How to Calculate the productivity multiplier?












1












$begingroup$


Given a Cobb Douglas



$Y_t = A (K_t^alpha L_t^{1-alpha}) $



$ K_{t+1} = sY_t + (1-delta) K_t$



How do we get the multiplier on productivity to be equal to $ frac{1}{1-alpha}$?
I understand that if productivity increases, output increases, thus we get more capital and thereby more output and so on. But I can't reach this multiplier.



My Attempt:



If have x increase in A,
then
$Y_t=(1+x)A (K_t^alpha L_t^{1-alpha}) $



That is an x increase in Y.



$ K_{t+1} = s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t$



$Y_{t+1}=(1+x)A * K_{t+1}^alpha *L_{t+1}^{1-alpha} $
$Y_{t+1}= (1+x)A * (s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t)^alpha * L_{t+1}^{1-alpha} $



I think the increase here should be $x * alpha $ but I can't see it.
so that for a unit increase in productivity i.e x=1;
$Delta Y = 1 + alpha + alpha^2 +... =frac{1}{1-alpha} $



Also in the steady state we have,
$ Y = A^frac{1}{1-alpha} * frac{s}{delta}^frac{alpha}{1-alpha} * L$
taking logs we have that percentage change in A increases Y by $frac{1}{1-alpha}$.






macroeconomics economic-growth cobb-douglas solow growth-accounting







share|improve this question









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  • 1




    $begingroup$
    You should show any work you've already done in attempting to solve this question.
    $endgroup$
    – Kenny LJ
    5 hours ago










  • $begingroup$
    @KennyLJ I updated with my attempt
    $endgroup$
    – Faye
    4 hours ago
















1












$begingroup$


Given a Cobb Douglas



$Y_t = A (K_t^alpha L_t^{1-alpha}) $



$ K_{t+1} = sY_t + (1-delta) K_t$



How do we get the multiplier on productivity to be equal to $ frac{1}{1-alpha}$?
I understand that if productivity increases, output increases, thus we get more capital and thereby more output and so on. But I can't reach this multiplier.



My Attempt:



If have x increase in A,
then
$Y_t=(1+x)A (K_t^alpha L_t^{1-alpha}) $



That is an x increase in Y.



$ K_{t+1} = s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t$



$Y_{t+1}=(1+x)A * K_{t+1}^alpha *L_{t+1}^{1-alpha} $
$Y_{t+1}= (1+x)A * (s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t)^alpha * L_{t+1}^{1-alpha} $



I think the increase here should be $x * alpha $ but I can't see it.
so that for a unit increase in productivity i.e x=1;
$Delta Y = 1 + alpha + alpha^2 +... =frac{1}{1-alpha} $



Also in the steady state we have,
$ Y = A^frac{1}{1-alpha} * frac{s}{delta}^frac{alpha}{1-alpha} * L$
taking logs we have that percentage change in A increases Y by $frac{1}{1-alpha}$.






macroeconomics economic-growth cobb-douglas solow growth-accounting







share|improve this question









New contributor




Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You should show any work you've already done in attempting to solve this question.
    $endgroup$
    – Kenny LJ
    5 hours ago










  • $begingroup$
    @KennyLJ I updated with my attempt
    $endgroup$
    – Faye
    4 hours ago














1












1








1





$begingroup$


Given a Cobb Douglas



$Y_t = A (K_t^alpha L_t^{1-alpha}) $



$ K_{t+1} = sY_t + (1-delta) K_t$



How do we get the multiplier on productivity to be equal to $ frac{1}{1-alpha}$?
I understand that if productivity increases, output increases, thus we get more capital and thereby more output and so on. But I can't reach this multiplier.



My Attempt:



If have x increase in A,
then
$Y_t=(1+x)A (K_t^alpha L_t^{1-alpha}) $



That is an x increase in Y.



$ K_{t+1} = s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t$



$Y_{t+1}=(1+x)A * K_{t+1}^alpha *L_{t+1}^{1-alpha} $
$Y_{t+1}= (1+x)A * (s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t)^alpha * L_{t+1}^{1-alpha} $



I think the increase here should be $x * alpha $ but I can't see it.
so that for a unit increase in productivity i.e x=1;
$Delta Y = 1 + alpha + alpha^2 +... =frac{1}{1-alpha} $



Also in the steady state we have,
$ Y = A^frac{1}{1-alpha} * frac{s}{delta}^frac{alpha}{1-alpha} * L$
taking logs we have that percentage change in A increases Y by $frac{1}{1-alpha}$.






macroeconomics economic-growth cobb-douglas solow growth-accounting







share|improve this question









New contributor




Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Given a Cobb Douglas



$Y_t = A (K_t^alpha L_t^{1-alpha}) $



$ K_{t+1} = sY_t + (1-delta) K_t$



How do we get the multiplier on productivity to be equal to $ frac{1}{1-alpha}$?
I understand that if productivity increases, output increases, thus we get more capital and thereby more output and so on. But I can't reach this multiplier.



My Attempt:



If have x increase in A,
then
$Y_t=(1+x)A (K_t^alpha L_t^{1-alpha}) $



That is an x increase in Y.



