Complete asymptotic expansions and the high-dimensional Bingham distributions

Abstract

For $d \ge 2$, let X be a random vector having a Bingham distribution on ${\mathcal {S}}^{d-1}$, the unit sphere centered at the origin in ${\mathbb {R}}^d$, and let $\Sigma$ denote the symmetric matrix parameter of the distribution. Let $\Psi (\Sigma )$ be the normalizing constant of the distribution and let $\nabla \Psi _d(\Sigma )$ be the matrix of first-order partial derivatives of $\Psi (\Sigma )$ with respect to the entries of $\Sigma$. We derive complete asymptotic expansions for $\Psi (\Sigma )$ and $\nabla \Psi _d(\Sigma )$, as $d \rightarrow \infty$; these expansions are obtained subject to the growth condition that $\Vert \Sigma \Vert$, the Frobenius norm of $\Sigma$, satisfies $\Vert \Sigma \Vert \le \gamma _0 d^{r/2}$ for all d, where $\gamma _0 > 0$ and $r \in [0,1)$. Consequently, we obtain for the covariance matrix of X an asymptotic expansion up to terms of arbitrary degree in $\Sigma$. Using a range of values of d that have appeared in a variety of applications of high-dimensional spherical data analysis, we tabulate the bounds on the remainder terms in the expansions of $\Psi (\Sigma )$ and $\nabla \Psi _d(\Sigma )$ and we demonstrate the rapid convergence of the bounds to zero as r decreases.

A Appendix: Proofs

1.1 A.1 The proofs of Proposition 2.1 and Corollary 2.2

Proof of Proposition 2.1

By (2.5),

$$\begin{aligned} \vert R_m(\Sigma ) \vert \le \sum _{k=m}^\infty \frac{(1/2)_k}{(d/2)_k} \frac{\vert C_{(k)}(\Sigma ) \vert }{k!}, \end{aligned}$$

(A.1)

and by (2.1),

$$\begin{aligned} \vert C_{(k)}(\Sigma ) \vert \le \frac{k!}{(1/2)_k} \ \sum _{i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k} \ \prod _{j=1}^k \frac{\vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j) \vert ^{i_j}}{i_j! \, (2j)^{i_j}}. \end{aligned}$$

(A.2)

For each vector of indices $(i_1,i_2,\ldots ,i_k)$ such that $i_1 + 2 i_2 + \cdots + k i_k = k$, we have

$$\begin{aligned} \prod _{j=1}^k \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j) \vert ^{i_j}&= \Vert \Sigma \Vert ^k \prod _{j=1}^k \frac{\vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j}}{\Vert \Sigma \Vert ^{ji_j}} \nonumber \\&= \Vert \Sigma \Vert ^k \prod _{j=1}^k \frac{\vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j}}{(\hspace{1.0pt}\textrm{tr}\,(\Sigma ^2))^{ji_j/2}}. \end{aligned}$$

(A.3)

Denote by $\lambda _1,\ldots ,\lambda _d$ the eigenvalues of $\Sigma$. Then for each $j=1,\ldots ,k$,

$$\begin{aligned} \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j} = \bigg \vert \sum _{i=1}^d \lambda _i^j\bigg \vert ^{i_j} \le \bigg (\sum _{i=1}^d \vert \lambda _i\vert ^j\bigg )^{i_j}, \end{aligned}$$

and by substituting this bound into (A.3) we obtain

$$\begin{aligned} \prod _{j=1}^k \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j) \vert ^{i_j} \le \Vert \Sigma \Vert ^k \prod _{j=1}^k \frac{\left( \sum _{i=1}^d \vert \lambda _i\vert ^j\right) ^{i_j}}{\left( \sum _{i=1}^d \vert \lambda _i\vert ^2\right) ^{ji_j/2}}. \end{aligned}$$

(A.4)

By a remarkable result of (Reznick 1983, p. 447, eq. (2.12)) we obtain, for each $j=1,\ldots ,k$ and all d,

$$\begin{aligned} \frac{\left( \sum _{i=1}^d \vert \lambda _i\vert ^j\right) ^{i_j}}{\left( \sum _{i=1}^d \vert \lambda _i\vert ^2\right) ^{ji_j/2}} \le d^{\max \{0,(2-j) i_j/2\}} = {\left\{ \begin{array}{ll} d^{i_1/2}, &{} j=1 \\ 1, &{} j \ge 2 \end{array}\right. }. \end{aligned}$$

(A.5)

