{"id":2018,"date":"2021-12-01T18:24:37","date_gmt":"2021-12-01T10:24:37","guid":{"rendered":"https:\/\/fanyuzhao.com\/?p=2018"},"modified":"2022-03-02T16:51:27","modified_gmt":"2022-03-02T08:51:27","slug":"more-about-solow-model","status":"publish","type":"post","link":"https:\/\/fanyuzhao.com\/?p=2018","title":{"rendered":"Something More about Solow Model"},"content":{"rendered":"\n

The current mostly used Solow model always have a depreciation term, and thus the law of motion becomes, \\(\\dot{K}=I-\\delta K\\).<\/p>\n\n\n\n

The mainstream model has different assumptions about the production function as well. For example, technological progress is generally added. 1. \\(Y=AF(K,L)\\) in which technology is exogenous, and it could be called Hicks-neutral; 2. \\(Y=F(K,AL)\\) that can represent the efficient workers, labour-augmented, or Harrow-neutral; 3. \\(Y=F(AK,L)\\) in which the technological progress is capital augmented.<\/p>\n\n\n\n

Applying for example the labour-augmented technology<\/strong> and \\( \\frac{\\dot{A}}{A}=g\\) , we can simply solve the Solow model as the following,<\/p>\n\n\n\n

$$k=\\frac{K}{AL}$$<\/p>\n\n\n\n

$$ \\frac{\\dot{k}}{k}= \\frac{\\dot{K}}{K}- \\frac{\\dot{A}}{A}- \\frac{\\dot{L}}{L} $$<\/p>\n\n\n\n

$$ \\frac{\\dot{k}}{k}= \\frac{sY-\\delta K}{K}- \\frac{\\dot{A}}{A}- \\frac{\\dot{L}}{L} $$<\/p>\n\n\n\n

$$ \\frac{\\dot{k}}{k}= \\frac{sY}{K}-\\delta-g- n $$<\/p>\n\n\n\n

$$ \\dot{k}=sy-(\\delta+g+n)k $$<\/p>\n\n\n\n

, where \\(y=\\frac{Y}{AL}\\) and \\(\\frac{K}{AL}\\) represent the output\/capital per efficient works. Therefore, if \\(\\dot{k}=0\\), then \\(sy=(\\delta+g+n)k\\).<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

The stable point of k is \\(k^*\\) in which \\(sf(k)=(\\delta+n+g)k\\).<\/p>\n\n\n\n

We always the Cobb-Douglas function to represent the production function, because it satisfies CRTS, increasing and diminishing assumptions, and the Inada conditions (\\(\\lim_{k\\rightarrow0}f'(k)=\\infty; \\lim_{k\\rightarrow \\infty}f'(k)=0\\), Inada, 1963 ).<\/p>\n\n\n\n

In the following, we would all analyse the model using efficient works to do analysis.<\/p>\n\n\n\n

Balance Growth Path<\/h4>\n\n\n\n

All the following is assuming the economy is at the steady state or stable point.<\/p>\n\n\n\n

For \\( \\frac{\\dot{K}}{K} \\),<\/strong><\/p>\n\n\n\n

$$ k=\\frac{K}{AL} $$<\/p>\n\n\n\n

By taking logritham,<\/p>\n\n\n\n

$$ ln(k)=ln(K)-ln(A)-ln(L) $$<\/p>\n\n\n\n

By taking differentiation and set \\(\\dot{k}=0\\) (based on our previous derivations of finding the steady state condition).<\/p>\n\n\n\n

$$ \\frac{\\dot{K}}{K} = \\frac{\\dot{A}}{A} + \\frac{\\dot{L}}{L} $$<\/p>\n\n\n\n

$$ \\frac{\\dot{K}}{K} = g+n $$<\/p>\n\n\n\n

For \\( \\frac{\\dot{Y}}{Y} \\),<\/strong> similar as the original Solow’s one.<\/p>\n\n\n\n

$$ln(Y)=ln(F(K,AL))$$<\/p>\n\n\n\n

Differentiate w.r.t. \\(t\\),<\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}=\\frac{ \\dot{K}F_1’+\\dot{A}LF_2’+ A\\dot{L}F_2′ }{F(K,AL)} $$<\/p>\n\n\n\n

