>With the numerical example I gave, and $P=0.5$, I have</p>

<ul>
<li>E[x | x_h == 0 \wedge M] = 1.57</li>
<li>P(x_h == 0| M) = 0.583</li>
</ul>

<p>and the total job-finding rate of the low-type is roughly 0.371</p>

<p>Is this the best way to proceed? I'm somewhat worried that my number is not clean (and I can't really represent it as a rational number), while the problem as I've set it up is pretty symmetrical. </p>

<p><strong>Check using corner-case</strong> When I set $P = 1$, the job-finding rate of the low-type becomes $0.5$. I think that's wrong, because 1/3 of applicants get a vacancy with 1 applicant, 1/3 with 2, and 1/3 with 3. That means, they have a 1/3(1+1/2 + 1/3) chance of matching, which is 11/18, roughly 0.6 - and not $0.5$.</p>

<p>Is there a different test I could do, or a different (simpler?) way to calculate the job-finding rate of the low type? Is my expression just blatantly false?</p>

CC BY-SA 3.0
2016-02-12T13:28:00.313Z
Just to clarify: what happens if two employers select the same candidate?
Oliv
CC BY-SA 3.0
2016-02-12T13:46:31.610Z
@Oliv Each candidate only has one application (one ball), hence that scenario does not apply.
FooBar
<labor-economics><probability><search-and-matching>
Job-finding rate in an urn-ball model with types
FooBar
# Job-finding rate in an urn-ball model with types

(Asked by: username_0 on 2016-02-12)

**Setup** Say you have two types of workers, high and low. The share of low-types among the unemployed population is $P$. I want to find the job-finding rate for these types.

**Matching** Matching is through an urn-ball model: Each unemployed has one single ball (application) that he tosses at random into an urn (vacancy). Each vacancy will then have $ x \in [0, \infty)$ many applications, mixed from both types. The vacancy will pick a high type (at random) whenever there is one, otherwise it will pick randomly among the whole (low-type) pool of its applicants. Denote by $g(x)$ the probability mass function of a vacancy having $x$ many applications.

**Job-finding rates** The average pool job-finding rate of each pool is equal to the likelihood of a specific unemployed of that pool of finding a job.

Consider the matching-rate of the low-types: Denote $P(M)$ probability of a specific low-type matching, $P(x\_h = 0)$ the probability of a vacancy unconditionally having zero high-types. Then, the low-type's job-finding rate is

$$ P(M) P(x\_h == 0| M) \frac{1}{E[x | x\_h == 0 \wedge M]}$$

A specific low-type will only find a job if

* he matches
* everyone where he matched was of the low-type (no high-type matched)
* and then, with the inverse rate of people that on average matched wherever he matched (if there's 10, he has a 1/10 chance of getting the job)

That's at least what I thought. Denote $P(M) = 1$, people toss a ball that always lands in one urn. Say $g(x) = 1/4$ for $x\in [0, 3]$: The tossing is in a way that applications are equally distributed over vacancies with boundaries 0, 3.

I think that

$$ E[x | x\_h == 0 \wedge M] = \sum\_x P(x | x\_h == 0 \wedge M) x \\
= \frac{1}{P(x\_h == 0 \wedge M)}\sum\_x P(x \wedge x\_h == 0 \wedge M) x$$

Where we can rewrite

$$ P(x\_h == 0 \wedge M) = \sum\_x P(x\_h == 0 \wedge M \wedge x)\\
= \sum\_x P(x\_h == 0 | \wedge M \wedge x)$$

As I said, $P(M) = 1$, and $P(x\_h == 0 | \wedge M \wedge x)$ is the binomial chance of drawing $0$ high-types out of $x-1$ applications with probability $1-P$. All sums start at $x=1$.

With the numerical example I gave, and $P=0.5$, I have

* E[x | x\_h == 0 \wedge M] = 1.57
* P(x\_h == 0| M) = 0.583

and the total job-finding rate of the low-type is roughly 0.371

Is this the best way to proceed? I'm somewhat worried that my number is not clean (and I can't really represent it as a rational number), while the problem as I've set it up is pretty symmetrical.

**Check using corner-case** When I set $P = 1$, the job-finding rate of the low-type becomes $0.5$. I think that's wrong, because 1/3 of applicants get a vacancy with 1 applicant, 1/3 with 2, and 1/3 with 3. That means, they have a 1/3(1+1/2 + 1/3) chance of matching, which is 11/18, roughly 0.6 - and not $0.5$.

Is there a different test I could do, or a different (simpler?) way to calculate the job-finding rate of the low type? Is my expression just blatantly false?

**Tags:** labor-economics, probability, search-and-matching
> **Comments:**
> **username_1**: Just to clarify: what happens if two employers select the same candidate?  
> **username_0**: @Oliv Each candidate only has one application (one ball), hence that scenario does not apply.  

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## 0 answers