Substituting the bound in (A.5) into (A.4), we obtain

$$\begin{aligned} \prod _{j=1}^k \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j} \le d^{i_1/2} \Vert \Sigma \Vert ^k \equiv d^{i_1/2} \Vert \Sigma \Vert ^{i_1+2i_2+3i_3+\cdots +ki_k}. \end{aligned}$$

(A.6)

On applying (A.6) to (A.2) we obtain

$$\begin{aligned} \vert C_{(k)}(\Sigma )\vert \le \frac{k!}{(1/2)_k} a_k \Vert \Sigma \Vert ^k, \end{aligned}$$

(A.7)

where

$$\begin{aligned} a_k = \sum _{i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k} d^{i_1/2} \prod _{j=1}^k \frac{1}{i_j! \, (2j)^{i_j}}. \end{aligned}$$

(A.8)

We will derive in Lemma A.1 a single-sum expression for the multiple sum $a_k$, and then we will deduce from that single-sum the inequality

$$\begin{aligned} a_k \le \frac{(d^{1/2}/2)_k}{k!}. \end{aligned}$$

(A.9)

Substituting (A.9) into (A.7) and recalling that $\Vert \Sigma \Vert \le \gamma _0 d^{r/2}$ for all d, we obtain

$$\begin{aligned} \vert C_{(k)}(\Sigma )\vert \le \frac{(d^{1/2}/2)_k}{(1/2)_k} \Vert \Sigma \Vert ^k \le \frac{(d^{1/2}/2)_k}{(1/2)_k} (\gamma _0 d^{r/2})^k. \end{aligned}$$

(A.10)

Inserting (A.10) into (A.1) we obtain

$$\begin{aligned} \vert R_m(\Sigma ) \vert \le \sum _{k=m}^\infty \frac{(d^{1/2}/2)_k}{(d/2)_k} \frac{\gamma _0 ^k d^{rk/2}}{k!}. \end{aligned}$$

(A.11)

As we shall show in Lemma A.2, with $\gamma _1 = \tfrac{1}{2}(\sqrt{3}+1) \simeq 1.366025$,

$$\begin{aligned} \frac{(d^{1/2}/2)_k}{(d/2)_k} \le \gamma _1^{k-1} ((k-1)!)^{1/2} d^{-k/2} \end{aligned}$$

(A.12)

for all $d \ge 1$, $k \ge 1$. Letting $\gamma _2 = \gamma _1 \gamma _0$ and substituting (A.12) into (A.11), we find that for all d,

$$\begin{aligned} \vert R_m(\Sigma )\vert \le \gamma _1^{-1} \sum _{k=m}^\infty \frac{((k-1)!)^{1/2}}{k!} \, \gamma _2^k d^{-k(1-r)/2}. \end{aligned}$$

(A.13)

By applying the ratio test, we find that the series (A.13) converges absolutely for all d such that $\gamma _2 d^{-(1-r)/2} < 1$, equivalently, $d > \gamma _2^{2/(1-r)}$.

Note that although the kth term in (A.13) is $O(d^{-k(1-r)/2})$, it is not evident that the sum of the resulting series is $O(d^{-m(1-r)/2})$ as $d \rightarrow \infty$. To establish that stated convergence rate we apply to (A.13) the Cauchy–Schwarz inequality, obtaining

$$\begin{aligned} \vert R_m(\Sigma ) \vert&\le \gamma _1^{-1} \Big (\sum _{k=m}^\infty \frac{(k-1)!}{(k!)^2}\Big )^{1/2} \Big (\sum _{k=m}^\infty \gamma _2^{2k} d^{-k(1-r)}\Big )^{1/2} \nonumber \\&= \gamma _1^{-1} \Big (\sum _{k=m}^\infty \frac{1}{k \cdot k!}\Big )^{1/2} \cdot \gamma _2^m d^{-m(1-r)/2} \big (1 - \gamma _2^2 d^{-(1-r)}\big )^{-1/2}. \end{aligned}$$

(A.14)

For $k \ge m$, it is straightforward that

$$\begin{aligned} \frac{m}{k} \le 1 \le \left( {\begin{array}{c}k\\ m\end{array}}\right) \equiv \frac{k!}{m! (k-m)!}, \end{aligned}$$

equivalently,

$$\begin{aligned} \frac{1}{k \cdot k!} \le \frac{1}{m \cdot m!} \cdot \frac{1}{(k-m)!}, \end{aligned}$$

also that

$$\begin{aligned} \frac{1}{m} \le \frac{2}{m+1}. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{k \cdot k!} \le \frac{1}{m \cdot m!} \cdot \frac{1}{(k-m)!} \le \frac{2}{(m+1)!} \cdot \frac{1}{(k-m)!}. \end{aligned}$$

