# Tag Info

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) ...