<p>Notice that the value of <span class="math-container">$w_r(n_r)$</span> will be such that <span class="math-container">$V(e_r) = V_r(e_0)$</span> so we can just put this equal to <span class="math-container">$V_r$</span>. Then the three conditions are
<span class="math-container">$$
\begin{align*}
&amp;V_r = w_r - e_r + \beta (q V_u + (1 - q) V_r) \tag{1}\\
&amp;V_r = w_r - e_0 + \beta (q V_u + (1 - q) V_r) - \beta \sigma(V_r - V_u) \tag{2}\\
&amp;V_u = \bar w_0 + \beta (n_r V_r + (1 - n_r) V_u) \tag{3}
\end{align*}
$$</span></p>
<p>Rewriting these we get:
<span class="math-container">$$
\begin{align*}
&amp;V_r(1 - \beta) = w_r - e_r + \beta q(V_u - V_r) \tag{1}\\
&amp;V_r(1 - \beta) = w_r - e_0 + \beta(q + \sigma)(V_u - V_r) \tag{2}\\
&amp;V_u(1 - \beta) = \bar w_0 - \beta n_r(V_u - V_r) \tag{3}
\end{align*}
$$</span>
Take the difference between <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> and between <span class="math-container">$(1)$</span> and <span class="math-container">$(3)$</span> gives:
<span class="math-container">$$
\begin{align*}
&amp;0 = e_0 - e_r - \beta \sigma(V_u - V_r) \tag{4}\\
&amp;(V_r - V_u)(1 - \beta) = w_r -e_r - \bar w_0 + \beta(q + n_r)(V_u - V_r) \tag{5}
\end{align*}
$$</span>
Equivalently:
<span class="math-container">$$
\begin{align*}
&amp;V_r - V_u = \frac{e_0 - e_r}{\beta \sigma} \tag{6}\\
&amp;(V_r - V_u)(1 - \beta(1 - q - n_r)) = w_r - e_r - \bar w_0 \tag{7}
\end{align*}
$$</span>
Substituting <span class="math-container">$(6)$</span> into <span class="math-container">$(7)$</span> gives:
<span class="math-container">$$
w_r = e_r + \bar w_0 + \frac{(1 - \beta(1 - q- n_r))(e_0 - e_r)}{\beta \sigma}
$$</span>
So except for the sign on the third part on the right hand side, this is the same (maybe I made a mistake?)</p>

CC BY-SA 4.0
2022-01-17T17:21:14.810Z
OK so the key is the first rewriting step here. Now I remember that when I learned the search model from the textbook there was always this rewriting step but I didn't notice the purpose so I totally forget it. Thanks.
Alalalalaki
CC BY-SA 4.0
2022-01-17T17:09:45.383Z
2022-01-17T17:09:45.383Z
tdm
<p>I am reading a paper using the common job search framework, which has functions  <span class="math-container">$\begin{aligned}
&amp;V_{r}\left(e_{r}\right)=w_{r}-e_{r}+\beta\left(q V_{u}+(1-q) V_{r}\left(e_{r}\right)\right) \\
&amp;V_{r}\left(e_{o}\right)=w_{r}-e_{o}+\beta\left(q V_{u}+(1-q) V_{r}\left(e_{o}\right)\right)-\beta \sigma\left(V_{r}\left(e_{o}\right)-V_{u}\right) \\ 
&amp; V_{u}=\bar{w}_{o}+\beta\left(n_{r} \max \left\{V_{r}\left(e_{r}\right), V_{r}\left(e_{o}\right)\right\}+\left(1-n_{r}\right) V_{u}\right)
\end{aligned}$</span>.</p>
<p>The paper then derives the minimal regular wage that elicits high effort <span class="math-container">$e_r$</span>: <span class="math-container">$\begin{aligned}
w_{r}\left(n_{r}\right) &amp;=\min \left\{w_{r} \mid V_{r}\left(e_{r}\right) \geq V_{r}\left(e_{o}\right)\right\} \\
&amp;=e_{r}+\bar{w}_{o}+\frac{\left(1-\beta\left(1-q-n_{r}\right)\right)\left(e_{r}-e_{o}\right)}{\beta \sigma} .
\end{aligned}$</span>.</p>
<p>The meaning of the model is not important here, I am just confused about the derivation here. It seems to be just three unknowns and three equations, but when I do it I find the equations soon go complicated and I cannot get the simple result as above. Is there any useful rule to do the derivation in this kind of situation?</p>