Summing this inequality over all $k \ge m$ we obtain

$$\begin{aligned} \sum _{k=m}^\infty \frac{1}{k \cdot k!} \le \frac{2}{(m+1)!} \sum _{k=m}^\infty \frac{1}{(k-m)!} = \frac{2e}{(m+1)!}. \end{aligned}$$

(A.15)

Noting that the inequality $d \ge (2\gamma _2^2)^{1/(1-r)}$ is equivalent to

$$\begin{aligned} \big (1 - \gamma _2^2 d^{-(1-r)}\big )^{-1} \le 2, \end{aligned}$$

(A.16)

and applying (A.15) and (A.16) to (A.14) we obtain, for all $d \ge (2\gamma _2^2)^{1/(1-r)}$, the inequality

$$\begin{aligned} \vert R_m(\Sigma ) \vert \le 2 \gamma _1^{-1} \Big (\frac{2e}{(m+1)!}\Big )^{1/2} \gamma _2^m d^{-m(1-r)/2}. \end{aligned}$$

(A.17)

This establishes (2.7) and proves that $R_m(\Sigma ) = O(d^{-m(1-r)/2})$ as $d \rightarrow \infty$, and the proof of the proposition now is complete. $\square$

Proof of Corollary 2.2

By (A.17) with $m=1$ we have, for all $d \ge (2\gamma _2^2)^{1/(1-r)}$,

$$\begin{aligned} \vert R_1(\Sigma ) \vert \le 2 \gamma _1^{-1} e^{1/2} \gamma _2 d^{-(1-r)/2}. \end{aligned}$$

(A.18)

For $d > (6\gamma _2^2)^{1/(1-r)}$, we use the values of the constants given in (2.6) to determine that the right-hand side of (A.18) is strictly less than 1. Then we apply the geometric series to obtain, for all $l \ge 2$,

$$\begin{aligned}{}[\Psi _d(\Sigma )]^{-1}&= [1 + R_1(\Sigma )]^{-1} \nonumber \\&= 1 - R_1(\Sigma ) + \sum _{j=2}^\infty (-1)^j \big (R_1(\Sigma )\big )^j \nonumber \\&= 1 - \bigg (\sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!} + R_l(\Sigma )\bigg ) + \sum _{j=2}^\infty (-1)^j \big (R_1(\Sigma )\big )^j. \end{aligned}$$

(A.19)

By Proposition 2.1, $R_l(\Sigma ) = O(d^{-l(1-r)/2})$. On applying (A.17) with $m=1$ we obtain, for all $d > (6\gamma _2^2)^{1/(1-r)}$,

$$\begin{aligned} \left| \sum _{j=2}^\infty (-1)^j \big (R_1(\Sigma )\big )^j\right|&\le \sum _{j=2}^\infty \vert R_1(\Sigma )\vert ^j \nonumber \\&\le \sum _{j=2}^\infty \big [2 \gamma _1^{-1} \gamma _2 e^{1/2} d^{-(1-r)/2}\big ]^j. \end{aligned}$$

(A.20)

By summing this geometric series, we deduce that (A.20) is $O(d^{-(1-r)})$ as $d \rightarrow \infty$. Therefore by (A.19), for all $l \ge 2$ and all $d > (6\gamma _2^2)^{1/(1-r)}$,

$$\begin{aligned}{}[\Psi _d(\Sigma )]^{-1} = 1 - \sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!} + O(d^{-l(1-r)/2}) + O(d^{-(1-r)}). \end{aligned}$$

On applying the well-known property,

$$\begin{aligned} O(d^{-a}) + O(d^{-b}) = O(d^{-\min (a,b)}) \end{aligned}$$

(A.21)

for $a,b > 0$, we obtain (2.8). $\square$

1.2 A.2 The proof of Proposition 3.2 and Theorem 3.3

Proof of Lemma 3.1

The result (3.2) follows by straightforwardly applying each element of the matrix $\nabla$ to the function $\exp \big (\hspace{1.0pt}\textrm{tr}\,(\Sigma H)\big )$.

The formula (3.3) follows from the chain rule:

$$\begin{aligned} \nabla [\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^k = k [\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^{k-1} \nabla \hspace{1.0pt}\textrm{tr}\,(\Sigma ) = k [\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^{k-1} I_d. \end{aligned}$$

As for (3.4), that result can be deduced by arguments similar to those given by (Sebastiani 1996, Lemmas 3.1 or 5.1). Also see (Dwyer 1948, p. 528, Section 14) for the derivatives of the trace of powers of square, non-symmetric matrices; calculations similar to theirs can also lead to (3.4).

