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Setup Blanchard-Diamond (1994) have a model where each unemployed has a single ball (application) that she tosses into urns (vacancies). With probability $a$ the application lands in the urn - the unemployed manages to fill out the form, and his application is "valid".

Vacancies are open for a fixed period and then evaluate their application pool, leading to multiple applications per vacancy. Under some conditions, the distribution of applications per vacancy (denoted as $g(x)$) will be Poisson with expected value $au/v$.

Firm-based job-finding rate The number of matches is hence given by the measure of vacancies that have at least one application, times the number of vacancies

$$M = (1-g(0))v$$

and the job-finding rate is given by

$$ M/u = (1-g(0))v/u$$

Unemployed-based job-finding rate Instead of looking at the firms, one can look at the unemployed. As firms randomly draw from the pool of applicants, the average chance of an unemployed matching is the probability of matching times the inverse of the number of contestants that he will face:

$$ f = a\frac{1}{E[x | x > 0]} \\ = a\frac{1}{\frac{E[x]}{1-g(0)}}\\ = a\frac{1-g(0)}{au/v}\\ = (1-g(0))v/u$$

Which is turns out to be the same job-finding rate as calculated from the firm's perspective. This makes sense, as it needs to be an identity.

Firm-state-based job-finding rate Finally, to the part where I am struggling: An alternative way of computing this job-finding rate should be

$$ \tilde f = a \sum_{x=1}^\infty Prob(\text{applicant matches with firm with $x$ many applications} | \text{ she matched}) \frac{1}{x}$$

Which basically has the same logic, but is more detailed. This is interesting when one wants to analyze the $x$-component to the job-finding rate (i.e.: How relevant are firms with 5 applicants to the job-finding rate). However, the probability appears to be complicated. How do I compute this job-finding rate?

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The total number of matched applications is

$$\sum_{x=1}^\infty g(x)xv = E[x]v$$

The number of applications matched with vacancies with $x$-many applications is

$$g(x)xv$$

Hence

$$Prob(\text{matching with x} | \text{matched} = \frac{g(x)x}{E[x]}$$

Using this in the equation in the question gives

$$ \tilde f = a \sum_{x=1}^\infty \frac{g(x)x}{E[x]}\frac{1}{x}\\ = E[x]a \sum_{x=1}^\infty g(x) \\ = \frac{v}{u} (1-g(0)) = f $$

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