Deriving the translog production function












1














Ive been having difficulty deriving the translog production function defined as:



$$ln y=alpha_0+sum_{i=1}^nalpha_i ln x_i+frac{1}{2}sum_{i=1}^nsum_{j=1}^n beta_{ij}ln x_iln x_j $$



I know we start with a log-log production function.
$$ln y=alpha_0+sum_{i=1}^nalpha_iln x_i$$



the next step from what i recall is to take the taylor series of this function around the point $x_i=0$. the reason why this is an issue is because $ln(0)$ is undefined.



How exactly is this function derived?






microeconomics econometrics production-function







share|improve this question



























    1














    Ive been having difficulty deriving the translog production function defined as:



    $$ln y=alpha_0+sum_{i=1}^nalpha_i ln x_i+frac{1}{2}sum_{i=1}^nsum_{j=1}^n beta_{ij}ln x_iln x_j $$



    I know we start with a log-log production function.
    $$ln y=alpha_0+sum_{i=1}^nalpha_iln x_i$$



    the next step from what i recall is to take the taylor series of this function around the point $x_i=0$. the reason why this is an issue is because $ln(0)$ is undefined.



    How exactly is this function derived?






    microeconomics econometrics production-function







    share|improve this question

























      1












      1








      1







      Ive been having difficulty deriving the translog production function defined as:



      $$ln y=alpha_0+sum_{i=1}^nalpha_i ln x_i+frac{1}{2}sum_{i=1}^nsum_{j=1}^n beta_{ij}ln x_iln x_j $$



      I know we start with a log-log production function.
      $$ln y=alpha_0+sum_{i=1}^nalpha_iln x_i$$



      the next step from what i recall is to take the taylor series of this function around the point $x_i=0$. the reason why this is an issue is because $ln(0)$ is undefined.



      How exactly is this function derived?






      microeconomics econometrics production-function







      share|improve this question













      Ive been having difficulty deriving the translog production function defined as:



      $$ln y=alpha_0+sum_{i=1}^nalpha_i ln x_i+frac{1}{2}sum_{i=1}^nsum_{j=1}^n beta_{ij}ln x_iln x_j $$



      I know we start with a log-log production function.
      $$ln y=alpha_0+sum_{i=1}^nalpha_iln x_i$$



      the next step from what i recall is to take the taylor series of this function around the point $x_i=0$. the reason why this is an issue is because $ln(0)$ is undefined.



      How exactly is this function derived?






      microeconomics econometrics production-function




      microeconomics econometrics production-function






      share|improve this question













      share|improve this question











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      asked 7 hours ago









      EconJohn♦

      3,3571938




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          The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as



          $$
          Y = A[alpha K^gamma + (1 - alpha)L^gamma]^{1/gamma} tag{1}
          $$



          in this case $X_1 = K$, $X_2 = L$. Now we expand $ln Y$ around $gamma = 0$




          $gamma^0$ term




          $$
          lim_{gamma to 0} ln Y = ln (A K^alpha L^{1 - alpha}) tag{2}
          $$




          $gamma^1$ term




          begin{eqnarray}
          lim_{gamma to 0} frac{partial ln Y}{partial gamma} &=& lim_{gamma to 0}frac{alpha K^{gamma } ln (L)+(1-alpha ) l^{gamma } log (L)}{gamma
          left(alpha K^{gamma }+(1-alpha ) L^{gamma }right)}-frac{ln left(alpha
          K^{gamma }+(1-alpha ) L^{gamma }right)}{gamma ^2}\
          &=& frac{1}{2} (1 - alpha) alpha (ln (K)-ln (L))^2 tag{3}
          end{eqnarray}



          Up to first order we have then



          begin{eqnarray}
          ln Y &approx& color{blue}{(ln Y)_{gamma = 0}} + color{red}{left(frac{partial ln Y}{partial gamma}right)_{gamma = 0} gamma} \
          &stackrel{(2),(3)}{=}& color{blue}{ln A + alpha ln K + (1 - alpha) ln L} + color{red}{frac{1}{2}alpha(1 - alpha)gamma [ln K - ln L]^2} \
          &=& ln A + alpha ln X_1 + (1 - alpha) ln X_2 + frac{alpha gamma (1 -alpha)}{2}left[ln^2 X_1 -2ln X_1ln X_2 + ln^2 X_2 right] \
          &=& alpha_0 + sum_{i = 1}^2 alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^2 beta_{ij}ln X_i ln X_j tag{3}
          end{eqnarray}



