Matrix calculus for statistics












2














I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.



Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).



For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.



But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).



I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.



How can I do this calculus from the variance-covariance matrix
$$
begin{pmatrix}
mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
end{pmatrix}
$$
[preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?



NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.






matrices descriptive-statistics







share|cite|improve this question





























    2














    I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.



    Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).



    For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.



    But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).



    I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.



    How can I do this calculus from the variance-covariance matrix
    $$
    begin{pmatrix}
    mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
    mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
    end{pmatrix}
    $$
    [preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?



    NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.






    matrices descriptive-statistics







    share|cite|improve this question



























      2












      2








      2







      I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.



      Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).



      For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.



      But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).



      I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.



      How can I do this calculus from the variance-covariance matrix
      $$
      begin{pmatrix}
      mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
      mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
      end{pmatrix}
      $$
      [preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?



      NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.






      matrices descriptive-statistics







      share|cite|improve this question















      I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.



      Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).



      For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.



      But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).



      I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.



      How can I do this calculus from the variance-covariance matrix
      $$
      begin{pmatrix}
      mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
      mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
      end{pmatrix}
      $$
      [preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?



      NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.






      matrices descriptive-statistics




      matrices descriptive-statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 '18 at 15:10









      Jean-Claude Arbaut

      14.7k63464




      14.7k63464










      asked Nov 22 '18 at 12:31









      Dan Chaltiel

      1134




      1134






















          2 Answers
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          3














          The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.



          Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is



          $$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
          =frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$






          share|cite|improve this answer























          • +1 Elegant solution
            – caverac
            Nov 22 '18 at 13:12










          • Great answer, thanks ! Learned a lot with this !
            – Dan Chaltiel
            Nov 22 '18 at 13:28



















          2














          The key point here is that



          $$
          mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
          $$



          so that you can express your first expression as



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
          &=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}



          In general



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}






          share|cite|improve this answer

















          • 1




            +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
            – Jean-Claude Arbaut
            Nov 22 '18 at 12:57











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.



          Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is



          $$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
          =frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$






          share|cite|improve this answer























          • +1 Elegant solution
            – caverac
            Nov 22 '18 at 13:12










          • Great answer, thanks ! Learned a lot with this !
            – Dan Chaltiel
            Nov 22 '18 at 13:28
















          3














          The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.



          Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is



          $$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
          =frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$






          share|cite|improve this answer























          • +1 Elegant solution
            – caverac
            Nov 22 '18 at 13:12










          • Great answer, thanks ! Learned a lot with this !
            – Dan Chaltiel
            Nov 22 '18 at 13:28














          3












          3








          3






          The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.



          Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is



          $$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
          =frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$






          share|cite|improve this answer














          The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.



          Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is



          $$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
          =frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 '18 at 13:25

























          answered Nov 22 '18 at 13:00









          Jean-Claude Arbaut

          14.7k63464




          14.7k63464












          • +1 Elegant solution
            – caverac
            Nov 22 '18 at 13:12










          • Great answer, thanks ! Learned a lot with this !
            – Dan Chaltiel
            Nov 22 '18 at 13:28


















          • +1 Elegant solution
            – caverac
            Nov 22 '18 at 13:12










          • Great answer, thanks ! Learned a lot with this !
            – Dan Chaltiel
            Nov 22 '18 at 13:28
















          +1 Elegant solution
          – caverac
          Nov 22 '18 at 13:12




          +1 Elegant solution
          – caverac
          Nov 22 '18 at 13:12












          Great answer, thanks ! Learned a lot with this !
          – Dan Chaltiel
          Nov 22 '18 at 13:28




          Great answer, thanks ! Learned a lot with this !
          – Dan Chaltiel
          Nov 22 '18 at 13:28











          2














          The key point here is that



          $$
          mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
          $$



          so that you can express your first expression as



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
          &=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}



          In general



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}






          share|cite|improve this answer

















          • 1




            +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
            – Jean-Claude Arbaut
            Nov 22 '18 at 12:57
















          2














          The key point here is that



          $$
          mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
          $$



          so that you can express your first expression as



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
          &=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}



          In general



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}






          share|cite|improve this answer

















          • 1




            +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
            – Jean-Claude Arbaut
            Nov 22 '18 at 12:57














          2












          2








          2






          The key point here is that



          $$
          mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
          $$



          so that you can express your first expression as



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
          &=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}



          In general



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}






          share|cite|improve this answer












          The key point here is that



          $$
          mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
          $$



          so that you can express your first expression as



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
          &=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
          &=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}



          In general



          begin{eqnarray}
          mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
          end{eqnarray}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 12:56









          caverac

          13.9k21130




          13.9k21130








          • 1




            +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
            – Jean-Claude Arbaut
            Nov 22 '18 at 12:57














          • 1




            +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
            – Jean-Claude Arbaut
            Nov 22 '18 at 12:57








          1




          1




          +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
          – Jean-Claude Arbaut
          Nov 22 '18 at 12:57




          +1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
          – Jean-Claude Arbaut
          Nov 22 '18 at 12:57


















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