Question: Suppose $C_t=(1-s)Y_t-\lambda G_t$ where $s>\sigma$ as in the basic Solow model. Out of the government expenditure , proportion $\phi$ is invested in public capital formation. Hence we assume $K_{t+1}=I_t+\phi G_t+(1-\delta)K_t$





$G_t=\sigma Y_t$

In what case will the steady state level of capital per capita increase in $\sigma$


step (1) $\frac{Y_t}{L_t}=\frac{K^{\alpha}_t}{L_t}\frac{L^{1-\alpha}_t}{L_t}$ which equals $y_t=k^{\alpha}_t$

Using the same process for the national identity we get

The evolution of capital per capita is given by the following equation:

$(1+n)k_{t+1}=i_t+\phi g_t+(1-\delta)k_t$ which then goes to:

$k_{t+1}\approx i_t+\phi g_t+(1-\delta-n)k_t$

$k_{t+1}=[k^{\alpha}_t-c_t-g_t]+\phi g_t+(1-\delta-n)k_t$

Now, we can subtract $k_t$ from $(1-\delta-n)k_t$ and we get:

$k_{t+1}-k_t=-(n+\delta)k_t+\phi g_t+k^{\alpha}_t-c_t-g_t$

=$(n+\delta)k_t+\phi g_t+k^{\alpha}_t - [(1-s)k^{\alpha}_t-\lambda g_t]-[\sigma k^{\alpha}_t]$

= $-(\delta +n)k_t+\phi [k^{\alpha}_t\sigma]+k^{\alpha}_t-[(1-s)k^{\alpha}_t-\lambda [k^{\alpha}_t \sigma]]$

We can simplify this algebraically to:

$k_{t+1}-k_t= \phi [k^{\alpha}_t \sigma] + sk^{\alpha}_t+ \lambda[k^{\alpha}_t\sigma]+(\delta+n)k_t$

Finally divide both sides by $k_t$ and set the LHS equal to 0 and get the steady state equilibrium as:

$0 = \frac{\phi[k^{\alpha}_t\sigma]}{k_t}+\frac{sk^{\alpha}_t}{k_t}+\frac{\lambda[k^{\alpha}_t\sigma]}{k_t}-(\delta+n)k_t$

Comments: I am pretty sure my calculus is correct (perhaps save for the last equation. Can anyone confirm this for me?

  • $\begingroup$ This is virtually the same question as in your previous answer here with just slightly different consumption function. I would be willing to help but can you please at least try to apply the steps you learned from the other answer before asking for help? Here you just can follow the same steps. $\endgroup$
    – 1muflon1
    Nov 30, 2020 at 17:24
  • $\begingroup$ Sorry about noteingewiesen clear enough with my work/ not trying enough. I did attempt the book exercise and it actually was easier than I thought. $\endgroup$
    – Tony456
    Nov 30, 2020 at 20:13
  • $\begingroup$ this is much better I upvoted the question. I am bit busy right now but I will definitely double check it to reward the effort either later today or tomorrow evening (unless someone gives you already answer first). Also yes, often problems are easier than they seem. You should always try them first not just to learn but also to build your own confidence $\endgroup$
    – 1muflon1
    Nov 30, 2020 at 20:16
  • $\begingroup$ Thank you very much on that end. I appreciate this more than you know. $\endgroup$
    – Tony456
    Nov 30, 2020 at 20:25

1 Answer 1


Overall it is almost correct but there are some small mistakes and it is incomplete.

1st Mistake:

in step (1) you write:


But it should actually be $$\frac{Y_t}{L_t}=\frac{K^{\alpha}_t L^{1-\alpha}_t}{L_t}$$

You cannot divide LHS of the equation by $L$ and RHS by $L L \implies L^2$ as that will generally violate the equality between $ Y= K^{\alpha}L^{1-\alpha}$.

However, this error had no impact on the rest of the problem because in spite of making this mistake you still used the correct definition for output per capita:  $y_t=k^{\alpha}_t$

2nd mistake

you say you divide both sides by $k_t$, but you forgot to do that for the last term. You write:

$0 = \frac{\phi[k^{\alpha}_t\sigma]}{k_t}+\frac{sk^{\alpha}_t}{k_t}+\frac{\lambda[k^{\alpha}_t\sigma]}{k_t}-(\delta+n)k_t$

correct steady state equation should be:

$$0 = \frac{\phi[k^{\alpha}_t\sigma]}{k_t}+\frac{sk^{\alpha}_t}{k_t}+\frac{\lambda[k^{\alpha}_t\sigma]}{k_t}-(\delta+n) \tag{*}$$

3rd 'mistake': It is unfinished

I am using here ‘mistake’ in quotes because this is not a mistake in the same sense as other ones but it is still not good practice to leave problem unfinished. If you would do it on an exam it would cost you some points (some professors might even refuse to give any points for unfinished work).

I get the impulse to just call it a day since you are in home stretch now but proper answer should have solution for $k^*$ (unless the problem specifically states it does not want full solution which I don’t see in this case in your Q).

Solving the equation $*$ for optimal $k^*$ yields:

$$ k^* =  \left( \frac{(\phi+\lambda)\sigma +s}{\delta +n} \right)^{\frac{1}{1-\delta}}$$

The above being said don’t get discouraged most of the attempt was correct. Just be more careful about the arithmetic and don’t stop just before finishing.

  • $\begingroup$ Thank you very much. I am actually starting to understand this topic much more. I really do need to grasp the rote arithmetic a bit more. (Kinda easy to make a mistake with all this math). Once again, thank you. $\endgroup$
    – Tony456
    Dec 1, 2020 at 2:42
  • $\begingroup$ @Tony456 you are welcome if you think this answer answered your question consider accepting it $\endgroup$
    – 1muflon1
    Dec 1, 2020 at 10:07

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