          This is naturally extended to $n > 2$ as



          $$
          ln Y = alpha_0 + sum_{i = 1}^n alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^n beta_{ij}ln X_i ln X_j
          $$






          share|improve this answer





















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            1 Answer
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            1 Answer
            1






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            active

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            2














            The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as



            $$
            Y = A[alpha K^gamma + (1 - alpha)L^gamma]^{1/gamma} tag{1}
            $$



            in this case $X_1 = K$, $X_2 = L$. Now we expand $ln Y$ around $gamma = 0$




            $gamma^0$ term




            $$
            lim_{gamma to 0} ln Y = ln (A K^alpha L^{1 - alpha}) tag{2}
            $$




            $gamma^1$ term




            begin{eqnarray}
            lim_{gamma to 0} frac{partial ln Y}{partial gamma} &=& lim_{gamma to 0}frac{alpha K^{gamma } ln (L)+(1-alpha ) l^{gamma } log (L)}{gamma
            left(alpha K^{gamma }+(1-alpha ) L^{gamma }right)}-frac{ln left(alpha
            K^{gamma }+(1-alpha ) L^{gamma }right)}{gamma ^2}\
            &=& frac{1}{2} (1 - alpha) alpha (ln (K)-ln (L))^2 tag{3}
            end{eqnarray}



            Up to first order we have then



            begin{eqnarray}
            ln Y &approx& color{blue}{(ln Y)_{gamma = 0}} + color{red}{left(frac{partial ln Y}{partial gamma}right)_{gamma = 0} gamma} \
            &stackrel{(2),(3)}{=}& color{blue}{ln A + alpha ln K + (1 - alpha) ln L} + color{red}{frac{1}{2}alpha(1 - alpha)gamma [ln K - ln L]^2} \
            &=& ln A + alpha ln X_1 + (1 - alpha) ln X_2 + frac{alpha gamma (1 -alpha)}{2}left[ln^2 X_1 -2ln X_1ln X_2 + ln^2 X_2 right] \
            &=& alpha_0 + sum_{i = 1}^2 alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^2 beta_{ij}ln X_i ln X_j tag{3}
            end{eqnarray}



            This is naturally extended to $n > 2$ as



            $$
            ln Y = alpha_0 + sum_{i = 1}^n alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^n beta_{ij}ln X_i ln X_j
            $$






            share|improve this answer


























              2














              The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as



              $$
              Y = A[alpha K^gamma + (1 - alpha)L^gamma]^{1/gamma} tag{1}
              $$



              in this case $X_1 = K$, $X_2 = L$. Now we expand $ln Y$ around $gamma = 0$




              $gamma^0$ term




              $$
              lim_{gamma to 0} ln Y = ln (A K^alpha L^{1 - alpha}) tag{2}
              $$




              $gamma^1$ term




              begin{eqnarray}
              lim_{gamma to 0} frac{partial ln Y}{partial gamma} &=& lim_{gamma to 0}frac{alpha K^{gamma } ln (L)+(1-alpha ) l^{gamma } log (L)}{gamma
              left(alpha K^{gamma }+(1-alpha ) L^{gamma }right)}-frac{ln left(alpha
              K^{gamma }+(1-alpha ) L^{gamma }right)}{gamma ^2}\
              &=& frac{1}{2} (1 - alpha) alpha (ln (K)-ln (L))^2 tag{3}
              end{eqnarray}



              Up to first order we have then



              begin{eqnarray}
              ln Y &approx& color{blue}{(ln Y)_{gamma = 0}} + color{red}{left(frac{partial ln Y}{partial gamma}right)_{gamma = 0} gamma} \
              &stackrel{(2),(3)}{=}& color{blue}{ln A + alpha ln K + (1 - alpha) ln L} + color{red}{frac{1}{2}alpha(1 - alpha)gamma [ln K - ln L]^2} \
              &=& ln A + alpha ln X_1 + (1 - alpha) ln X_2 + frac{alpha gamma (1 -alpha)}{2}left[ln^2 X_1 -2ln X_1ln X_2 + ln^2 X_2 right] \
              &=& alpha_0 + sum_{i = 1}^2 alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^2 beta_{ij}ln X_i ln X_j tag{3}
              end{eqnarray}



