Question #228351
1) suppose output is growing at 3 percenst per year and capital’s and labor's share of income are. 3 and. 7 respectively.

a. if the labour and capital grow at 1 percent per year what would the growth rate of total factor productivity have to be?
b.what if both the labor and stocks are fixed?
2) Consider the following production function: Y=K^0.5(AN)^0.5, where both the production and pool of labor are growing at a rate n =.07, the capital stock is depreciating at a rate d =.03 and A is normalized to 1.
a. What are capital's and labor's share of income?
b. What is the form of this production function?
c. Find the steady state values of k and y when s=.20.
d. At what rate is per capita output growing at the steady state? At what rate is total output growing? What if total factor productivity is increasing at a rate of 2 percent per year (g=.02)?
1
Expert's answer
2021-08-23T13:13:46-0400

2.

a. Capital's share of income = 0.5 (given by the index of capital

and labor's share of income = (given by the index pf labor)

b. the form of this production function = cobb-douglas

c.Y=K0.5(AN)0.5;A=1Y=K0.5N0.5    y=YN=K0.5;K=KLAt steady state,8.y=(σ+n)K    0.20K0.5=(0.03+0.07)K    K0.5=0.03+0.070.20    Kα=0.25    yα=(0.25)0.5=05.c.\\Y=K^{0.5}(AN)^{0.5};A=1\\Y=K^{0.5}N^{0.5}\\\implies y=\frac{Y}{N}=K^{0.5};K=\frac{K}{L}\\At \space steady\space state,\\8.y=(\sigma+n)K\\\implies0.20K^{0.5}=(0.03+0.07)K\\\implies K^{0.5}=\frac{0.03+0.07}{0.20}\\\implies K^{\alpha}=0.25\\\implies y^{\alpha}=(0.25)^{0.5}=05.


d. at steady state, per capita output is growing at the rate of n=0.07 and at steady state, total output is growing at the rate of n+d=0.07+0.03=0.10n+d =0.07+0.03=0.10


Now, total factor productivity is increasing at a rate of 2 percent per year (g=0.02), then the new steady state equilibrium and growth rates are as follows.

steady state

0.2K0.5=0.03+0.07+0.020.200.2K^{0.5}=\frac{0.03+0.07+0.02}{0.20}\\


K0.5=0.03+0.07+0.020.20K^{0.5}=\frac{0.03+0.07+0.02}{0.20}

Kα=0.36Yα=(0.36)0.5=0.6K^{\alpha}=0.36\\Y^{\alpha}=(0.36)^{0.5}=0.6


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