Total Variation Approximation of Random Orthogonal Matrices by Gaussian Matrices

Abstract

The topic of this paper is the asymptotic distribution of the entries of random orthogonal matrices distributed according to Haar measure. We examine the total variation distance between the joint distribution of the entries of \(W_n\), the \(p_n \times q_n\) upper-left block of a Haar-distributed matrix, and that of \(p_nq_n\) independent standard Gaussian random variables and show that the total variation distance converges to zero when \(p_nq_n = o(n)\).

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Acknowledgements

The results of this paper are part of a Ph.D. thesis written under the direction of Elizabeth Meckes; the author very much thanks her for many helpful conversations.

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Correspondence to Kathryn Stewart.

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Supported in part by NSF Grant DMS 1612589 to Elizabeth Meckes.

Appendix

Appendix

The following theorem gives the limit of the maximum eigenvalues of the sample covariance matrix.

Lemma 4

(Bai–Silverstein, [2]) Assume that the entries of \(\{ x_{ij} \}\) are a double array of iid random variables with mean zero, variance \(\sigma ^2\), and finite fourth moment. Let \(X_p = (x_{ij}; i \le p, j \le q)\) be the \(p \times q\) matrix of the upper-left corner of the double array. Let \(S_p = \frac{1}{p}X_p^TX_p\). If \(q/p \rightarrow y \in (0, 1]\), then, with probability 1, we have

$$\begin{aligned} \begin{aligned} \lim _{p \rightarrow \infty } \lambda _{max}(S_p) = \sigma ^2(1 + \sqrt{y})^2. \end{aligned} \end{aligned}$$

For a proof, see [2] Theorem 5.11.

As a direct consequence, we get the following,

Lemma 5

Let \(\{ p_n: n \ge 1 \}\) and \(\{ q_n: n \ge 1 \}\) be two sequences of positive integers such that \(q_n \le p_n\). For each n, let \(X_n = (x_{ij})\) be a \(p_n \times q_n\) matrix where \(x_{ij}\) are independent standard Gaussian random variables. If \(\lambda _{max}(n)\) denotes the largest eigenvalue of the matrix \(X_n^TX_n\), then, with probability 1,

$$\begin{aligned} \limsup _{n \rightarrow \infty } \frac{\lambda _{max}(n)}{p_n} \le 4. \end{aligned}$$

Proof

If \(p_n = q_n\), the lemma follows immediately from Lemma 4. If \(p_n > q_n\) and Y is an arbitrary \(p_n \times (p_n-q_n)\) matrix, then the singular values of \(X_n\) are no more than the singular values of the square matrix \((X_n, Y)\). The lemma then follows in general. \(\square \)

The following uniform version of Stirling’s approximation is used in Lemmas 2 and 3.

Lemma 6

For each positive integer n,

$$\begin{aligned} \sqrt{2\pi }n^{n+\frac{1}{2}}e^{-n} \le n! \le en^{n+\frac{1}{2}}e^{-n}. \end{aligned}$$

The proof follows from equation (9.15) in [8].

Lemma 7

Let \(p_nq_n = o(n)\) and \(l = \frac{\log p_n}{\log (n/p_nq_n)}\). Then, \(\sum _{j=1}^l \Big ( \frac{Cp_nq_n}{n} \Big )^j \frac{1}{j}\) converges to 0 as \(n \rightarrow \infty \) for any constant C.

Proof

Since \(\frac{Cpq}{n} \rightarrow 0\),

$$\begin{aligned} \sum _{j=1}^l \Bigg ( \frac{Cpq}{n} \Bigg )^j \frac{1}{j}&\le \sum _{j=1}^{\infty } \Bigg ( \frac{Cpq}{n} \Bigg )^j \frac{1}{j} = -\log \left( 1 - \frac{Cpq}{n} \right) \rightarrow 0, \end{aligned}$$

so that \(\sum _{j=1}^l \Big ( \frac{Cpq}{n} \Big )^j \frac{1}{j}\) converges to 0 as \(n \rightarrow \infty \) for any constant C. \(\square \)

