# Why is matching function increasing in arguments?

In matching theory, the matching function $m_t=m(u_t,v_t)$, where $m_t$ is the number of matches(new hires) between unemployed looking for work($u_t$) and vacancies($v_t$).

This matching function is usually assumed to be increasing in its arguments, i.e. partial derivatives are positive.

What's the rationale for this?

Is there any data allowing us to conclude that it should be so?

Any help would be appreciated.

## 2 Answers

Think about it this way. The function is measuring the number of new matches being created. If we increase either the vacancies or the amount of unemployed, the function will generate more matches. Therefore, the partial derivatives are expected to be positive.

For example consider the matching function with something other than employment. If we consider a marriage game where we have a pool of single men and women, increasing either the amount of single men or the amount of single women, the amount of new matches created should increase. Note, that while increasing one of the two arguments might not actually increase the amount of matches, at worst it has no impact (such as in the edge case where we have only men and add another man). Since our function only considers new matches created in time T, increasing either argument should have a partial derivative of greater than or equal to zero. i.e. creating more matches.

Imagine now that instead of single men we have active job seekers (analogue to unemployment), increasing unemployment causes an increase in new matches.

• 'New matches'? I think $m$ is measuring matches...which is precisely why the partial derivatives are non-negative, there might be new matches if you increase the number of either party. Nov 4 '16 at 14:54
• Lee Sin, thanks for the answer. I'm interested in knowing why that happens... In a recession, I tend to think of as unemployment increasing and matching decreasing. Nov 4 '16 at 14:55
• @Anoldmaninthesea. That is not a valid comparison because you (obviously) do not hold $v_t$ constant. Nov 4 '16 at 15:07
• Ok let's use a parallel example. Boys and girls at a dance. If we have 10 boys and 10 girls, then within our matching function we'll create a set amount of matches (not necessarily 10). If we add an additional boy does it make sense that there are less matches intuitively? If we add in 100 of both? Raising unemployment is like increasing the number of boys at the dance. Nov 4 '16 at 15:20
• Also yes. The matching function measures new matches, not total matches. That's why it's mt and not m. Nov 4 '16 at 15:23

Let's refresh the definition of a partial derivative.

partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary).

Example: Let $f(t, x) = 3t + x^2$.

The (total derivative) is $\frac{d}{dt}f(t, x) = 3 + 2x\frac{dx}{dt}$

Whereas the partial derivative is $\frac{\partial}{\partial t}f(t, x) = 3$ since we are forcing $\frac{dx}{dt}$ to be zero.

So why are there

1. More new hires when there are more unemployed (people without work actively looking for some) given the same number of vacancies?
2. More new hires when there are more vacancies given the same number of unemployed?

The unemployed want to find the vacancies and the companies want to fill them. So the more potential vacancies there are, the easier it is to find one (maybe there is one just down the street) and vice versa. Imagine walking through the streets of a city taking random turns at each intersection, the more jobs ads there are the faster you will come across one.

Unless I misremember, the matching functions are parts of models where unemployment is caused by some kind of frictions so maybe reading more on that will help build intuition why the matching function makes sense (or doesn't).