A succinct derivation of (3.4) is obtained from the Taylor expansion (3.1), as follows. By using the commutativity property of the trace, i.e., $\hspace{1.0pt}\textrm{tr}\,(\Sigma H) = \hspace{1.0pt}\textrm{tr}\,(H \Sigma )$, we find that

$$\begin{aligned} \hspace{1.0pt}\textrm{tr}\,[(\Sigma +H)^k]&= \hspace{1.0pt}\textrm{tr}\,[(\Sigma +H)(\Sigma +H)\cdots (\Sigma +H)] \nonumber \\&= \hspace{1.0pt}\textrm{tr}\,(\Sigma ^k) + k \hspace{1.0pt}\textrm{tr}\,(H \Sigma ^{k-1}) + O(\Vert H\Vert ^2) \nonumber \\&\equiv \hspace{1.0pt}\textrm{tr}\,(\Sigma ^k) + \langle H,k\Sigma ^{k-1}\rangle + O(\Vert H\Vert ^2) \end{aligned}$$

(A.22)

as $H \rightarrow 0$. Setting $f(\Sigma ) = \hspace{1.0pt}\textrm{tr}\,(\Sigma ^k)$ in (3.1) and comparing the result with (A.22), we obtain (3.4). $\square$

Proof of Proposition 3.2:

By the subadditivity property of the Frobenius norm, we have

$$\begin{aligned} \Vert \nabla R_m(\Sigma )\Vert&= \bigg \Vert \sum _{k=m}^\infty \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert \\&\le \sum _{k=m}^\infty \bigg \Vert \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert . \end{aligned}$$

For $k \ge 2$, it follows from (2.1) that

$$\begin{aligned} C_{(k)}(\Sigma )&= \frac{k!}{(1/2)_k} \Bigg [\frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^k}{k! \, 2^k} + \sum _{\begin{array}{c} i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \ \prod _{j=1}^k \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)]^{i_j}}{i_j! \, (2j)^{i_j}}\Bigg ] \nonumber \\&\equiv \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^k}{(1/2)_k \, 2^k} + \frac{k!}{(1/2)_k} p_k(\Sigma ), \end{aligned}$$

(A.23)

where

$$\begin{aligned} p_k(\Sigma ) = \sum _{\begin{array}{c} i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \ \prod _{j=1}^k \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)]^{i_j}}{i_j! \, (2j)^{i_j}} \end{aligned}$$

(A.24)

is a polynomial in $\{\hspace{1.0pt}\textrm{tr}\,(\Sigma ),\hspace{1.0pt}\textrm{tr}\,(\Sigma ^2),\ldots ,\hspace{1.0pt}\textrm{tr}\,(\Sigma ^k)\}$. It follows from (A.24) that $p_k(\Sigma )$ is homogeneous of degree k in $\Sigma$ and its coefficients do not depend on d; moreover, because of the restriction $i_1 \le k-2$, d the highest power of $\hspace{1.0pt}\textrm{tr}\,(\Sigma )$ which can appear in $p_k(\Sigma )$ is $k-2$.

By (3.3) and (A.23),

$$\begin{aligned} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}&= \frac{(1/2)_k}{(d/2)_k \, k!} \nabla \bigg [\frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^k}{(1/2)_k \, 2^k} + \frac{k!}{(1/2)_k} p_k(\Sigma )\bigg ] \nonumber \\&= \frac{1}{(d/2)_k}\bigg [\frac{1}{(k-1)! \, 2^k} [\hspace{1.0pt}\textrm{tr}\,(\Sigma )]^{k-1} I_d + \nabla p_k(\Sigma )\bigg ]. \end{aligned}$$

(A.25)

On applying to (A.25) the subadditivity property of the Frobenius norm we obtain

$$\begin{aligned} \bigg \Vert \frac{(1/2)_k}{(d/2)_k}&\frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert \nonumber \\&\le \frac{1}{(d/2)_k} \bigg [\frac{1}{(k-1)! \, 2^k} \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma )\vert ^{k-1} \Vert I_d\Vert + \Vert \nabla p_k(\Sigma )\Vert \bigg ] \nonumber \\&= \frac{1}{(d/2)_k} \bigg [\frac{d^{1/2}}{(k-1)! \, 2^k} \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma )\vert ^{k-1} + \Vert \nabla p_k(\Sigma )\Vert \bigg ]. \end{aligned}$$

(A.26)