Lemma 8

Let \(p_nq_n = o(n)\). Set

$$\begin{aligned} K_n = \left( \frac{2}{n} \right) ^{p_nq_n/2} \prod _{j=1}^{q_n} \frac{\varGamma ((n-j+1)/2)}{\varGamma ((n-p_n-j+1)/2)}. \end{aligned}$$

Then,

$$\begin{aligned} \log (K_n) = - \frac{p_nq_n^2}{4n} - \sum _{k=1}^{l-1} \frac{p_n^{k+1}q_n}{2(k+1)kn^k} - \frac{p_n^{l+1}q_n}{2ln^l}+ o(1), \end{aligned}$$

as n tends to infinity.

Proof

Rewriting,

$$\begin{aligned} \begin{aligned} \log (K_n)&= \frac{pq}{2} \log \left( \frac{2}{n} \right) + \sum _{j=1}^q \log \left( \frac{ \varGamma ((n-j+1)/2)}{\varGamma ((n-p-j+1)/2)} \right) \end{aligned} \end{aligned}$$

Lemma 5.1 in [11] gives the following

$$\begin{aligned} \begin{aligned} \log (K_n)&= \frac{pq}{2}\log \left( \frac{2}{n} \right) \\&\quad +\,\sum _{j=1}^q \bigg [ \frac{n-j+1}{2} \log \left( \frac{n-j+1}{2} \right) \\&\quad -\, \frac{n-p-j+1}{2} \log \left( \frac{n-p-j+1}{2} \right) - \frac{p}{2}\\&\quad -\, \frac{p}{2(n-p-j+1)} + O \left( \frac{p^2 + 4}{(n-p-j+1)^2} \right) \bigg ]. \end{aligned} \end{aligned}$$

Using the fact that \(pq = o(n)\) for the last two terms,

$$\begin{aligned} \begin{aligned} \log (K_n)&= \frac{pq}{2}\log \left( \frac{2}{n} \right) \\&\quad +\, \sum _{j=1}^q \bigg [ \left( \frac{n-p-j+1}{2} + \frac{p}{2} \right) \log \left( \frac{n-j+1}{2} \right) \\&\quad -\, \frac{n-p-j+1}{2} \log \left( \frac{n-p-j+1}{2} \right) - \frac{p}{2} \bigg ] + o(1)\\&= \frac{-pq}{2} + \frac{pq}{2}\log \left( \frac{2}{n} \right) \\&\quad +\, \sum _{j=1}^q \bigg [ \frac{n-p-j+1}{2} \log \left( \frac{n-j+1}{n-p-j+1} \right) \\&\quad +\, \frac{p}{2} \log \left( \frac{n-j+1}{2} \right) \bigg ] + o(1). \\ \end{aligned} \end{aligned}$$

Bringing \(\frac{pq}{2}\log \left( \frac{2}{n} \right) \) inside the sum yields

$$\begin{aligned} \begin{aligned} \log (K_n)&= \frac{-pq}{2} + \sum _{j=1}^q \bigg [ \frac{n-p-j+1}{2} \log \left( \frac{n-j+1}{n-p-j+1} \right) \\&\quad +\, \frac{p}{2} \log \left( \frac{n-j+1}{n} \right) \bigg ] + o(1). \\ \end{aligned} \end{aligned}$$

Rewriting the logarithms

$$\begin{aligned} \begin{aligned} \log \left( \frac{n-j+1}{n-p-j+1} \right)&= \log \left( \frac{1 + \frac{p(j-1)}{n(n-p-j+1)}}{1 - \frac{p}{n}} \right) \\&= - \log \left( 1 - \frac{p}{n} \right) + \log \left( 1 + \frac{p(j-1)}{n(n-p-j+1)} \right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \log \left( \frac{n-j+1}{n} \right) = \log \left( 1 - \frac{j-1}{n} \right) \end{aligned}$$

yields

$$\begin{aligned} \begin{aligned} \log (K_n)&= \frac{-pq}{2} \\&\quad +\, \sum _{j=1}^q \bigg [ -\frac{n-p-j+1}{2} \log \left( 1 - \frac{p}{n} \right) \\&\quad +\, \frac{n-p-j+1}{2}\log \left( 1 + \frac{p(j-1)}{n(n-p-j+1)} \right) \\&\quad +\, \frac{p}{2}\log \left( 1 - \frac{j-1}{n} \right) \bigg ] + o(1). \end{aligned} \end{aligned}$$