<labor-economics>
Derive Value Functions in Job Search Model
Alalalalaki
# Derive Value Functions in Job Search Model

(Asked by: username_0 on 2022-01-17)

I am reading a paper using the common job search framework, which has functions $\begin{aligned}
&V\_{r}\left(e\_{r}\right)=w\_{r}-e\_{r}+\beta\left(q V\_{u}+(1-q) V\_{r}\left(e\_{r}\right)\right) \\
&V\_{r}\left(e\_{o}\right)=w\_{r}-e\_{o}+\beta\left(q V\_{u}+(1-q) V\_{r}\left(e\_{o}\right)\right)-\beta \sigma\left(V\_{r}\left(e\_{o}\right)-V\_{u}\right) \\
& V\_{u}=\bar{w}\_{o}+\beta\left(n\_{r} \max \left\{V\_{r}\left(e\_{r}\right), V\_{r}\left(e\_{o}\right)\right\}+\left(1-n\_{r}\right) V\_{u}\right)
\end{aligned}$.

The paper then derives the minimal regular wage that elicits high effort $e\_r$: $\begin{aligned}
w\_{r}\left(n\_{r}\right) &=\min \left\{w\_{r} \mid V\_{r}\left(e\_{r}\right) \geq V\_{r}\left(e\_{o}\right)\right\} \\
&=e\_{r}+\bar{w}\_{o}+\frac{\left(1-\beta\left(1-q-n\_{r}\right)\right)\left(e\_{r}-e\_{o}\right)}{\beta \sigma} .
\end{aligned}$.

The meaning of the model is not important here, I am just confused about the derivation here. It seems to be just three unknowns and three equations, but when I do it I find the equations soon go complicated and I cannot get the simple result as above. Is there any useful rule to do the derivation in this kind of situation?

**Tags:** labor-economics


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## 1 answers

### Answer by username_1 on 2022-01-17. Score: 3 | Accepted answer

Notice that the value of $w\_r(n\_r)$ will be such that $V(e\_r) = V\_r(e\_0)$ so we can just put this equal to $V\_r$. Then the three conditions are
$$
\begin{align\*}
&V\_r = w\_r - e\_r + \beta (q V\_u + (1 - q) V\_r) \tag{1}\\
&V\_r = w\_r - e\_0 + \beta (q V\_u + (1 - q) V\_r) - \beta \sigma(V\_r - V\_u) \tag{2}\\
&V\_u = \bar w\_0 + \beta (n\_r V\_r + (1 - n\_r) V\_u) \tag{3}
\end{align\*}
$$

Rewriting these we get:
$$
\begin{align\*}
&V\_r(1 - \beta) = w\_r - e\_r + \beta q(V\_u - V\_r) \tag{1}\\
&V\_r(1 - \beta) = w\_r - e\_0 + \beta(q + \sigma)(V\_u - V\_r) \tag{2}\\
&V\_u(1 - \beta) = \bar w\_0 - \beta n\_r(V\_u - V\_r) \tag{3}
\end{align\*}
$$
Take the difference between $(1)$ and $(2)$ and between $(1)$ and $(3)$ gives:
$$
\begin{align\*}
&0 = e\_0 - e\_r - \beta \sigma(V\_u - V\_r) \tag{4}\\
&(V\_r - V\_u)(1 - \beta) = w\_r -e\_r - \bar w\_0 + \beta(q + n\_r)(V\_u - V\_r) \tag{5}
\end{align\*}
$$
Equivalently:
$$
\begin{align\*}
&V\_r - V\_u = \frac{e\_0 - e\_r}{\beta \sigma} \tag{6}\\
&(V\_r - V\_u)(1 - \beta(1 - q - n\_r)) = w\_r - e\_r - \bar w\_0 \tag{7}
\end{align\*}
$$
Substituting $(6)$ into $(7)$ gives:
$$
w\_r = e\_r + \bar w\_0 + \frac{(1 - \beta(1 - q- n\_r))(e\_0 - e\_r)}{\beta \sigma}
$$
So except for the sign on the third part on the right hand side, this is the same (maybe I made a mistake?)

> **Comments:**
> **username_0**: OK so the key is the first rewriting step here. Now I remember that when I learned the search model from the textbook there was always this rewriting step but I didn't notice the purpose so I totally forget it. Thanks.  

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