              This is naturally extended to $n > 2$ as



              $$
              ln Y = alpha_0 + sum_{i = 1}^n alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^n beta_{ij}ln X_i ln X_j
              $$






              share|improve this answer
























                2












                2








                2






                The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as



                $$
                Y = A[alpha K^gamma + (1 - alpha)L^gamma]^{1/gamma} tag{1}
                $$



                in this case $X_1 = K$, $X_2 = L$. Now we expand $ln Y$ around $gamma = 0$




                $gamma^0$ term




                $$
                lim_{gamma to 0} ln Y = ln (A K^alpha L^{1 - alpha}) tag{2}
                $$




                $gamma^1$ term




                begin{eqnarray}
                lim_{gamma to 0} frac{partial ln Y}{partial gamma} &=& lim_{gamma to 0}frac{alpha K^{gamma } ln (L)+(1-alpha ) l^{gamma } log (L)}{gamma
                left(alpha K^{gamma }+(1-alpha ) L^{gamma }right)}-frac{ln left(alpha
                K^{gamma }+(1-alpha ) L^{gamma }right)}{gamma ^2}\
                &=& frac{1}{2} (1 - alpha) alpha (ln (K)-ln (L))^2 tag{3}
                end{eqnarray}



                Up to first order we have then



                begin{eqnarray}
                ln Y &approx& color{blue}{(ln Y)_{gamma = 0}} + color{red}{left(frac{partial ln Y}{partial gamma}right)_{gamma = 0} gamma} \
                &stackrel{(2),(3)}{=}& color{blue}{ln A + alpha ln K + (1 - alpha) ln L} + color{red}{frac{1}{2}alpha(1 - alpha)gamma [ln K - ln L]^2} \
                &=& ln A + alpha ln X_1 + (1 - alpha) ln X_2 + frac{alpha gamma (1 -alpha)}{2}left[ln^2 X_1 -2ln X_1ln X_2 + ln^2 X_2 right] \
                &=& alpha_0 + sum_{i = 1}^2 alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^2 beta_{ij}ln X_i ln X_j tag{3}
                end{eqnarray}



                This is naturally extended to $n > 2$ as



                $$
                ln Y = alpha_0 + sum_{i = 1}^n alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^n beta_{ij}ln X_i ln X_j
                $$






                share|improve this answer












                The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as



                $$
                Y = A[alpha K^gamma + (1 - alpha)L^gamma]^{1/gamma} tag{1}
                $$



                in this case $X_1 = K$, $X_2 = L$. Now we expand $ln Y$ around $gamma = 0$




                $gamma^0$ term




                $$
                lim_{gamma to 0} ln Y = ln (A K^alpha L^{1 - alpha}) tag{2}
                $$




                $gamma^1$ term




                begin{eqnarray}
                lim_{gamma to 0} frac{partial ln Y}{partial gamma} &=& lim_{gamma to 0}frac{alpha K^{gamma } ln (L)+(1-alpha ) l^{gamma } log (L)}{gamma
                left(alpha K^{gamma }+(1-alpha ) L^{gamma }right)}-frac{ln left(alpha
                K^{gamma }+(1-alpha ) L^{gamma }right)}{gamma ^2}\
                &=& frac{1}{2} (1 - alpha) alpha (ln (K)-ln (L))^2 tag{3}
                end{eqnarray}



                Up to first order we have then



                begin{eqnarray}
                ln Y &approx& color{blue}{(ln Y)_{gamma = 0}} + color{red}{left(frac{partial ln Y}{partial gamma}right)_{gamma = 0} gamma} \
                &stackrel{(2),(3)}{=}& color{blue}{ln A + alpha ln K + (1 - alpha) ln L} + color{red}{frac{1}{2}alpha(1 - alpha)gamma [ln K - ln L]^2} \
                &=& ln A + alpha ln X_1 + (1 - alpha) ln X_2 + frac{alpha gamma (1 -alpha)}{2}left[ln^2 X_1 -2ln X_1ln X_2 + ln^2 X_2 right] \
                &=& alpha_0 + sum_{i = 1}^2 alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^2 beta_{ij}ln X_i ln X_j tag{3}
                end{eqnarray}



                This is naturally extended to $n > 2$ as



                $$
                ln Y = alpha_0 + sum_{i = 1}^n alpha_i ln X_i + frac{1}{2}sum_{i, j = 1}^n beta_{ij}ln X_i ln X_j
                $$







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 5 hours ago









                caverac

                7721214




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