I have a question about a derivation in Abraham (1987)'s simple job market matching model (equations 3 through 7):

She begins by writing down tautologies:

J - V = L - U = E

where J is the number of jobs, V is the number of vacancies, L is the total labor force, U is the total unemployed, and E is the total employed. Consider a probability function that a particular vacancy will be filled:

p(V/U) where $p_u > 0, p_v <0, p_{uu} < 0, p_{vv} > 0$.

Now, letting the total quits and firings from the current stock of employees be sE we must have: \begin{equation} p(V/U)V - sE =0 \end{equation} Letting U/E = u, V/E = v, and dividing through by E we get:

\begin{equation} p(v/u)v = s \end{equation}

Now things get tricky. She writes "along the set of points satisfying (the above)" we have:

\begin{equation} \frac{dv}{du} = - \frac{p_u}{p+p_v v} \end{equation}

My question is: how? Suppose I treat this as an implicit differentiation problem. Write $F(v,u) = p(v/u)v-s = 0$ then we know we know:

\begin{equation} \frac{d v}{d u} = - \frac{\partial F}{du} / \frac{\partial F}{dv} \end{equation}

I obtain $F_v = p + (v/u)p_v $ and $F_u = - (\frac{v}{u})^2 p_{u}$ which yields:

\begin{equation} \frac{dv}{du} = \frac{(\frac{v}{u})^2 p_{u}}{p + (v/u)p_v} \end{equation}

This doesn't look very close. Am I missing something? What's going on here?


Inspired by the answer below, I agree it is as simple as totally differentiating P(u,v) = p(v/u).

\begin{equation} p(v/u) = P(v,u) \rightarrow P(v,u)v - s = 0 \end{equation}

Now, let's totally differentiate this expression with respect to v,u, and s.

\begin{equation} \left[P_v(v,u)v + P\right] dv + v P_u du - ds = 0 \end{equation}

Consider $ds = 0 $ e.g. no change in the separation rate. This yields:

\begin{equation} \frac{dv}{du} = - \frac{vP_u}{vP_v + P} \end{equation}

Substituting $p(v/u)$ for $P(v,u)$ gives the expression off by a factor v.


1 Answer 1


Note that function $p$ actually depends upon a single variable $u/v$ and I prefer avoiding the abuse of notation (source of confusion) and write $p(u/v) = P(u,v)$. This implies:

\begin{equation} p'(u/v) = P_u(u,v)v = P_v(u,v) \cdot (-v^2/u). \end{equation}

So if $P(u,v)v = s$ total differentiation along $ds=0$ yields

\begin{equation} P_u(u,v)vdu + (P+P_v(u,v)v)dv = 0, \end{equation}

or equivalently

\begin{equation} \frac{dv}{du} = - \frac{P_uv}{P+P_v v}. \end{equation}

This result is still slighly different from those given in your question... Who can help further?

  • $\begingroup$ By Euler's theorem, $P_uu + P_vv = 0$ but this is not helpful either... $\endgroup$
    – Bertrand
    Aug 16, 2019 at 22:19
  • $\begingroup$ My original post had typos which I've now corrected. $\endgroup$
    – random
    Aug 16, 2019 at 22:38
  • $\begingroup$ Additionally, I do not see how you obtain the first line's equality. $\endgroup$
    – random
    Aug 17, 2019 at 0:04
  • $\begingroup$ The claim follows from partial derivation. If $P(u,v) = p(u/v)$ then $P_u(u,v) = p'(u/v)/v$ and $P_v(u,v) = p'(u/v) \cdot (-u/v^2).$ $\endgroup$
    – Bertrand
    Aug 17, 2019 at 9:36
  • $\begingroup$ Yes, but what you’ve just typed is different from the first line of your actual answer! $\endgroup$
    – random
    Aug 17, 2019 at 13:38

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