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

<p><span class="math-container">\begin{equation}
p'(u/v) = P_u(u,v)v = P_v(u,v) \cdot (-v^2/u). 
\end{equation}</span></p>

<p>So if <span class="math-container">$P(u,v)v = s$</span> total differentiation along <span class="math-container">$ds=0$</span> yields</p>

<p><span class="math-container">\begin{equation}
P_u(u,v)vdu + (P+P_v(u,v)v)dv = 0, 
\end{equation}</span></p>

<p>or equivalently </p>

<p><span class="math-container">\begin{equation}
\frac{dv}{du} = - \frac{P_uv}{P+P_v v}. 
\end{equation}</span></p>

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

CC BY-SA 4.0
2019-08-16T22:19:58.897Z
By Euler's theorem, $P_uu + P_vv = 0$ but this is not helpful either...
Bertrand
CC BY-SA 4.0
2019-08-16T22:38:59.557Z
My original post had typos which I've now corrected.
random
CC BY-SA 4.0
2019-08-17T00:04:43.470Z
Additionally, I do not see how you obtain the first line's equality.
random
CC BY-SA 4.0
2019-08-17T09:36:09.740Z
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
CC BY-SA 4.0
2019-08-17T13:38:44.207Z
Yes, but what you’ve just typed is different from the first line of your actual answer!
random
CC BY-SA 4.0
2019-08-17T17:27:36.817Z
Different, but equivalent.
Bertrand
CC BY-SA 4.0
2019-08-17T17:38:20.983Z
Aha, now I see. p’ is p differentiated with respect to u/v treated as a single variable. Yes, I now strongly suspect there is a typo in the paper. Thanks!
random
CC BY-SA 4.0
2019-08-16T22:14:10.223Z
2019-08-18T15:44:48.887Z
2019-08-18T15:44:48.887Z
Bertrand
<p>I have a question about a derivation in Abraham (1987)'s simple job market matching model (equations 3 through 7):</p>

<p>She begins by writing down tautologies:</p>

<p>J - V = L - U = E</p>

<p>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>

<p>p(V/U) where <span class="math-container">$p_u &gt; 0, p_v &lt;0, p_{uu} &lt; 0, p_{vv} &gt; 0$</span>.</p>

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

<p><span class="math-container">\begin{equation}
p(v/u)v = s
\end{equation}</span></p>

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

<p><span class="math-container">\begin{equation}
\frac{dv}{du} = - \frac{p_u}{p+p_v v} 
\end{equation}</span></p>

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

<p><span class="math-container">\begin{equation}
\frac{d v}{d u} = - \frac{\partial F}{du} / \frac{\partial F}{dv} 
\end{equation}</span></p>

<p>I obtain <span class="math-container">$F_v = p + (v/u)p_v $</span> and <span class="math-container">$F_u =  - (\frac{v}{u})^2 p_{u}$</span> which yields:</p>

<p><span class="math-container">\begin{equation}
\frac{dv}{du} =  \frac{(\frac{v}{u})^2 p_{u}}{p + (v/u)p_v} 
\end{equation}</span></p>

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

<p>EDIT:</p>

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

<p><span class="math-container">\begin{equation} 
p(v/u) = P(v,u) \rightarrow P(v,u)v - s = 0 
\end{equation}</span></p>

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

<p><span class="math-container">\begin{equation} 
\left[P_v(v,u)v + P\right] dv + v P_u du - ds = 0
\end{equation}</span></p>

<p>Consider <span class="math-container">$ds = 0 $</span> e.g. no change in the separation rate. This yields:</p>

<p><span class="math-container">\begin{equation} 
\frac{dv}{du} = - \frac{vP_u}{vP_v + P}
\end{equation}</span></p>

<p>Substituting <span class="math-container">$p(v/u)$</span> for <span class="math-container">$P(v,u)$</span> gives the expression off by a factor v.</p>

<mathematical-economics><search-and-matching>
Abraham (1987) Simple Job Market Matching Model
random
# Abraham (1987) Simple Job Market Matching Model

(Asked by: username_0 on 2019-08-15)

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.

**Tags:** mathematical-economics, search-and-matching


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## 1 answers

### Answer by username_1 on 2019-08-16. Score: 1 | Accepted answer

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?

> **Comments:**
> **username_1**: By Euler's theorem, $P_uu + P_vv = 0$ but this is not helpful either...  
> **username_0**: My original post had typos which I've now corrected.  
> **username_0**: Additionally, I do not see how you obtain the first line's equality.  
> **username_1**: The claim follows from partial derivation. If $P(u,v) = p(u/v)$ then $P_u(u,v) = p'(u/v)/v$ and $P_v(u,v) = p'(u/v) \cdot (-u/v^2).$  
> **username_0**: Yes, but what you’ve just typed is different from the first line of your actual answer!  
> **username_1**: Different, but equivalent.  
> **username_0**: Aha, now I see. p’ is p differentiated with respect to u/v treated as a single variable. Yes, I now strongly suspect there is a typo in the paper. Thanks!  

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