Why does this expression simplify as such?












3












$begingroup$


I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



$$E[(b-beta)e']=E[(X'X)^{-1}epsilonepsilon'M_{[X]}]. $$



In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.






regression multiple-regression linear-model residuals







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



    $$E[(b-beta)e']=E[(X'X)^{-1}epsilonepsilon'M_{[X]}]. $$



    In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.






    regression multiple-regression linear-model residuals







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



      $$E[(b-beta)e']=E[(X'X)^{-1}epsilonepsilon'M_{[X]}]. $$



      In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.






      regression multiple-regression linear-model residuals







      share|cite|improve this question











      $endgroup$




      I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



      $$E[(b-beta)e']=E[(X'X)^{-1}epsilonepsilon'M_{[X]}]. $$



      In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.






      regression multiple-regression linear-model residuals




      regression multiple-regression linear-model residuals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 21 at 22:21









      Benjamin Christoffersen

      1,282520




      1,282520










      asked Mar 21 at 21:18









      DavidDavid

      25011




      25011






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^{-1}$.



          Start with the definition of $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hat{Y}=Y-Xb=Y-X(X'X)^{-1}X'Y.$$



          Notice $X(X'X)^{-1}X'$ is the projection matrix for $X$, which we will denote with $P_{[X]}$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_{[X]})Y=M_{[X]}Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_{[X]}(Xbeta+epsilon)=M_{[X]}epsilon,$$
          since $M_{[X]}X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36



















          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$begin{equation} begin{aligned}
          b-beta
          &= (X'X)^{-1} X'y - beta \[6pt]
          &= (X'X)^{-1} X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^{-1} (X'X) beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= (X'X)^{-1} X' epsilon. \[6pt]
          end{aligned} end{equation}$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_{[X]} Y = M_{[X]} epsilon$. Substituting this vector gives:



          $$begin{equation} begin{aligned}
          (b-beta) e'
          &= (X'X)^{-1} X' epsilon (M_{[X]} epsilon)' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}. \[6pt]
          end{aligned} end{equation}$$



          (The last step follows from the fact that $M_{[X]}$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13











          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: "65"
          };
          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
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f398797%2fwhy-does-this-expression-simplify-as-such%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^{-1}$.



          Start with the definition of $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hat{Y}=Y-Xb=Y-X(X'X)^{-1}X'Y.$$



          Notice $X(X'X)^{-1}X'$ is the projection matrix for $X$, which we will denote with $P_{[X]}$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_{[X]})Y=M_{[X]}Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_{[X]}(Xbeta+epsilon)=M_{[X]}epsilon,$$
          since $M_{[X]}X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36
















          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^{-1}$.



          Start with the definition of $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hat{Y}=Y-Xb=Y-X(X'X)^{-1}X'Y.$$



          Notice $X(X'X)^{-1}X'$ is the projection matrix for $X$, which we will denote with $P_{[X]}$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_{[X]})Y=M_{[X]}Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_{[X]}(Xbeta+epsilon)=M_{[X]}epsilon,$$
          since $M_{[X]}X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36














          3












          3








          3





          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^{-1}$.



          Start with the definition of $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hat{Y}=Y-Xb=Y-X(X'X)^{-1}X'Y.$$



          Notice $X(X'X)^{-1}X'$ is the projection matrix for $X$, which we will denote with $P_{[X]}$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_{[X]})Y=M_{[X]}Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_{[X]}(Xbeta+epsilon)=M_{[X]}epsilon,$$
          since $M_{[X]}X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$






          share|cite|improve this answer









          $endgroup$



          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^{-1}$.



          Start with the definition of $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hat{Y}=Y-Xb=Y-X(X'X)^{-1}X'Y.$$



          Notice $X(X'X)^{-1}X'$ is the projection matrix for $X$, which we will denote with $P_{[X]}$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_{[X]})Y=M_{[X]}Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_{[X]}(Xbeta+epsilon)=M_{[X]}epsilon,$$
          since $M_{[X]}X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 21 at 22:57









          dlnBdlnB

          92212




          92212












          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36


















          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36
















          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          Mar 21 at 23:36




          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          Mar 21 at 23:36













          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$begin{equation} begin{aligned}
          b-beta
          &= (X'X)^{-1} X'y - beta \[6pt]
          &= (X'X)^{-1} X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^{-1} (X'X) beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= (X'X)^{-1} X' epsilon. \[6pt]
          end{aligned} end{equation}$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_{[X]} Y = M_{[X]} epsilon$. Substituting this vector gives:



          $$begin{equation} begin{aligned}
          (b-beta) e'
          &= (X'X)^{-1} X' epsilon (M_{[X]} epsilon)' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}. \[6pt]
          end{aligned} end{equation}$$



          (The last step follows from the fact that $M_{[X]}$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13
















          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$begin{equation} begin{aligned}
          b-beta
          &= (X'X)^{-1} X'y - beta \[6pt]
          &= (X'X)^{-1} X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^{-1} (X'X) beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= (X'X)^{-1} X' epsilon. \[6pt]
          end{aligned} end{equation}$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_{[X]} Y = M_{[X]} epsilon$. Substituting this vector gives:



          $$begin{equation} begin{aligned}
          (b-beta) e'
          &= (X'X)^{-1} X' epsilon (M_{[X]} epsilon)' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}. \[6pt]
          end{aligned} end{equation}$$



          (The last step follows from the fact that $M_{[X]}$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13














          3












          3








          3





          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$begin{equation} begin{aligned}
          b-beta
          &= (X'X)^{-1} X'y - beta \[6pt]
          &= (X'X)^{-1} X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^{-1} (X'X) beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= (X'X)^{-1} X' epsilon. \[6pt]
          end{aligned} end{equation}$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_{[X]} Y = M_{[X]} epsilon$. Substituting this vector gives:



          $$begin{equation} begin{aligned}
          (b-beta) e'
          &= (X'X)^{-1} X' epsilon (M_{[X]} epsilon)' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}. \[6pt]
          end{aligned} end{equation}$$



          (The last step follows from the fact that $M_{[X]}$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$



          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$begin{equation} begin{aligned}
          b-beta
          &= (X'X)^{-1} X'y - beta \[6pt]
          &= (X'X)^{-1} X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^{-1} (X'X) beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= beta + (X'X)^{-1} X' epsilon - beta \[6pt]
          &= (X'X)^{-1} X' epsilon. \[6pt]
          end{aligned} end{equation}$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_{[X]} Y = M_{[X]} epsilon$. Substituting this vector gives:



          $$begin{equation} begin{aligned}
          (b-beta) e'
          &= (X'X)^{-1} X' epsilon (M_{[X]} epsilon)' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}' \[6pt]
          &= (X'X)^{-1} X' epsilon epsilon' M_{[X]}. \[6pt]
          end{aligned} end{equation}$$



          (The last step follows from the fact that $M_{[X]}$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 21 at 23:07









          BenBen

          26.9k230124




          26.9k230124








          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13














          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13








          2




          2




          $begingroup$
          Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
          $endgroup$
          – dlnB
          Mar 21 at 23:11




          $begingroup$
          Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
          $endgroup$
          – dlnB
          Mar 21 at 23:11




          1




          1




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          Mar 21 at 23:13




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          Mar 21 at 23:13


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Cross Validated!


          • 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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f398797%2fwhy-does-this-expression-simplify-as-such%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