By Euler’s Theorem to the demoninator (see math tools),<\/p>\n\n\n\n

P.S. differentiate \\(tY=F(tK,tAL)\\) w.r.t. \\(t\\), then we get \\(Y=F’_1 K+F’_2 AL\\).<\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}=\\frac{ \\dot{K}F_1’+\\dot{A}LF_2’+ A\\dot{L}F_2′ }{ F’_1 K+F’_2 AL } $$<\/p>\n\n\n\n

Devide both numerator and demoninator by \\(KAL\\),<\/p>\n\n\n\n

\\frac{\\dot{Y}}{Y}=\\frac{ \\frac{\\dot{K}}{KAL}F_1’+\\frac{\\dot{A}L}{KAL}F_2’+\\frac{ A\\dot{L}}{KAL}F_2′ }{ \\frac{F’_1 K}{KAL}+\\frac{F’_2 AL}{KAL} }<\/span> <\/p>\n\n\n\n

\\frac{\\dot{Y}}{Y}=\\frac{ \\frac{\\dot{K}}{K}\\frac{F_1′}{AL}+\\frac{\\dot{A}}{A}\\frac{F_2′}{K}+\\frac{ \\dot{L}}{L}\\frac{F_2′}{K} }{ \\frac{F’_1 }{AL}+\\frac{F’_2 }{K} }= \\frac{ \\frac{\\dot{K}}{K}\\frac{F_1′}{AL}+(\\frac{\\dot{A}}{A}+\\frac{\\dot{L}}{L})\\frac{F_2′}{K} }{ \\frac{F’_1 }{AL}+\\frac{F’_2 }{K} }<\/span> <\/p>\n\n\n\n

\\frac{\\dot{Y}}{Y}= (n+g)\\frac{ \\frac{F_1′}{AL}+\\frac{F_2′}{K} }{ \\frac{F’_1 }{AL}+\\frac{F’_2 }{K} } = (\\frac{\\dot{K}}{K})\\frac{ \\frac{F_1′}{AL}+\\frac{F_2′}{K} }{ \\frac{F’_1 }{AL}+\\frac{F’_2 }{K} }<\/span> <\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}=n+g = \\frac{\\dot{K}}{K} $$<\/p>\n\n\n\n

For \\( \\frac{\\dot{y}}{y} \\),<\/strong> (as \\(y=\\frac{Y}{AL})<\/p>\n\n\n\n

$$ln(y)=ln(Y)-ln(A)-ln(L)$$<\/p>\n\n\n\n

$$ \\frac{\\dot{y}}{y}= \\frac{\\dot{Y}}{Y}- \\frac{\\dot{A}}{A}- \\frac{\\dot{L}}{L} =(n+g)-n-g $$<\/p>\n\n\n\n

$$ \\frac{\\dot{y}}{y} =0$$<\/p>\n\n\n\n

Similarly, for per capita terms,<\/p>\n\n\n\n

For \\( \\frac{\\dot{K\/L}}{K\/L} \\)<\/strong>, per capita capital,<\/p>\n\n\n\n

$$\\frac{\\dot{K\/L}}{K\/L}=\\frac{ \\frac{\\dot{K}L-K\\dot{L}}{L^2} }{K\/L}$$<\/p>\n\n\n\n

$$\\frac{\\dot{K\/L}}{K\/L}=\\frac{\\dot{K}L-K\\dot{L}}{KL}= \\frac{\\dot{K}}{K}- \\frac{\\dot{L}}{L} $$<\/p>\n\n\n\n

\\frac{\\dot{K\/L}}{K\/L} =(g+n)-n=g<\/span><\/p>\n\n\n\n

For \\( \\frac{\\dot{Y\/L}}{Y\/L} \\)<\/strong> (per capita output) we apply the same transformation as K\/L, <\/p>\n\n\n\n

\\frac{\\dot{Y\/L}}{Y\/L}= \\frac{\\dot{Y}}{Y}- \\frac{\\dot{L}}{L} =g<\/span><\/p>\n\n\n\n

In summary, the BGP is a situation in which each variable of the model is growing at a constant rate. On the balanced growth path, the growth rate of output per worker is determined solely by the rate of growth of technology.<\/p>\n\n\n\n

