Reference request: Convergence of singular values of tall random matrix
$begingroup$
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
asked Jan 15 at 19:54
LudwigLudwig
813715
$endgroup$
add a comment |
$begingroup$
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
asked Jan 15 at 19:54
LudwigLudwig
813715
$endgroup$
add a comment |
$begingroup$
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
asked Jan 15 at 19:54
LudwigLudwig
813715
$endgroup$
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
asked Jan 15 at 19:54
LudwigLudwig
813715
asked Jan 15 at 19:54
LudwigLudwig
813715
asked Jan 15 at 19:54
LudwigLudwig
813715
asked Jan 15 at 19:54
LudwigLudwig
813715
asked Jan 15 at 19:54
LudwigLudwig
813715
813715
add a comment |
add a comment |
1 Answer
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oldest
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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.
answered Jan 15 at 20:54
SongSong
18.5k21651
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
active
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active
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votes
$begingroup$
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.
answered Jan 15 at 20:54
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 15 at 20:54
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 15 at 20:54
SongSong
18.5k21651
$endgroup$
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.
answered Jan 15 at 20:54
SongSong
18.5k21651
edited Jan 21 at 19:00
answered Jan 15 at 20:54
SongSong
18.5k21651
answered Jan 15 at 20:54
SongSong
18.5k21651
answered Jan 15 at 20:54
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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