Find how quickly $y$ converges to $y^*$ in the vicinity of the balanced growth path $$ y=Y/AL=f(k); \; y^*=f(k^*) $$ They suggest the hint, write $k=g(y)$, where $g(\bullet)=f^{-1}(\bullet)$.)

$Y$ is product output; $AL$ is effective labour; $K$ is capital; lowercase letters are divided by $AL$ (e.g. $y=Y/AL$); $k^*$ is the balanced growth path of capital.

This is problem $1.11$ in Advanced Macro by Romer. I have tried looking at Lagrange Inversion Theorem, and applying the above get to the first order Taylor series centered around $k^*$

$$ \displaystyle g(y) = k^* + \lim_{k \to k^*} \left( \frac{y-f(k^*)}{1!} \right) \frac{d^0}{dw^0} \left( \frac{k-k^*}{f(k)-f(k^*)} \right) $$

but since $y=f(k)$ and the zeroth derivative is the function itself, we get the following cancelation:

$$ = k^* + \lim_{k \to k^*} \frac{f(k)-f(k^*)}{f(k)-f(k^*)}(k-k^*) $$

$$ = k^* + \lim_{k \to k^*}(k - k^*) = k^* $$

Which doesn't tell us anything ... Do we need to look at the second order derivative?

  • $\begingroup$ Time as a variable does not appear in any of your equations. Perhaps you neglected to type in relevant information? $\endgroup$
    – Giskard
    Jan 22, 2016 at 10:33
  • $\begingroup$ Ahh, okay thank you, I think I have it now .. should I post the answer? $\endgroup$
    – Sunhwa
    Jan 22, 2016 at 15:41
  • $\begingroup$ If you have the time, sure! $\endgroup$
    – Giskard
    Jan 22, 2016 at 16:07

2 Answers 2


I believe there is a quicker way here. By the inverse function theorem, we have

$$y = f(k) \implies k = f^{-1}(y) = f^{-1}[f(k)] \implies \frac {\partial f^{-1}(y)}{\partial y} = \frac {1}{f'(k)}$$

Keeping all these relations in mind, we also have

$$\dot y = f'(k)\cdot \big[sf(k) - (n+g+\delta)\cdot f^{-1}(y)\big]$$

$$\implies \frac{\partial \dot y}{\partial y} \bigg|_{y=y^*} = \frac {\partial f'(k^*)}{\partial y}\cdot \big[sf(k^*) - (n+g+\delta)\cdot k^*\big]\\ + f'(k^*)\cdot \left[s - (n+g+\delta)\cdot \frac {1}{f'(k^*)} \right]$$

$$\implies \frac{\partial \dot y}{\partial y} \bigg|_{y=y^*} = 0 + sf'(k^*) - (n+g+\delta)$$


  • $\begingroup$ Wow absolutely! Reading my answer again is so messy .. Thanks @Alecos_Papadopoulos ! $\endgroup$
    – Sunhwa
    Jan 23, 2016 at 4:25

Okay so, in the book they looked at the first order taylor series of $\dot{k}$, so as @denesp said, we need the time here too! So rather here we should look at the taylor series around $\dot{y}$

$$ \displaystyle \dot{y}(y) \simeq \left[ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} \right] (y(k) - y(k^*)) $$

$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \left( \frac{\partial \dot{y}(y)}{\partial k(t)} \bigg|_{y=y^*} \right) \left( \frac{\partial k(t)}{\partial y(k)} \bigg|_{k=k^*} \right) $$

First we look at:

$$ y=f(k) $$

$$\Rightarrow \dot{y} = \frac{d}{dt} f(k) = \frac{d f}{d k} \frac{dk}{dt} = f'(k) \dot{k} $$

We know the key equation of the Solow model is:

$$ \dot{k}(t) = s f(k(t)) - (n+g+\delta)k(t) $$

$$ \Rightarrow \dot{y} = f'(k) \left[ s f(k(t)) - (n+g+\delta)k(t) \right] $$

We take the derivative of this with respect to capital:

$$ \frac{\partial \dot{y}}{\partial k} = f''(k)\left[ s f(k(t)) - (n+g+\delta)k(t) \right] + f'(k) \left[ s f'(k) - (n+g+\delta) \right]$$

The value $k^*$ is the golden-rule level of the capital stock so:

$$ s f(k^*) = (n+g+\delta)k^* $$

And hence

$$ \left( \frac{\partial \dot{y}}{\partial k} \bigg|_{y=y^*} \right) = f''(k^*)*(0) + f'(k^*)\left[ s f'(k^*) - (n+g+\delta) \right] $$

Next we use the hint (not much of a hint, and rather misleading to begin with in my opinion)

$$ \left( \frac{\partial k(y)}{\partial y(t)} \bigg|_{k=k^*} \right) = \left( \frac{\partial y(k)}{\partial k(t)} \bigg|_{y=y^*} \right)^{-1} = f'(k^*)^{-1} = g'(y^*) $$

Plugging both of these in to our first order partial derivative:

$$ \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \left( f'(k^*)\left[ s f'(k^*) - (n+g+\delta) \right] \right) * \left( f'(k^*)^{-1} \right) $$

$$ = s f'(k^*) - (n+g+\delta) $$

since around the balanced growth path $s = (n+g+\delta)k^*/f(k^*)$ and putting

$$ - \frac{\partial \dot{y}(y)}{\partial y(k)} \bigg|_{y=y^*} = \lambda $$

$$ \lambda = (n+g+\delta) - \frac{(n+g+\delta)(k^*)*f'(k^*)}{f(k^*)} $$

With $\alpha_k = \frac{k*f'(k)}{f(k)} $ being the elasticity of output with respect to capital we obtain:

$$ \lambda = (n+g+\delta)(1-\alpha_k(k^*)) $$

So $y$ converges to its balanced growth-path value rate $\lambda$ the same as $k$ converges to $k^*$


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