# Derive Value Functions in Job Search Model

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?

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?)