Linearization and the effect of a change

Question

There is the following system of four equations and four endogenous variables $(K,L,w,q)$. Assume $F$ is a concave function.

$\partial F(K,L)/\partial K = r + (1-p)$

$\partial F(K,L)/\partial L = w$

$F(K,L)-rK-wL-q-(1-p)K = 0$

$v(w,p)=\bar{v}$

It seems to linearize the system around the equilibrium values to get $\frac{\partial K}{\partial p}$ and $\frac{\partial L}{\partial p}$ and it seems that these equations become

$\frac{\partial K}{\partial p} = \frac{-F_{LL}-F_{KL}\partial w/\partial p}{(F_{KK}F_{LL}-F^2_{KL})}$

$\frac{\partial L}{\partial p} = \frac{F_{KK}\partial w/\partial p + F_{KL}}{(F_{KK}F_{LL}-F^2_{KL})}$

I tried to Taylar expansion but I could not get these equations. How do we get these last equations? These equations are from p.85 Kousky et al. (2006)

Answer 1

Start from the first order conditions: $$ \begin{align*} &F_k = r + (1- p)\\ &F_L = w\\ \end{align*} $$ Now totally differentiate with respect to $p$: $$ \begin{align*} &F_{KK} dK + F_{KL} dL = - dp\\ &F_{LK} dK + F_{LL} dL = \frac{\partial w}{\partial p} dp\\ \end{align*} $$ This is equivalent to: $$ \begin{align*} &dK = \frac{- dp - F_{KL} dL}{F_{KK}}\\ &dL = \frac{\frac{\partial w}{\partial p}dp - F_{LK} dK}{F_{LL}} \end{align*} $$ substituting the second into the first gives: $$ \begin{align*} &dK = \frac{(- dp)F_{LL} - F_{KL}(\frac{\partial w}{\partial p} dp - F_{LK} dK)}{F_{KK} F_{LL}},\\ \to &\left(1 - \frac{F_{KL} F_{LK}}{F_{LL} F_{KK}}\right) dK = \frac{-F_{LL}dp- F_{KL}\frac{\partial w}{\partial p} dp}{F_{KK} F_{LL}},\\ \to &dK = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2} dp \end{align*} $$ This gives: $$ \frac{\partial K}{\partial p} = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2} $$ The expression for $\frac{\partial L}{\partial p}$ can be derived in a similar way.