Reference request: Convergence of singular values of tall random matrix












0












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Let $X_ninmathbb{R}^{ntimes m}$ be a matrix whose entries are i.i.d. random variables with zero mean and variance $sigma^2$. Let $m$ be a fixed integer and $|cdot|$ denote the 2-norm of a matrix. I would like to prove that
$$
left|frac{1}{n} X_n^top X_n -sigma^2 Iright| overset{P}{to} 0 text{ as } ntoinfty
$$
where $overset{P}{to} 0$ denotes convergence in probability.



I believe that this should be a well-known fact, which should follow from the fact that high-dimensional random vectors are almost orthogonal as their dimension increases (see e.g. here). However, I couldn't find a reference with a rigorous proof. Hence, I would really appreciate any comment with pointers to the literature. Thanks!






linear-algebra probability reference-request convergence random-matrices







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    0












    $begingroup$


    Let $X_ninmathbb{R}^{ntimes m}$ be a matrix whose entries are i.i.d. random variables with zero mean and variance $sigma^2$. Let $m$ be a fixed integer and $|cdot|$ denote the 2-norm of a matrix. I would like to prove that
    $$
    left|frac{1}{n} X_n^top X_n -sigma^2 Iright| overset{P}{to} 0 text{ as } ntoinfty
    $$
    where $overset{P}{to} 0$ denotes convergence in probability.



    I believe that this should be a well-known fact, which should follow from the fact that high-dimensional random vectors are almost orthogonal as their dimension increases (see e.g. here). However, I couldn't find a reference with a rigorous proof. Hence, I would really appreciate any comment with pointers to the literature. Thanks!






    linear-algebra probability reference-request convergence random-matrices







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X_ninmathbb{R}^{ntimes m}$ be a matrix whose entries are i.i.d. random variables with zero mean and variance $sigma^2$. Let $m$ be a fixed integer and $|cdot|$ denote the 2-norm of a matrix. I would like to prove that
      $$
      left|frac{1}{n} X_n^top X_n -sigma^2 Iright| overset{P}{to} 0 text{ as } ntoinfty
      $$
      where $overset{P}{to} 0$ denotes convergence in probability.



      I believe that this should be a well-known fact, which should follow from the fact that high-dimensional random vectors are almost orthogonal as their dimension increases (see e.g. here). However, I couldn't find a reference with a rigorous proof. Hence, I would really appreciate any comment with pointers to the literature. Thanks!






      linear-algebra probability reference-request convergence random-matrices







      share|cite|improve this question









      $endgroup$




      Let $X_ninmathbb{R}^{ntimes m}$ be a matrix whose entries are i.i.d. random variables with zero mean and variance $sigma^2$. Let $m$ be a fixed integer and $|cdot|$ denote the 2-norm of a matrix. I would like to prove that
      $$
      left|frac{1}{n} X_n^top X_n -sigma^2 Iright| overset{P}{to} 0 text{ as } ntoinfty
      $$
      where $overset{P}{to} 0$ denotes convergence in probability.



      I believe that this should be a well-known fact, which should follow from the fact that high-dimensional random vectors are almost orthogonal as their dimension increases (see e.g. here). However, I couldn't find a reference with a rigorous proof. Hence, I would really appreciate any comment with pointers to the literature. Thanks!






      linear-algebra probability reference-request convergence random-matrices




      linear-algebra probability reference-request convergence random-matrices






      share|cite|improve this question













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      asked Jan 15 at 19:54









      LudwigLudwig

      813715




      813715






















          1 Answer
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          In fact, since we are already assuming i.i.d. sequence, it is a direct consequence of the famous Strong (Weak) law of large numbers. (We don't need any advanced theorems other than SLLN.) The proof goes like this. Let $X_{k,j}$ $(k=1,2,ldots, n$ ,$j=1,2,ldots m)$ denote $(k,j)$-th entry of $X_n$. Observe that $(i,j)$-th entry of the matrix $frac{1}{n}X_n^top X_n$ is given by
          $$
          frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}.
          $$           Note that for each $(i,j)$, $(X_{ki}X_{kj})_{kinBbb N}$ is an i.i.d. sequence of random variables. Thus by Strong law of large numbers, we have
          $$frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}to_{text{a.s.}} E[X_{1i}X_{1j}]=begin{cases}sigma^2,quad i=j\0,quad ine jend{cases}.
          $$           This implies that each entry of $mtimes m$ matrix $M_n:=frac{1}{n}X_n^top X_n-sigma^2I$ converges to $0$ almost surely. Now, since
          $$
          |M_n|_2^2= sum_{i,j=1}^m |M_{i,j}|^2,
          $$           it follows that
          $$
          lim_{ntoinfty}|M_n|_2=0
          $$           almost surely. It follows that
          $$
          |frac{1}{n}X_n^top X_n-sigma^2I|_2to_p 0,
          $$           since almost sure convergence implies convergence in probability.






