Implicit Function Theorem Help in a Macroeconomic Model

Question

I'm having trouble with this problem:

"A macroeconomic model contains the following equations characterizing its steady state:

$1=B(f'(K)+1-Y)$

$C+YK=f(K)$

$B$ and $Y$ are parameters between $0$ and $1$.

The question is, "do these equations implicitly define $C$ and $K$ as $C^1$ functions of $Y$ in an open ball around $(C^*,K^*)$? If so find the derivatives of these implicit functions with respect to $Y$.

Now I have no problem doing implicit function theorem problems in my textbook. But I don't know where to start with this one. Help! Can I just take the derivative of $f(k)$ and plug it into the first function?

EDIT: Here is what I know about $f$: $$f:\mathbb{R}^+ \to \mathbb{R}^+ , f \in C^2 \text{ and } f' > 0, f'' < 0.$$

Answer 1

One (standard) method is to differentiate the system. Since $B$ is held constant, we can set ${\rm d}B=0$. This should yield $$0=f''(K){\rm d}K-{\rm d}Y$$ $${\rm d}C+K{\rm d}Y+Y{\rm d}K=f'(K){\rm d}K$$ This is a linear system of equations in the differentials, $$\pmatrix{0 & f''(K) \cr 1 & Y-f'(K)}\pmatrix{{\rm d}C \cr {\rm d}K}=\pmatrix{{\rm d}Y \cr -K{\rm d}Y}$$ The condition in the implicit function theorem is the same that the coefficient matrix in this system is invertible. This is OK since $f''<0$, and then you can solve for ${\rm d}C$ and ${\rm d}K$ by linear algebra.

Answer 2

First of all, I think that your confusion might be caused (at least partially) by denominating $Y$ as a parameter. In fact, to use the Implicit Function Theorem, it should be ssen as a variable (it's implied in the problem). :D

If that's not the case, here's my answer, as fully detailed as I could do it (so you can see where you are confused):

To use the Implicit Function Theorem, the first thing you need is to have functions that define a relation between all your variables. These functions must be constant (and not other variable), and it's commonly regrouped so the functions are xero-valued (not that it really mattters).

Let's reorder the equations to be expressions equal to 0, and then name the equations as functions of $(Y,C,K)$: $$\begin{array}{rcl} F(Y,C,K)&=&B(f'(K)+1-Y)-1=0\\ G(Y,C,K)&=&C+YK-f(K)=0 \end{array}$$

Now you have two functions, $F$ and $G$, that define a relation between $C$, $K$ and $Y$. With 2 functions and 3 variables, you could say that two of those variables can be expressed as a $C^1$ function of the remaining one in an open ball, provided certain conditions are met.

The condition for $C$ and $K$ to be functions of $Y$ in an open ball around $(C*,K*)$ is that the Jacobian of the functions with respect to the variables is invertible, namely $$\det\biggl(J_{(C,K)}^{(F,G)}\biggr)\neq 0$$ where $$ J_{(C,K)}^{(F,G)}=\frac{\partial (F,G)}{\partial (C,K)}=\Biggl[\begin{array}{cc}\frac{\partial F}{\partial C}&\frac{\partial F}{\partial K}\\\frac{\partial G}{\partial C}&\frac{\partial G}{\partial K}\end{array}\Biggr]$$ is the Jacobian of the functions $F$ and $G$ (seen as a vector function) with respect to the variables $C$ and $K$.

Let's calculate the Jacobian: $$\begin{array}{rcl} \frac{\partial F}{\partial C} &=&0\\ \frac{\partial F}{\partial K} &=&Bf''(K)\\ \frac{\partial G}{\partial C} &=&1\\ \frac{\partial G}{\partial K} &=&Y-f'(K)\\ \end{array}$$

Then $$\begin{array}{rcl} \det\biggl(J_{(C,K)}^{(F,G)}\biggr)&=&\frac{\partial F}{\partial C}\cdot \frac{\partial G}{\partial K}-\frac{\partial F}{\partial K}\cdot\frac{\partial G}{\partial C}\\ &=&0-1\cdot(Bf''(K))\\ &=&-Bf''(K) \end{array}$$

We know that $f''<0$ everywhere in its domain, so the only way that the expression is zero is that $B=0$. But $B\neq0$, because that would imply $1=0$ in the expression $F(Y,C,K)$.

So, the determinant of the Jacobian is NOT zero, and the variables $C$ and $K$ can be expressed as $C^1$ functions of $Y$ in an open ball.

Now, we need to find the expressions of the partial derivatives $C_Y$ and $K_Y$. This can be done by calculating $$ C_Y=-\frac{J_{(Y,K)}^{(F,G)}}{J_{(C,K)}^{(F,G)}}$$ $$ K_Y=-\frac{J_{(C,Y)}^{(F,G)}}{J_{(C,K)}^{(F,G)}}$$

Note that the derivatives are calculated using the Jacobian of the functions $(F,G)$ with respect to all variables $C$, $K$ and $Y$. The numerator and denominator switch their variables: in the demoninator, you use the "dependent" variables, and in the numerator you switch the wanted dependent variable with the requested independent variable.

We need to find

$$\begin{array}{rcl} \Rightarrow J_{(Y,K)}^{(F,G)}&=&F_YG_K-F_KG_Y\\ &=&(-B)\cdot(Y-f'(K))-(Bf''(K))\cdot(K)\\ &=&BY+Bf'(K)-Bf''(K) &&\\ &&\\ \Rightarrow J_{(C,Y)}^{(F,G)}&=&F_CG_Y-F_YG_C\\ &=&(0)\cdot(K)-(-B)\cdot(1)\\ &=&B \end{array}$$

Finally, $$\begin{array}{rcl} C_Y&=&-\frac{J_{(Y,K)}^{(F,G)}}{J_{(C,K)}^{(F,G)}}\\ &=&-\frac{BY+Bf'(K)-Bf''(K)}{-Bf''(K)}\\ &=&\frac{Y+f'(K)-f''(K)}{f''(K)}\\ &&\\ &&\\ K_Y&=&-\frac{J_{(C,Y)}^{(F,G)}}{J_{(C,K)}^{(F,G)}}\\ &=&-\frac{B}{-Bf''(K)}\\ &=&-\frac{1}{f''(K)} \end{array}$$