By (A.24) and the product rule for derivatives,

$$\begin{aligned} \nabla p_k(\Sigma )&= \sum _{\begin{array}{c} i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \, \nabla \prod _{j=1}^k \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)]^{i_j}}{i_j! \, (2j)^{i_j}} \\&= \sum _{\begin{array}{c} i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \, \sum _{l=1}^k \Big (\prod _{\begin{array}{c} j=1 \\ j \ne l \end{array}}^k \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)]^{i_j}}{i_j! \, (2j)^{i_j}}\Big ) \cdot \frac{\nabla [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)]^{i_l}}{i_l! \, (2l)^{i_l}}. \end{aligned}$$

By (3.4) and the chain rule,

$$\begin{aligned} \nabla [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)]^{i_l} = i_l [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)]^{i_l - 1} \nabla [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)] = l \, i_l [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)]^{i_l - 1} \Sigma ^{l-1}; \end{aligned}$$

hence,

$$\begin{aligned} \nabla p_k(\Sigma ) = \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \sum _{l=1}^k \frac{l \, i_l}{i_l! (2l)^{i_l}} \Big (\prod _{\begin{array}{c} j=1 \\ j \ne l \end{array}}^k \frac{[\hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)]^{i_j}}{i_j! (2j)^{i_j}}\Big ) [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)]^{i_l - 1} \Sigma ^{l-1}, \end{aligned}$$

which also reveals that $\nabla p_k(\Sigma )$ is homogeneous of degree $k-1$. On applying the subadditivity property of the Frobenius norm, we obtain

$$\begin{aligned}&\Vert \nabla p_k(\Sigma )\Vert \\&\ \le \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \bigg (\prod _{j=1}^k \frac{1}{i_j! (2j)^{i_j}}\bigg ) \sum _{l=1}^k l \, i_l \Big (\prod _{\begin{array}{c} j=1 \\ j \ne l \end{array}}^k \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j}\Big ) \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)\vert ^{i_l - 1} \Vert \Sigma ^{l-1}\Vert . \end{aligned}$$

On applying (A.6) we have, for all d,

$$\begin{aligned} \Big (\prod _{\begin{array}{c} j=1 \\ j \ne l \end{array}}^k \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^j)\vert ^{i_j}\Big ) \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma ^l)\vert ^{i_l - 1} \le d^{i_1/2} \Vert \Sigma \Vert ^{i_1+2i_2+3i_3+\cdots +ki_k-l} = d^{i_1/2} \Vert \Sigma \Vert ^{k-l}. \end{aligned}$$

By also applying the submultiplicative inequality, $\Vert \Sigma ^{l-1}\Vert \le \Vert \Sigma \Vert ^{l-1}$, we obtain

$$\begin{aligned} \Vert \nabla p_k(\Sigma )\Vert&\le \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} \bigg (\prod _{j=1}^k \frac{1}{i_j! \, (2j)^{i_j}}\bigg ) \sum _{l=1}^k l i_l \, d^{i_1/2} \Vert \Sigma \Vert ^{k-l} \Vert \Sigma \Vert ^{l-1} \nonumber \\&= \Vert \Sigma \Vert ^{k-1} \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} d^{i_1/2}\bigg (\prod _{j=1}^k \frac{1}{i_j! \, (2j)^{i_j}}\bigg ) \sum _{l=1}^k l \, i_l \nonumber \\&= k \Vert \Sigma \Vert ^{k-1} \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} d^{i_1/2} \prod _{j=1}^k \frac{1}{i_j! \, (2j)^{i_j}}. \end{aligned}$$

(A.27)

By (A.8) and (A.9),

$$\begin{aligned} \sum _{\begin{array}{c} i_1 + 2 i_2 + \cdots + k i_k = k \\ i_1 \le k-2 \end{array}} d^{i_1/2} \prod _{j=1}^k \frac{1}{i_j! \, (2j)^{i_j}}&\equiv a_k - \frac{d^{k/2}}{k! \, 2^k} \\&\le \frac{1}{k!}\big [(d^{1/2}/2)_k - (d^{1/2}/2)^k\big ], \end{aligned}$$

and by substituting this bound into (A.27), we obtain

$$\begin{aligned} \Vert \nabla p_k(\Sigma )\Vert \le \frac{1}{(k-1)!}\big [(d^{1/2}/2)_k - (d^{1/2}/2)^k\big ] \Vert \Sigma \Vert ^{k-1}. \end{aligned}$$

On applying the latter inequality at (A.26), we find that

$$\begin{aligned}{} & {} \bigg \Vert \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert \\{} & {} \quad \le \frac{1}{(d/2)_k \, (k-1)!} \bigg [\frac{d^{1/2}}{2^k} \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma )\vert ^{k-1} + \big [(d^{1/2}/2)_k - (d^{1/2}/2)^k\big ] \Vert \Sigma \Vert ^{k-1}\bigg ]. \end{aligned}$$