For the first term

$$\begin{aligned} \begin{aligned} -\log&\left( 1 - \frac{p}{n} \right) \sum _{j=1}^q \frac{n-p-j+1}{2} = - \log \left( 1 - \frac{p}{n} \right) \left( \frac{(n-p)q}{2} - \frac{q(q-1)}{4} \right) . \end{aligned} \end{aligned}$$

By Taylor’s Theorem there exists some \(\xi _p \in (0,p)\) such that

$$\begin{aligned} \begin{aligned} - \log \left( 1 - \frac{p}{n} \right)&= \sum _{k=1}^l \frac{1}{k} \left( \frac{p}{n} \right) ^k + \frac{p^{l+1}}{(l+1)(\xi _p - n)^{l+1}}. \end{aligned} \end{aligned}$$

Now, \(\xi _p = o(n)\) so the error term in the expansion is

$$\begin{aligned} \begin{aligned} \left( \frac{(n-p)q}{2} - \frac{q(q-1)}{4} \right) \left( \frac{p^{l+1}}{(l+1)(\xi _p - n)^{l+1}} \right)&\le \frac{p^{l+1}(n-p)q}{(l+1)(\xi _p - n)^{l+1}} \\&\le \frac{C p^{l+1}q}{(l+1)(\xi _p - n)^{l}}\\&= o(1) \end{aligned} \end{aligned}$$

by choice of l. (See Eq. (1) for a similar computation.) The first term is thus

$$\begin{aligned} \begin{aligned}&\left( \frac{(n-p)q}{2} - \frac{q(q-1)}{4} \right) \sum _{k=1}^l \frac{1}{k} \left( \frac{p}{n} \right) ^k + o(1) \\&\quad = \sum _{k=1}^l \frac{p^kq}{2kn^{k-1}} - \sum _{k=1}^l \frac{p^{k+1}q}{2kn^k} - \sum _{k=1}^l \frac{q^2}{4k} \left( \frac{p}{n} \right) ^k + \sum _{k=1}^l \frac{q}{4k} \left( \frac{p}{n} \right) ^k + o(1). \end{aligned} \end{aligned}$$

Notice that

$$\begin{aligned} \begin{aligned} \sum _{k=2}^l \frac{q^2}{4k}&\left( \frac{p}{n} \right) ^k + \sum _{k=1}^l \frac{q}{4k} \left( \frac{p}{n} \right) ^k \le \frac{1}{2} \sum _{k=1}^{\infty } \frac{1}{k} \left( \frac{pq}{n} \right) ^k = o(1) \end{aligned} \end{aligned}$$

by Lemma 7. Therefore, the first term is

$$\begin{aligned} \begin{aligned}&\frac{pq}{2} + \sum _{k=1}^{l-1} \frac{p^{k+1}q}{2(k+1)n^k} - \sum _{k=1}^l \frac{p^{k+1}q}{2kn^k} -\frac{pq^2}{4n} + o(1) \\&\quad = \frac{pq}{2} - \frac{pq^2}{4n} - \sum _{k=1}^{l-1} \frac{p^{k+1}q}{2(k+1)kn^k} - \frac{p^{l+1}q}{2ln^l}+ o(1). \end{aligned} \end{aligned}$$

For the second term there exist \(\xi _{p,j} \in \left( 0, \frac{p(j-1)}{n-p-j+1} \right) \) so that

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^q \frac{n-p-j+1}{2}\log \left( 1 + \frac{p(j-1)}{n(n-p-j+1)} \right) \\&\quad = \sum _{j=1}^q \frac{n-p-j+1}{2} \bigg [ \sum _{k=1}^l (-1)^{k+1} \frac{1}{k} \left( \frac{p(j-1)}{n(n-p-j+1)} \right) ^k \\&\qquad +\, \frac{ p^{l+1}(j-1)^{l+1}}{(l+1) (n-p-j+1)^{l+1} (\xi _{p,j} - n)^{l+1}} \bigg ]. \end{aligned} \end{aligned}$$