$ K_{t+1} = s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t$



$Y_{t+1}=(1+x)A * K_{t+1}^alpha *L_{t+1}^{1-alpha} $
$Y_{t+1}= (1+x)A * (s(1+x)A (K_t^alpha L_t^{1-alpha}) + (1-delta) K_t)^alpha * L_{t+1}^{1-alpha} $



I think the increase here should be $x * alpha $ but I can't see it.
so that for a unit increase in productivity i.e x=1;
$Delta Y = 1 + alpha + alpha^2 +... =frac{1}{1-alpha} $



Also in the steady state we have,
$ Y = A^frac{1}{1-alpha} * frac{s}{delta}^frac{alpha}{1-alpha} * L$
taking logs we have that percentage change in A increases Y by $frac{1}{1-alpha}$.






macroeconomics economic-growth cobb-douglas solow growth-accounting




macroeconomics economic-growth cobb-douglas solow growth-accounting






share|improve this question









New contributor




Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 3 hours ago









Kenny LJ

4,86421644




4,86421644






New contributor




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Check out our Code of Conduct.









asked 5 hours ago









FayeFaye

1084




1084




New contributor




Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Faye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    You should show any work you've already done in attempting to solve this question.
    $endgroup$
    – Kenny LJ
    5 hours ago










  • $begingroup$
    @KennyLJ I updated with my attempt
    $endgroup$
    – Faye
    4 hours ago














  • 1




    $begingroup$
    You should show any work you've already done in attempting to solve this question.
    $endgroup$
    – Kenny LJ
    5 hours ago










  • $begingroup$
    @KennyLJ I updated with my attempt
    $endgroup$
    – Faye
    4 hours ago








1




1




$begingroup$
You should show any work you've already done in attempting to solve this question.
$endgroup$
– Kenny LJ
5 hours ago




$begingroup$
You should show any work you've already done in attempting to solve this question.
$endgroup$
– Kenny LJ
5 hours ago












$begingroup$
@KennyLJ I updated with my attempt
$endgroup$
– Faye
4 hours ago




$begingroup$
@KennyLJ I updated with my attempt
$endgroup$
– Faye
4 hours ago










1 Answer
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oldest

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2












$begingroup$

Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^alpha$.



Steady state is given by: $$k^*=sy^*+(1-delta)k^*,$$ or $$k^*=sA(k^*)^alpha+(1-delta)k^*.$$



Doing the algebra: $$k^*=left(frac{sA}{delta}right)^{frac{1}{1-alpha}}.$$



And: $$y^*=Aleft(frac{sA}{delta}right)^{frac{alpha}{1-alpha}}=A^{frac{1}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$frac{partial y^*}{partial A}=frac{1}{1-alpha}A^{frac{alpha}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$frac{partial y^*}{partial A}div frac{y^*}{A}=frac{1}{1-alpha}.$$






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    $begingroup$

    Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^alpha$.



    Steady state is given by: $$k^*=sy^*+(1-delta)k^*,$$ or $$k^*=sA(k^*)^alpha+(1-delta)k^*.$$



    Doing the algebra: $$k^*=left(frac{sA}{delta}right)^{frac{1}{1-alpha}}.$$



    And: $$y^*=Aleft(frac{sA}{delta}right)^{frac{alpha}{1-alpha}}=A^{frac{1}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



    I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$frac{partial y^*}{partial A}=frac{1}{1-alpha}A^{frac{alpha}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



    However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$frac{partial y^*}{partial A}div frac{y^*}{A}=frac{1}{1-alpha}.$$






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^alpha$.



      Steady state is given by: $$k^*=sy^*+(1-delta)k^*,$$ or $$k^*=sA(k^*)^alpha+(1-delta)k^*.$$



      Doing the algebra: $$k^*=left(frac{sA}{delta}right)^{frac{1}{1-alpha}}.$$



      And: $$y^*=Aleft(frac{sA}{delta}right)^{frac{alpha}{1-alpha}}=A^{frac{1}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



      I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$frac{partial y^*}{partial A}=frac{1}{1-alpha}A^{frac{alpha}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



      However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$frac{partial y^*}{partial A}div frac{y^*}{A}=frac{1}{1-alpha}.$$






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^alpha$.



        Steady state is given by: $$k^*=sy^*+(1-delta)k^*,$$ or $$k^*=sA(k^*)^alpha+(1-delta)k^*.$$



        Doing the algebra: $$k^*=left(frac{sA}{delta}right)^{frac{1}{1-alpha}}.$$



        And: $$y^*=Aleft(frac{sA}{delta}right)^{frac{alpha}{1-alpha}}=A^{frac{1}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



        I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$frac{partial y^*}{partial A}=frac{1}{1-alpha}A^{frac{alpha}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



        However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$frac{partial y^*}{partial A}div frac{y^*}{A}=frac{1}{1-alpha}.$$






        share|improve this answer









        $endgroup$



        Let $y=Y/L$ and $k=K/L$ be the per-worker levels of output and capital. Observe that $y=Ak^alpha$.



        Steady state is given by: $$k^*=sy^*+(1-delta)k^*,$$ or $$k^*=sA(k^*)^alpha+(1-delta)k^*.$$



        Doing the algebra: $$k^*=left(frac{sA}{delta}right)^{frac{1}{1-alpha}}.$$



        And: $$y^*=Aleft(frac{sA}{delta}right)^{frac{alpha}{1-alpha}}=A^{frac{1}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



        I am not sure what is meant by the "multiplier on productivity". I would interpret this term to be the answer to the question, "Given a small unit change in $A$, what is the resultant change in $y^*$?" That is, the following expression: $$frac{partial y^*}{partial A}=frac{1}{1-alpha}A^{frac{alpha}{1-alpha}}left(frac{s}{delta}right)^{frac{alpha}{1-alpha}}.$$



        However, clearly, this does not correspond to your desired answer. So I suspect what is really meant is the elasticity of $y^*$ with respect to $A$: $$frac{partial y^*}{partial A}div frac{y^*}{A}=frac{1}{1-alpha}.$$







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        Kenny LJKenny LJ

        4,86421644




        4,86421644






















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