Setting $(i_1,i_2,\ldots ,i_k) = (1,0,\ldots ,0)$ in (A.6), we obtain

$$\begin{aligned} \vert \hspace{1.0pt}\textrm{tr}\,(\Sigma )\vert \le d^{1/2} [\hspace{1.0pt}\textrm{tr}\,(\Sigma ^2)]^{1/2} = d^{1/2} \Vert \Sigma \Vert ; \end{aligned}$$

hence

$$\begin{aligned} \bigg \Vert&\frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert \\&\le \frac{1}{(d/2)_k \, (k-1)!} \bigg [\frac{d^{1/2}}{2^k} (d^{1/2}\Vert \Sigma \Vert )^{k-1} + \big [(d^{1/2}/2)_k - (d^{1/2}/2)^k\big ] \Vert \Sigma \Vert ^{k-1}\bigg ] \\&= \frac{(d^{1/2}/2)_k}{(d/2)_k} \frac{\Vert \Sigma \Vert ^{k-1}}{(k-1)!}. \end{aligned}$$

On applying the inequalities $\Vert \Sigma \Vert \le \gamma _0 d^{r/2}$ and (A.32) we obtain

$$\begin{aligned} \bigg \Vert \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert&\le \frac{(d^{1/2}/2)_k}{(d/2)_k} \frac{(\gamma _0 d^{r/2})^{k-1}}{(k-1)!} \\&\le \gamma _1^{k-1} ((k-1)!)^{1/2} d^{-k/2} \frac{(\gamma _0 d^{r/2})^{k-1}}{(k-1)!} \\&= d^{-1/2} \frac{(\gamma _2 d^{-(1-r)/2})^{k-1}}{((k-1)!)^{1/2}}. \end{aligned}$$

Therefore

$$\begin{aligned} \Vert \nabla R_m(\Sigma )\Vert&\le \sum _{k=m}^\infty \bigg \Vert \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert \nonumber \\&\le d^{-1/2} \sum _{k=m}^\infty \frac{(\gamma _2 d^{-(1-r)/2})^{k-1}}{((k-1)!)^{1/2}}, \end{aligned}$$

(A.28)

and, by applying the ratio test, we find that the latter series converges for all $d > \gamma _2^{2/(1-r)}$.

On applying the Cauchy–Schwarz inequality to (A.28), we obtain

$$\begin{aligned} \Vert \nabla R_m(\Sigma )\Vert \le d^{-1/2} \Big (\sum _{k=m}^\infty \frac{1}{(k-1)!}\Big )^{1/2} \Big (\sum _{k=m}^\infty (\gamma _2^2 d^{-(1-r)})^{k-1}\Big )^{1/2}. \end{aligned}$$

For any nonnegative integers a and b, it is elementary that

$$\begin{aligned} \frac{1}{(a+b)!} \le \frac{1}{a! b!}; \end{aligned}$$

therefore

$$\begin{aligned} \sum _{k=m}^\infty \frac{1}{(k-1)!} = \sum _{k=0}^\infty \frac{1}{(k+m-1)!} \le \sum _{k=0}^\infty \frac{1}{k! \, (m-1)!} = \frac{e}{(m-1)!}. \end{aligned}$$

Also, for all d such that $\gamma _2^2 d^{-(1-r)} \le 1/2$, equivalently, $d \ge (2\gamma _2^2)^{1/(1-r)}$, we have

$$\begin{aligned} \sum _{k=m}^\infty (\gamma _2^2 d^{-(1-r)})^{k-1}&= (\gamma _2^2 d^{-(1-r)})^{m-1} (1 - \gamma _2^2 d^{-(1-r)})^{-1} \\&\le 2 (\gamma _2^2 d^{-(1-r)})^{m-1}, \end{aligned}$$

and then we obtain

$$\begin{aligned} \Vert \nabla R_m(\Sigma )\Vert&\le d^{-1/2} \Big (\frac{e}{(m-1)!}\Big )^{1/2} \big (2 (\gamma _2^2 d^{-(1-r)})^{m-1}\big )^{1/2} \\&= (2e)^{1/2} [(m-1)!]^{-1/2} \gamma _2^{m-1} d^{-[1+(m-1)(1-r)]/2}. \end{aligned}$$

Therefore $\Vert \nabla R_m(\Sigma )\Vert = O(d^{-[1+(m-1)(1-r)]/2})$ as $d \rightarrow \infty$. $\square$

Proof of Theorem 3.3:

Since $\nabla C_{(0)}(\Sigma ) = 0$ then by the zonal polynomial expansion (2.3) of $\Psi _d(\Sigma )$,

$$\begin{aligned} \nabla \Psi _d(\Sigma ) = \sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!} + \nabla R_m(\Sigma ) \end{aligned}$$

(A.29)

and, by Proposition 3.2, $\Vert \nabla R_m(\Sigma )\Vert = O(d^{-[1+(m-1)(1-r)]/2})$ as $d \rightarrow \infty$.

