Understanding Variance-Covariance Matrix












4












$begingroup$


Suppose data set is expressed by the matrix $X inmathbb R^{n times d}$



where $n =$ Number of samples and $d =$ dimension/features of each sample



Then what does $operatorname{Cov}(X) inmathbb R^{d times d}$ (Variance-Covariance matrix of $X$) represent. Does below interpretation would be right




Variance-Covariance matrix of $X$ represents covariance between every pair of dimension/feature for all samples.











share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Suppose data set is expressed by the matrix $X inmathbb R^{n times d}$



    where $n =$ Number of samples and $d =$ dimension/features of each sample



    Then what does $operatorname{Cov}(X) inmathbb R^{d times d}$ (Variance-Covariance matrix of $X$) represent. Does below interpretation would be right




    Variance-Covariance matrix of $X$ represents covariance between every pair of dimension/feature for all samples.











    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Suppose data set is expressed by the matrix $X inmathbb R^{n times d}$



      where $n =$ Number of samples and $d =$ dimension/features of each sample



      Then what does $operatorname{Cov}(X) inmathbb R^{d times d}$ (Variance-Covariance matrix of $X$) represent. Does below interpretation would be right




      Variance-Covariance matrix of $X$ represents covariance between every pair of dimension/feature for all samples.











      share|cite|improve this question











      $endgroup$




      Suppose data set is expressed by the matrix $X inmathbb R^{n times d}$



      where $n =$ Number of samples and $d =$ dimension/features of each sample



      Then what does $operatorname{Cov}(X) inmathbb R^{d times d}$ (Variance-Covariance matrix of $X$) represent. Does below interpretation would be right




      Variance-Covariance matrix of $X$ represents covariance between every pair of dimension/feature for all samples.








      linear-algebra matrices covariance






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      edited Dec 31 '18 at 21:10









      Davide Giraudo

      128k17156268




      128k17156268










      asked Dec 31 '18 at 9:44









      AtineshAtinesh

      4694822




      4694822






















          1 Answer
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          $begingroup$

          Let's review how the covariance matrix is computed in this context. Let $mu_j$ denote the mean of the $j$th column, and let $mu$ denote the row-vector $mu = (mu_1,mu_2,dots,mu_d)$. Then
          $$
          operatorname{cov}(X) = (X - mu 1_n)(X - mu 1_n)^T
          $$

          where $1_n$ denotes the column vector $(1,dots,1)^T$ of length $n$.



          With this in mind, the $i,j$ entry of the covariance matrix is given by
          $$
          operatorname{cov}(X)[i,j] = sum_{k=1}^n (x_{ik} - mu_k)(x_{jk} - mu_k)
          $$

          So, $frac 1n operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $frac 1n operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.






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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes









            3












            $begingroup$

            Let's review how the covariance matrix is computed in this context. Let $mu_j$ denote the mean of the $j$th column, and let $mu$ denote the row-vector $mu = (mu_1,mu_2,dots,mu_d)$. Then
            $$
            operatorname{cov}(X) = (X - mu 1_n)(X - mu 1_n)^T
            $$

            where $1_n$ denotes the column vector $(1,dots,1)^T$ of length $n$.



            With this in mind, the $i,j$ entry of the covariance matrix is given by
            $$
            operatorname{cov}(X)[i,j] = sum_{k=1}^n (x_{ik} - mu_k)(x_{jk} - mu_k)
            $$

            So, $frac 1n operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $frac 1n operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Let's review how the covariance matrix is computed in this context. Let $mu_j$ denote the mean of the $j$th column, and let $mu$ denote the row-vector $mu = (mu_1,mu_2,dots,mu_d)$. Then
              $$
              operatorname{cov}(X) = (X - mu 1_n)(X - mu 1_n)^T
              $$

              where $1_n$ denotes the column vector $(1,dots,1)^T$ of length $n$.



              With this in mind, the $i,j$ entry of the covariance matrix is given by
              $$
              operatorname{cov}(X)[i,j] = sum_{k=1}^n (x_{ik} - mu_k)(x_{jk} - mu_k)
              $$

              So, $frac 1n operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $frac 1n operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Let's review how the covariance matrix is computed in this context. Let $mu_j$ denote the mean of the $j$th column, and let $mu$ denote the row-vector $mu = (mu_1,mu_2,dots,mu_d)$. Then
                $$
                operatorname{cov}(X) = (X - mu 1_n)(X - mu 1_n)^T
                $$

                where $1_n$ denotes the column vector $(1,dots,1)^T$ of length $n$.



                With this in mind, the $i,j$ entry of the covariance matrix is given by
                $$
                operatorname{cov}(X)[i,j] = sum_{k=1}^n (x_{ik} - mu_k)(x_{jk} - mu_k)
                $$

                So, $frac 1n operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $frac 1n operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.






                share|cite|improve this answer









                $endgroup$



                Let's review how the covariance matrix is computed in this context. Let $mu_j$ denote the mean of the $j$th column, and let $mu$ denote the row-vector $mu = (mu_1,mu_2,dots,mu_d)$. Then
                $$
                operatorname{cov}(X) = (X - mu 1_n)(X - mu 1_n)^T
                $$

                where $1_n$ denotes the column vector $(1,dots,1)^T$ of length $n$.



                With this in mind, the $i,j$ entry of the covariance matrix is given by
                $$
                operatorname{cov}(X)[i,j] = sum_{k=1}^n (x_{ik} - mu_k)(x_{jk} - mu_k)
                $$

                So, $frac 1n operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $frac 1n operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 31 '18 at 19:50









                OmnomnomnomOmnomnomnom

                129k794188




                129k794188






























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