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Consider an economy with production function Y = K^α*T^β*(AL)^1−α−β where K, A, L are, as usual, capital, technological progress and labour and T is the stock of land. α, β > 0. A grows at rate g and L grows at rate n. The stock of land is fixed though. Aggregate saving equals a fraction s of aggregate output. Assuming K depreciates at rate δ Derive the steady state growth rate of capital a) K./K = (1-α-β)(g+n)/(1-β) b) K./K = (1-α-β)(n)/(1-β) c) K./K = (1-α-β)(g+n)/(1-α) d) None of these Derive the condition when steady state growth rate of output per worker is positive. a) (1-α-β)g > βn b) (1-α-β)g < βn c) (1-α-β)g = βn d) None of these
Akshay Jain
Masters in Economics Delhi School of Economics 2013-15 |
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1. c
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In reply to this post by Akshay Jain
I m also getting c for 1.....
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In reply to this post by Akshay Jain
Nd a for 2..
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In reply to this post by Akshay Jain
Raj,Vaibhav...how did you get c....plss help
MA Economics
DSE 2014-16 |
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Just write steady state eq first interms of per labour output nd per labour capital.....den apply log nd differentiate it with respect to t nd solve for dk/dt*k..
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The correct answers are 'c' for 1st nd 'a' for 2nd
Akshay Jain
Masters in Economics Delhi School of Economics 2013-15 |
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In reply to this post by Akshay Jain
Vaibhav my answer is not matching .Please help me with solution.
MA Economics
DSE 2014-16 |
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This post was updated on Jun 18, 2014; 6:03pm.
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Thanks vaibhav . Please explain second part !
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Dreyfus
Thanxx vaibhav..
MA Economics
DSE 2014-16 |
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In reply to this post by Anjali
Steady state growth rate of k obtained in 1 is embedded with growth rate of labour.....thus the steady state growth rate of output per capita is the steady state growth rate of k less growth rate of labour,n, and for this growth rate rate to b positive use inequality
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Yeah the 1st one is inclusive of technology also . But Iam not getting the question - why to reduce only growth labour parameter , why not technology parameter ?
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Akshay Jain
Akshay plz help me with the second part !
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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after getting the steady state level of capital/worker k*, compute steady state output per worker y* same as what u do in solow model.
Take logs both sides and dn differentiate it w.r.t. time t...you vl get (dy*/dt)/y*=growth rate of output per worker in steady state=some expression.....now this expresssion must be greater dn 0 for growth rate to be positive......u vl get condition "a" after solving...try this out else vl upload.....:)
Akshay Jain
Masters in Economics Delhi School of Economics 2013-15 |
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See first we will take the steady state equation :
Change in k = s * y-( dep+n+g ) *k So at steady state change in k will be 0 Then we will take log and differentiate , that will give us growth rate Now my equation is coming as Growth rate of steady state capital = growth of output per worker I guess Iam going wrong ! ![]()
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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Akshay Jain
Masters in Economics Delhi School of Economics 2013-15 |
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In reply to this post by Akshay Jain
Thanx Akshay :)
MA Economics
DSE 2014-16 |
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