On applying the asymptotic expansion of $[\Psi _d(\Sigma )]^{-1}$ given in (2.8), and using (A.29), we obtain

$$\begin{aligned} \textrm{Cov}(X)&= [\Psi _d(\Sigma )]^{-1} \nabla \Psi _d(\Sigma ) \\&= \bigg (1 - \sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!} + O(d^{-(1-r)})\bigg ) \\&\qquad \cdot \bigg (\sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!} + O(d^{-[1+(m-1)(1-r)]/2})\bigg ). \end{aligned}$$

Expanding this product, we obtain

$$\begin{aligned} \textrm{Cov}(X)&= \bigg (1 - \sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!}\bigg ) \bigg (\sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg ) \nonumber \\&\qquad + O(d^{-(1-r)}) \cdot \bigg (\sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg ) \nonumber \\&\qquad + 1 \cdot O(d^{-[1+(m-1)(1-r)]/2}) \nonumber \\&\qquad + \bigg (\sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!}\bigg ) \cdot O(d^{-[1+(m-1)(1-r)]/2}) \nonumber \\&\qquad + O(d^{-(1-r)}) \cdot O(d^{-[1+(m-1)(1-r)]/2}). \end{aligned}$$

(A.30)

On applying Proposition 3.2, we obtain

$$\begin{aligned} \bigg \Vert \sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert&= \Vert \nabla R_1(\Sigma ) - \nabla R_m(\Sigma )\Vert \\&\le \Vert \nabla R_1(\Sigma )\Vert + \Vert \nabla R_m(\Sigma )\Vert \\&= O(d^{-1/2}) + O(d^{-[1+(m-1)(1-r)]/2}) \\&= O(d^{-1/2}); \end{aligned}$$

therefore

$$\begin{aligned} O(d^{-(1-r)}) \cdot \bigg \Vert \sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg \Vert&= O(d^{-(1-r)}) \cdot O(d^{-1/2}) \\&= O(d^{-(3-2r)/2}). \end{aligned}$$

Similarly, by Proposition 2.1,

$$\begin{aligned} \sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!}&= R_1(\Sigma ) - R_l(\Sigma ) \\&= O(d^{-(1-r)/2}) + O(d^{-l(1-r)/2}) \\&= O(d^{-(1-r)/2}), \end{aligned}$$

and therefore

$$\begin{aligned} \bigg (\sum _{j=1}^{l-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!}&\bigg ) \cdot O(d^{-[1+(m-1)(1-r)]/2}) \\&= O(d^{-(1-r)/2}) \cdot O(d^{-[1+(m-1)(1-r)]/2}) \\&= O(d^{-[1+m(1-r)]/2}). \end{aligned}$$

The last $O(\cdot )$ term in (A.30) is

$$\begin{aligned} O(d^{-(1-r)}) \cdot O(d^{-[1+(m-1)(1-r)]/2}) = O(d^{-[1+(m+1)(1-r)]/2}). \end{aligned}$$

Collecting all $O(\cdot )$ terms in (A.30), we obtain

$$\begin{aligned} \textrm{Cov}(X)&= \bigg (1 - \sum _{j=1}^{m-1} \frac{(1/2)_j}{(d/2)_j} \frac{C_{(j)}(\Sigma )}{j!}\bigg ) \bigg (\sum _{k=1}^{m-1} \frac{(1/2)_k}{(d/2)_k} \frac{\nabla C_{(k)}(\Sigma )}{k!}\bigg ) \nonumber \\&\qquad + O(d^{-(3-2r)/2}) + O(d^{-[1+(m-1)(1-r)]/2}) \\&\qquad + O(d^{-[1+m(1-r)]/2}) + O(d^{-[1+(m+1)(1-r)]/2}). \end{aligned}$$

On applying (A.21), we obtain (3.7). $\square$

1.3 A.3 The proofs of (A.9) and (A.12)

First, we establish the inequality (A.9) for the coefficients $a_k$ defined in (A.8).