Notice that

$$\begin{aligned} \begin{aligned} \sum _{j=1}^q \frac{ p^{l+1}(j-1)^{l+1}}{(l+1) (n-p-j+1)^{l+1} (\xi _{p,j} - n)^{l+1}}&= o(1) \end{aligned} \end{aligned}$$

by choice of l. The second term is thus

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^q \frac{p(j-1)}{2n} + \sum _{j=1}^q \frac{n-p-j+1}{2} \sum _{k=2}^l (-1)^{k+1} \frac{1}{k} \left( \frac{p(j-1)}{n(n-p-j+1)} \right) ^k + o(1) \\&\quad = \frac{pq^2}{4n} + o(1). \end{aligned} \end{aligned}$$

For the third term, there exist \(\xi _{j-1} \in (0, j-1)\) such that

$$\begin{aligned} \begin{aligned} \sum _{j=1}^q\frac{p}{2}\log \left( 1 - \frac{j-1}{n} \right)&= - \sum _{j=1}^q \frac{p}{2} \left[ \sum _{k=1}^l \frac{1}{k} \left( \frac{j-1}{n} \right) ^k + \frac{(j-1)^{l+1}}{(l+1)(\xi _{j-1} -n)^{l+1}} \right] \\&= - \sum _{j=1}^q \frac{p}{2} \frac{j-1}{n} - \sum _{j=1}^q\frac{p}{2}\sum _{k=2}^l \frac{1}{k} \left( \frac{j-1}{n} \right) ^k \\&\quad +\, \frac{p}{2}\sum _{j=1}^q \frac{(j-1)^{l+1}}{(l+1)(\xi _{j-1} -n)^{l+1}}. \end{aligned} \end{aligned}$$

Now, observe that

$$\begin{aligned} \begin{aligned} \sum _{j=1}^q\frac{p}{2}\sum _{k=2}^l \frac{1}{k} \left( \frac{j-1}{n} \right) ^k&\le \frac{pq}{2} \sum _{k=2}^l \frac{1}{k} \left( \frac{q}{n} \right) ^k \\&\le \frac{p^2}{2} \sum _{k=2}^l \frac{1}{k} \left( \frac{q}{n} \right) ^k \\&\le \sum _{k=2}^l \frac{1}{2k} \left( \frac{pq}{n} \right) ^k \\&= o(1) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{p}{2} \sum _{j=1}^q \frac{(j-1)^{l+1}}{(l+1)(\xi _{j-1} -n)^{l+1}}&\le \frac{p}{2} \sum _{j=1}^q \frac{q^{l+1}}{(l+1)(\xi _{j-1} -n)^{l+1}} \\&= o(1). \end{aligned} \end{aligned}$$

Thus, the third term is

$$\begin{aligned} \begin{aligned} \sum _{j=1}^q\frac{p}{2}\log \left( 1 - \frac{j-1}{n} \right)&= \frac{-pq^2}{4n} +o(1). \end{aligned} \end{aligned}$$

Putting everything together,

$$\begin{aligned} \begin{aligned} \log (K_n) = - \frac{pq^2}{4n} - \sum _{k=1}^{l-1} \frac{p^{k+1}q}{2(k+1)kn^k} - \frac{p^{l+1}q}{2ln^l}+ o(1). \end{aligned} \end{aligned}$$

