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











      share|improve this question




      share|improve this question










      asked 7 hours ago









      EconJohn♦

      3,3571938




      3,3571938






















          1 Answer
          1






          active

          oldest

          votes


















          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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "591"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2feconomics.stackexchange.com%2fquestions%2f26144%2fderiving-the-translog-production-function%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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




                7721214






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Economics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2feconomics.stackexchange.com%2fquestions%2f26144%2fderiving-the-translog-production-function%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                    Read more

                    Puebla de Zaragoza

                    Image
                    Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                    Read more

                    Musa

                    Image
                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                    Read more