P.S. Technology Independent of Labour And Capital<\/strong><\/p>\n\n\n\n

Applying for example the Type 1 case<\/strong> and \\( \\frac{\\dot{A}}{A}=g\\) , we can simply solve the Solow model as the following, <\/p>\n\n\n\n

We would not use capital per efficient worker here, because labour is not technology-augmented by assumption. Instead, we simply assume capital per capita, \\(k=\\frac{K}{L}\\). We can easily get the relationship,<\/p>\n\n\n\n

\\frac{\\dot{k}}{k}= \\frac{\\dot{K}}{K}- \\frac{\\dot{L}}{L}<\/span> <\/p>\n\n\n\n

By setting \\(\\dot{k}=0\\), we can find \\( \\frac{\\dot{K}}{K}=\\frac{\\dot{L}}{L}=n \\), which is same as Solow’s original works. <\/p>\n\n\n\n

However, the difference is when we deal with the output. As the output is now \\(Y=AF(K,L)\\), so the changes in outputs (numerator<\/strong>) are,<\/p>\n\n\n\n

$$ \\dot{Y}=\\dot{A}F(K,L)+AF’_1\\dot{K}+AF’_2\\dot{L} $$<\/p>\n\n\n\n

We expand output per se<\/em> (demoninator<\/strong>) by Euler’s Theorem \\(Y=AF’_1K+AF’_2L\\) (A is now outside the production function), and then calculate the percentage changes of outputs,<\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}=\\frac{ \\dot{A}F(K,L)+AF’_1\\dot{K}+AF’_2\\dot{L} }{ AF’_1K+AF’_2L } $$<\/p>\n\n\n\n

Devided both demoninator and numerator by AKL,<\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}=\\frac{ \\dot{A}F(K,L) }{ AF(K,L) }+\\frac{\\frac{F’_1}{L}\\frac{\\dot{K}}{K}+\\frac{F’_2}{K}\\frac{\\dot{L}}{L} }{ \\frac{F’_1}{L}+\\frac{F’_2}{K} } $$<\/p>\n\n\n\n

$$ \\frac{\\dot{Y}}{Y}= \\frac{\\dot{A}}{A}+ \\frac{\\dot{L}}{L}=g+n $$<\/p>\n\n\n\n

Saving Rates<\/h4>\n\n\n\n

We now consider first how does changes in the saving rate affect those factors.<\/p>\n\n\n\n

The determinants of saving rate are, for example, uncertainty or decrease in expected income, and required pension rate.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

See the following figures,<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

An increase in the saving rate would result in an increase in the investment curve. \\(\\dot{K}=I-\\delta K\\) tells that there would be a huge increase in \\(\\dot{K}\\) initially, and by the shape of production function, the difference diminishes until achieving the new stable point \\(k^*_{new}\\).<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

As \\(\\dot{k}\\) is a derivative of \\(k\\) w.r.t. \\(t\\), we can easily get the time path of \\(k\\) as the following,<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Another important factor is the growth rate of output per capita,<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Also \\(ln(Y\/L)\\),<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

For this one, we can prove that the slope of \\(ln(Y\/L)\\) is \\(\\dot{ln(Y\/L)}=\\frac{\\partial}{\\partial t}[ln(Y)-ln(L)]=(g+n)-n=g\\), so it grows constantly at rate “g” before \\(t_0\\). Later growth rate of Y jumps makes the slope of \\(ln(Y\/L)\\) increases, but \\(ln(Y\/L)=g\\) when achieves a new steady state and \\(ln(Y\/L)\\) keeps growing at “g” in the long run.<\/p>\n\n\n\n

The Speed of Convergence<\/h4>\n\n\n\n

Way 1<\/strong><\/p>\n\n\n\n

We follow our Solow model with labour-augmented technology. The time path of changes of capital per efficient works is,<\/p>\n\n\n\n

$$ \\dot{k}=sy-(\\delta+n+g)k$$<\/p>\n\n\n\n

$$ \\dot{k}=sy-(\\delta+n+g)k$$<\/p>\n\n\n\n

At the steady state, \\(\\dot{k}=0\\), so \\( sy-(\\delta+n+g)k \\). We then plug in the Cobb-Doglas production function and denote \\(y=\\frac{Y}{AL}=\\frac{K^{\\alpha}(AL)^{1-\\alpha}}{AL}=k^{\\alpha}\\), we can find the \\(k^*\\),<\/p>\n\n\n\n