Recall that the notation \((x)_a\) represents the falling factorial, which is defined by

$$\begin{aligned} (x)_a = x (x-1)(x-2) \cdots (x-a+1). \end{aligned}$$

Lemma 9

Let \( \{ p_n : n \ge 1 \}\) and \( \{q_n : n \ge 1 \}\) be two sequences of positive integers such that \(p_nq_n = o(n)\) and \(q_n \le p_n\). For each n, let \(X_n = (x_{ij})\) be a \(p_n \times q_n\) matrix where \(x_{ij}\) are independent standard Gaussian random variables. Let

$$\begin{aligned} E_j = \mathbb {E} \bigg [ \frac{1}{n^j} \bigg ( \frac{p_n+q_n+1}{2j} {\hbox {tr}}(X^TX)^j - \frac{1}{2(j+1)} {\hbox {tr}}(X^TX)^{j+1} \bigg ) \bigg ] \end{aligned}$$

when \(j \le l-1\) and

$$\begin{aligned} E_l = \mathbb {E} \bigg [ \frac{1}{n^l} \frac{p_n+q_n+1}{2l} {\hbox {tr}}(X^TX)^l \bigg ]. \end{aligned}$$

Then,

$$\begin{aligned} \sum _{j=1}^l E_j = \frac{p_nq_n^2}{4n}+\sum _{j=1}^{l-1} \frac{(p_n)_{j}p_nq_n}{2j(j+1)n^j} + \frac{(p_n)_{l}p_nq_n}{2ln^l} + o(1) . \end{aligned}$$

Proof

When \(j=1\),

$$\begin{aligned} \begin{aligned} E_1&= \frac{p+q+1}{2n}pq - \frac{p^2q + pq^2 - 2pq}{4n} \\&= \frac{p^2q + pq^2}{4n} + o(1), \end{aligned} \end{aligned}$$

since \(pq = o(n)\). When \(j=l\):

$$\begin{aligned} \begin{aligned} E_l&= \frac{p+q+1}{2ln^l} \sum _{s=0}^{l-1} (p)_{l-s}(q)_{s+1}\frac{1}{s+1} \left( {\begin{array}{c}l\\ s\end{array}}\right) \left( {\begin{array}{c}l-1\\ s\end{array}}\right) \\&= \frac{(p)_{l}pq}{2ln^l} + o(1), \end{aligned} \end{aligned}$$

by computations similar to Eq. (3) in Sect. 3. For \(1< j < l\),

$$\begin{aligned} E_j&= \mathbb {E} \bigg [ \frac{1}{n^j} \bigg ( \frac{p+q+1}{2j} {\hbox {tr}}(X^TX)^j - \frac{1}{2(j+1)} {\hbox {tr}}(X^TX)^{j+1} \bigg ) \bigg ] \\&= \frac{p+q+1}{2jn^j} \mathbb {E} [{\hbox {tr}}(X^TX)^j] - \frac{1}{2(j+1)n^j} \mathbb {E}[{\hbox {tr}}(X^TX)^{j+1}] \\&= \frac{p+q+1}{2jn^j}(p)_jq - \frac{1}{2(j+1)n^j}(p)_{j+1}q+ \frac{p+q+1}{2jn^j}S_1 - \frac{1}{2(j+1)n^j}S_2, \end{aligned}$$

where

$$\begin{aligned} 0 \le S_1&= \sum _{s=1}^{j-1} (p)_{j-s}(q)_{s+1} \frac{1}{s+1} \left( {\begin{array}{c}j\\ s\end{array}}\right) \left( {\begin{array}{c}j-1\\ s\end{array}}\right) \le \frac{Cp^{j-1}q^22^{2j}}{j^{3/2}} \end{aligned}$$

and

$$\begin{aligned} 0 \le S_2&= \sum _{s=1}^j (p)_{j+1-s}(q)_{s+1}\frac{1}{s+1} \left( {\begin{array}{c}j+1\\ s\end{array}}\right) \left( {\begin{array}{c}j\\ s\end{array}}\right) \le \frac{Cp^jq^22^{2j+2}}{(j+1)^{3/2}}, \end{aligned}$$

by computations similar to Eq. (3). Therefore,

$$\begin{aligned} E_j&= (p)_jq \left( \frac{p+q+1}{2jn^j} - \frac{p-j}{2(j+1)n^j} \right) + \frac{p+q+1}{2jn^j}S_1 - \frac{1}{2(j+1)n^j}S_2 \\&= \frac{(p)_{j}pq}{2j(j+1)n^j} + D_j, \end{aligned}$$

