# Abraham (1987) Simple Job Market Matching Model

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?

EDIT:

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.

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?

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