$$ k^*=(\\frac{s}{\\delta+g+n})^{\\frac{1}{1-\\alpha}} $$<\/p>\n\n\n\n

And get the path of k,<\/p>\n\n\n\n

$$ \\frac{\\dot{k}}{k}=sk^{\\alpha-1}-(\\delta+g+n) :=G(k)$$<\/p>\n\n\n\n

To find the speed of convergence, we would focus on the time path of k around \\(k^*\\). Or approximate the time-path by taking first-order Taylor expansion<\/strong> around \\(k^*\\)<\/strong> to approximate,<\/p>\n\n\n\n

$$ G(k)\\approx G(k^*)+G'(k^*)(k-k^*) $$<\/p>\n\n\n\n

As \\(G(k^*)=0\\) by our proof of steady state condition, thus,<\/p>\n\n\n\n

$$ G(k)\\approx (\\alpha-1)s {k^*}^{\\alpha-1}\\frac{k-k^*}{k^*} $$<\/p>\n\n\n\n

We plug the steady state \\(k^*\\) back into the above equation and get,<\/p>\n\n\n\n

$$ G(k)=-(1-\\alpha)(\\delta+g+n)\\frac{k-k^*}{k^*} $$<\/p>\n\n\n\n

Therefore, we find the mathematic expression of the convergence speed, \\( (1-\\alpha)(\\delta+g+n) \\). It is the measure of how quickly k changes when k diviates from \\(k^*\\). Also, we find that the growth rate \\( G(k)=\\frac{\\dot{k}}{k} \\) depends on both the convergence speed and \\( \\frac{k-k^*}{k^*} \\), which is how far k deviates from its steady state level.<\/p>\n\n\n\n

Take also a Taylor expansion to \\(ln(k)\\) at \\(k^*\\), we would get,<\/p>\n\n\n\n

$$ G(k)=-(1-\\alpha)(\\delta+g+n)(ln(k)-ln(k^*)) :=g_k$$<\/p>\n\n\n\n

Then, to find the convergence speed of outputs, we apply \\(y=k^{\\alpha}\\) and take logritham \\(ln(y)=\\alpha ln(k)\\). Differentiate w.r.t. \\(t\\),<\/p>\n\n\n\n

$$ \\frac{\\dot{y}}{y} =\\alpha\\frac{\\dot{k}}{k} $$<\/p>\n\n\n\n

$$ g_y:=\\frac{\\dot{y}}{y} =\\alpha( -(1-\\alpha)(\\delta+g+n)(ln(k)-ln(k^*))) \\\\= -(1-\\alpha)(\\delta+g+n)(ln(y)-ln(y^*)) $$<\/p>\n\n\n\n

So we get \\(g_y=\\alpha g_k\\), and \\(\\beta= (1-\\alpha)(\\delta+g+n) \\) is the speed of convergence. It measures how quickly \\(y\\) increases when \\(y<y^*\\). The growth rate of y depends on the speed of convergence, \\(\\beta\\), and the log-difference between \\(y\\) and \\(y^*\\).<\/p>\n\n\n\n

Way 2<\/strong><\/p>\n\n\n\n

We take first order Taylor approximation to \\(f(k)=\\dot{k}\\) around \\(k=k^*\\).<\/p>\n\n\n\n

$$ \\dot{k} \\approx \\dot{k}|_{k=k^*}+\\frac{\\partial \\dot{k}}{\\partial k}|_{k=k^*}(k-k^*) $$<\/p>\n\n\n\n