where

$$\begin{aligned} D_j&= (p)_jq \frac{q + 1 + j^2 +qj + j}{2j(j+1)n^j} + \frac{p+q+1}{2jn^j}S_1 - \frac{1}{2(j+1)n^j}S_2, \end{aligned}$$

and thus

$$\begin{aligned} |D_j|&\le \frac{q + 1 + j^2 +qj + j}{2j(j+1)n^j} + \frac{Cp^{j}q^22^{2j}}{j^{5/2}n^j} + \frac{Cp^jq^22^{2j}}{(j+1)^{5/2}n^j}. \end{aligned}$$

Summing over \(1< j <l\) and using Lemma 7,

$$\begin{aligned} \sum _{j=2}^{l-1} E_j&= \sum _{j=2}^{l-1} \left( \frac{(p)_{j}pq}{2j(j+1)n^j} + D_j \right) \\&= \sum _{j=2}^{l-1} \frac{(p)_{j}pq}{2j(j+1)n^j} + o(1). \end{aligned}$$

\(\square \)

Lemma 10

Let \( \{ p_n : n \ge 1 \}\) and \( \{q_n : n \ge 1 \}\) be two sequences of positive integers such that \(p_nq_n = o(n)\) and \(q_n \le p_n\). For each n, let \(X_n = (x_{ij})\) be a \(p_n \times q_n\) matrix where \(x_{ij}\) are independent standard Gaussian random variables. Let

$$\begin{aligned} E_j = \mathbb {E} \bigg [ \frac{1}{n^j} \bigg ( \frac{p_n+q_n+1}{2j} {\hbox {tr}}(X^TX)^j - \frac{1}{2(j+1)} {\hbox {tr}}(X^TX)^{j+1} \bigg ) \bigg ] \end{aligned}$$

when \(j \le l-1\) and

$$\begin{aligned} E_l = \mathbb {E} \bigg [ \frac{1}{n^l} \frac{p_n+q_n+1}{2l} {\hbox {tr}}(X^TX)^l \bigg ]. \end{aligned}$$

Let

$$\begin{aligned} K_n = \left( \frac{2}{n} \right) ^{p_nq_n/2} \prod _{j=1}^{q_n} \frac{\varGamma ((n-j+1)/2)}{\varGamma ((n-p_n-j+1)/2)}. \end{aligned}$$

Then,

$$\begin{aligned} \left( \log (K_n) + \sum _{j=1}^l E_j \right) = o(1). \end{aligned}$$

Proof

By Lemmas 8 and 9,

$$\begin{aligned} \begin{aligned}&\log (K_n) + \sum _{j=1}^l E_j \\&\quad = \sum _{j=1}^{l-1} \left( \frac{(p)_jpq}{2j(j+1)n^j} - \frac{p^{j+1}q}{2j(j+1)n^j} \right) \\&\qquad +\, \frac{(p)_l pq}{2ln^l} - \frac{p^{l+1}q}{2ln^l} + o(1). \end{aligned} \end{aligned}$$

Expanding the falling factorials,

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^{l-1} \left( \frac{(p)_jpq}{2j(j+1)n^j} - \frac{p^{j+1}q}{2j(j+1)n^j} \right) \\&\quad = \sum _{j=1}^{l-1} \frac{pq^2}{2j(j+1)n^j} \left( (p-1)(p-2) \cdots (p-l+1) - p^{l+1} \right) \\&\quad = o(1). \end{aligned} \end{aligned}$$

Similarly,

$$\begin{aligned} \frac{(p)_l pq}{2ln^l} - \frac{p^{l+1}q}{2ln^l} = o(1). \end{aligned}$$

\(\square \)

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Stewart, K. Total Variation Approximation of Random Orthogonal Matrices by Gaussian Matrices. J Theor Probab 33, 1111–1143 (2020). https://doi.org/10.1007/s10959-019-00900-5

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Keywords

  • Random orthogonal matrix
  • Central limit theorem
  • Wishart matrices
  • Moments

Mathematics Subject Classification (2010)

  • 60F05
  • 60C05