          share|cite|improve this answer











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            1 Answer
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            1 Answer
            1






            active

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            active

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            active

            oldest

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            1












            $begingroup$

            In fact, since we are already assuming i.i.d. sequence, it is a direct consequence of the famous Strong (Weak) law of large numbers. (We don't need any advanced theorems other than SLLN.) The proof goes like this. Let $X_{k,j}$ $(k=1,2,ldots, n$ ,$j=1,2,ldots m)$ denote $(k,j)$-th entry of $X_n$. Observe that $(i,j)$-th entry of the matrix $frac{1}{n}X_n^top X_n$ is given by
            $$
            frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}.
            $$             Note that for each $(i,j)$, $(X_{ki}X_{kj})_{kinBbb N}$ is an i.i.d. sequence of random variables. Thus by Strong law of large numbers, we have
            $$frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}to_{text{a.s.}} E[X_{1i}X_{1j}]=begin{cases}sigma^2,quad i=j\0,quad ine jend{cases}.
            $$             This implies that each entry of $mtimes m$ matrix $M_n:=frac{1}{n}X_n^top X_n-sigma^2I$ converges to $0$ almost surely. Now, since
            $$
            |M_n|_2^2= sum_{i,j=1}^m |M_{i,j}|^2,
            $$             it follows that
            $$
            lim_{ntoinfty}|M_n|_2=0
            $$             almost surely. It follows that
            $$
            |frac{1}{n}X_n^top X_n-sigma^2I|_2to_p 0,
            $$             since almost sure convergence implies convergence in probability.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              In fact, since we are already assuming i.i.d. sequence, it is a direct consequence of the famous Strong (Weak) law of large numbers. (We don't need any advanced theorems other than SLLN.) The proof goes like this. Let $X_{k,j}$ $(k=1,2,ldots, n$ ,$j=1,2,ldots m)$ denote $(k,j)$-th entry of $X_n$. Observe that $(i,j)$-th entry of the matrix $frac{1}{n}X_n^top X_n$ is given by
              $$
              frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}.
              $$               Note that for each $(i,j)$, $(X_{ki}X_{kj})_{kinBbb N}$ is an i.i.d. sequence of random variables. Thus by Strong law of large numbers, we have
              $$frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}to_{text{a.s.}} E[X_{1i}X_{1j}]=begin{cases}sigma^2,quad i=j\0,quad ine jend{cases}.
              $$               This implies that each entry of $mtimes m$ matrix $M_n:=frac{1}{n}X_n^top X_n-sigma^2I$ converges to $0$ almost surely. Now, since
              $$
              |M_n|_2^2= sum_{i,j=1}^m |M_{i,j}|^2,
              $$               it follows that
              $$
              lim_{ntoinfty}|M_n|_2=0
              $$               almost surely. It follows that
              $$
              |frac{1}{n}X_n^top X_n-sigma^2I|_2to_p 0,
              $$               since almost sure convergence implies convergence in probability.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                In fact, since we are already assuming i.i.d. sequence, it is a direct consequence of the famous Strong (Weak) law of large numbers. (We don't need any advanced theorems other than SLLN.) The proof goes like this. Let $X_{k,j}$ $(k=1,2,ldots, n$ ,$j=1,2,ldots m)$ denote $(k,j)$-th entry of $X_n$. Observe that $(i,j)$-th entry of the matrix $frac{1}{n}X_n^top X_n$ is given by
                $$
                frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}.
                $$                 Note that for each $(i,j)$, $(X_{ki}X_{kj})_{kinBbb N}$ is an i.i.d. sequence of random variables. Thus by Strong law of large numbers, we have
                $$frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}to_{text{a.s.}} E[X_{1i}X_{1j}]=begin{cases}sigma^2,quad i=j\0,quad ine jend{cases}.
                $$                 This implies that each entry of $mtimes m$ matrix $M_n:=frac{1}{n}X_n^top X_n-sigma^2I$ converges to $0$ almost surely. Now, since
                $$
                |M_n|_2^2= sum_{i,j=1}^m |M_{i,j}|^2,
                $$                 it follows that
                $$
                lim_{ntoinfty}|M_n|_2=0
                $$                 almost surely. It follows that
                $$
                |frac{1}{n}X_n^top X_n-sigma^2I|_2to_p 0,
                $$                 since almost sure convergence implies convergence in probability.






                share|cite|improve this answer











                $endgroup$



                In fact, since we are already assuming i.i.d. sequence, it is a direct consequence of the famous Strong (Weak) law of large numbers. (We don't need any advanced theorems other than SLLN.) The proof goes like this. Let $X_{k,j}$ $(k=1,2,ldots, n$ ,$j=1,2,ldots m)$ denote $(k,j)$-th entry of $X_n$. Observe that $(i,j)$-th entry of the matrix $frac{1}{n}X_n^top X_n$ is given by
                $$
                frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}.
                $$                 Note that for each $(i,j)$, $(X_{ki}X_{kj})_{kinBbb N}$ is an i.i.d. sequence of random variables. Thus by Strong law of large numbers, we have
                $$frac{1}{n}sum_{k=1}^n X_{ki}X_{kj}to_{text{a.s.}} E[X_{1i}X_{1j}]=begin{cases}sigma^2,quad i=j\0,quad ine jend{cases}.
                $$                 This implies that each entry of $mtimes m$ matrix $M_n:=frac{1}{n}X_n^top X_n-sigma^2I$ converges to $0$ almost surely. Now, since
                $$
                |M_n|_2^2= sum_{i,j=1}^m |M_{i,j}|^2,
                $$                 it follows that
                $$
                lim_{ntoinfty}|M_n|_2=0
                $$                 almost surely. It follows that
                $$
                |frac{1}{n}X_n^top X_n-sigma^2I|_2to_p 0,
                $$                 since almost sure convergence implies convergence in probability.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 21 at 19:00

























                answered Jan 15 at 20:54









                SongSong

                18.5k21651




                18.5k21651






























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