By definition of steady state condition, the first term of RHS is zero. So,<\/p>\n\n\n\n

$$ \\dot{k}\\approx -\\lambda \\cdot (k-k^*) $$<\/p>\n\n\n\n

We denote \\(-\\frac{\\partial \\dot{k}}{\\partial k}|_{k=k^*}\\:=\\lambda\\) as the speed of convergence. As \\(\\dot{k}=sy-(\\delta+g+n)k=sk^{\\alpha}- (\\delta+g+n)k\\), so,<\/p>\n\n\n\n

$$ \\lambda=-s\\alpha {k^*}^{\\alpha-1}- (\\delta+g+n) $$<\/p>\n\n\n\n

Plug \\(k^*\\) into, we get,<\/p>\n\n\n\n

$$ \\lambda=(1-\\alpha)(\\delta+g+n) $$<\/p>\n\n\n\n

To see why we denote \\(\\lambda\\)as the speed of convergence, solve the differential equation, \\( \\dot{k}\\approx -\\lambda \\cdot (k-k^*) \\), by restrict time from 0 to t.<\/p>\n\n\n\n

$$ \\dot{k}=\\frac{\\partial k}{\\partial t} \\approx -\\lambda \\cdot (k-k^*) $$<\/p>\n\n\n\n

$$ \\frac{1}{k-k^*} dk=-\\lambda dt $$<\/p>\n\n\n\n

$$\\int_{k(0)}^{k(t)} \\frac{1}{k-k^*} dk=\\int_{0}^t -\\lambda dt $$<\/p>\n\n\n\n

$$ [ln(k-k^*)]|^{k(t)}_{k(0)}=-\\lambda t|_0^t $$<\/p>\n\n\n\n

$$ ln(k(t)-k^*)=-\\lambda t|_0^t+ ln(k(0)-k^*) $$<\/p>\n\n\n\n

Finally,<\/p>\n\n\n\n

$$ k(t)=k^*+e^{-\\lambda t}[ k(0)-k^* ] $$<\/p>\n\n\n\n

Or in other form,<\/p>\n\n\n\n

$$ ln(\\frac{k(t)-k^*}{k(0)-k^*})=-\\lambda t $$<\/p>\n\n\n\n

Solow Residuals<\/h4>\n\n\n\n

Recall our labour-augmented production function, \\(Y(t)=F(K(t),A(t)L(t))\\).<\/p>\n\n\n\n

$$ \\dot{Y}=\\frac{\\partial Y}{\\partial t}=F’_1\\dot{K}+ F’_2\\dot{A}+ F’_2\\dot{L} $$<\/p>\n\n\n\n

$$\\frac{ \\dot{Y}}{Y}=\\frac{\\partial Y(t)}{\\partial K(t)}\\dot{K(t)}+ \\frac{\\partial Y(t)}{\\partial L(t)}\\dot{L(t)}+ \\frac{\\partial Y(t)}{\\partial A(t)}\\dot{A(t)} $$<\/p>\n\n\n\n

Then, applying the replacement equation into the above equation,<\/p>\n\n\n\n

$$ \\frac{\\partial Y(t)}{\\partial L(t)}=\\frac{\\partial Y(t)}{\\partial A(t)L(t)}\\cdot A(t) $$<\/p>\n\n\n\n

$$ \\frac{\\partial Y(t)}{\\partial A(t)}=\\frac{\\partial Y(t)}{\\partial A(t)L(t)}\\cdot L(t) $$<\/p>\n\n\n\n

Then we get,<\/p>\n\n\n\n

$$ \\frac{ \\dot{Y}}{Y}=\\frac{Y(t)}{K(t)}\\frac{\\dot{K(t)}}{K(t)}\\frac{K(t)}{Y(t)}+ \\frac{Y(t)}{L(t)}\\frac{\\dot{L(t)}}{L(t)}\\frac{L(t)}{Y(t)}+ \\frac{Y(t)}{A(t)}\\frac{\\dot{A(t)}}{A(t)}\\frac{A(t)}{Y(t)}\\\\=\\epsilon(t)_{Y,K}\\frac{\\dot{K(t)}}{K(t)}+ \\epsilon(t)_{Y,L}\\frac{\\dot{L(t)}}{L(t)}+R(t) $$<\/p>\n\n\n\n

,where we denote \\(R(t)\\) as the Solow Residuals.<\/p>\n\n\n\n

$$ R(t) = \\frac{Y(t)}{A(t)}\\frac{\\dot{A(t)}}{A(t)}\\frac{A(t)}{Y(t)} $$<\/p>\n\n\n\n

Solow Residuals represent the residuals unexplained by growth of capital and labours.<\/p>\n\n\n\n

Golden Rule Saving Rate (Phelps)<\/h4>\n\n\n\n

To be continued.<\/p>\n\n\n\n

Reference<\/h4>\n\n\n\n