Lemma A.1

For $k = 0,1,2,\ldots$,

$$\begin{aligned} a_k = \sum _{l=0}^k \frac{\big ((d^{1/2}-1)/2\big )^l}{l!} \frac{(1/2)_{k-l}}{(k-l)!} \le \frac{(d^{1/2}/2)_k}{k!}. \end{aligned}$$

(A.31)

Proof

First, we follow the approach of (Comtet 1974, p. 97, Eq. 2d) to derive an explicit formula for $a_k$.

Let $t = (t_1,t_2,t_3,\ldots )$ be a vector of indeterminates, and define

$$\begin{aligned} a_k(t) = \sum _{i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k} \prod _{j=1}^k \frac{t_j^{i_j}}{i_j! \, (2j)^{i_j}}, \end{aligned}$$

$k \ge 0$. For an indeterminate u, a formal generating-function for the sequence $\{a_k(t), k=0,1,2,\ldots \}$ is

$$\begin{aligned} G_t(u) = \sum _{k=0}^\infty a_k(t) u^k&= \sum _{k=0}^\infty u^k \sum _{i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k} \prod _{j=1}^k \frac{t_j^{i_j}}{i_j! \, (2j)^{i_j}} \\&= \sum _{k=0}^\infty \sum _{i_1 + 2 i_2 + 3 i_3 + \cdots + k i_k = k} \prod _{j=1}^k \frac{t_j^{i_j} u^{ji_j}}{i_j! \, (2j)^{i_j}} \\&= \prod _{j=1}^\infty \bigg (\sum _{i_j=0}^\infty \frac{t_j^{i_j} u^{ji_j}}{i_j! \, (2j)^{i_j}}\bigg ) \\&= \prod _{j=1}^\infty \exp \Big (\frac{t_j u^j}{2j}\Big ) = \exp \bigg (\sum _{j=1}^\infty \frac{t_j u^j}{2j}\bigg ). \end{aligned}$$

Now set $t_1 = d^{1/2}$ and $t_j = 1$ for all $j \ge 2$. Then $a_k(t)$ reduces to $a_k$; $G_t(u)$ reduces to G(u), the formal generating-function for the sequence $\{a_k\}$; and we obtain

$$\begin{aligned} G(u) = \sum _{k=0}^\infty a_k u^k&= \exp \bigg (\tfrac{1}{2} d^{1/2} u + \tfrac{1}{2} \sum _{j=2}^\infty \frac{u^j}{j}\bigg ) \\&= \exp \bigg (\tfrac{1}{2} (d^{1/2}-1) u + \tfrac{1}{2} \sum _{j=1}^\infty \frac{u^j}{j}\bigg ) \\&= \exp \big (\tfrac{1}{2} (d^{1/2}-1) u - \tfrac{1}{2} \log (1-u)\big ) \\&= \exp \big ((d^{1/2}-1) u/2\big ) \cdot (1-u)^{-1/2}. \end{aligned}$$

Expanding both of the latter functions in infinite series, we obtain

$$\begin{aligned} G(u)&= \bigg (\sum _{l=0}^\infty \frac{\big ((d^{1/2}-1)/2\big )^l}{l!} u^l\bigg ) \bigg (\sum _{m=0}^\infty \frac{(1/2)_m}{m!} u^m\bigg ) \\&= \sum _{k=0}^\infty u^k \sum _{l=0}^k \frac{\big ((d^{1/2}-1)/2\big )^l}{l!} \frac{(1/2)_{k-l}}{(k-l)!}. \end{aligned}$$

By comparing the coefficients of $u^k$ in $G_t(u)$ and G(u), we obtain the equality in (A.31).

Since $\big ((d^{1/2}-1)/2\big )^l \le \big ((d^{1/2}-1)/2\big )_l$ then, by applying the well-known convolution identity,

$$\begin{aligned} \sum _{l=0}^k \frac{(v_1)_l}{l!} \frac{(v_2)_{k-l}}{(k-l)!} = \frac{(v_1 + v_2)_k}{k!}, \end{aligned}$$

$v_1, v_2 \in {\mathbb {R}}$, we obtain

$$\begin{aligned} a_k \le \sum _{l=0}^k \frac{\big ((d^{1/2}-1)/2\big )_l}{l!} \frac{(1/2)_{k-l}}{(k-l)!} = \frac{(d^{1/2}/2)_k}{k!}. \end{aligned}$$

The proof of (A.31) now is complete. $\square$

Next, we establish (A.12).

Lemma A.2

Let $\gamma _1 = \tfrac{1}{2}(\sqrt{3}+1)$. Then, for all $k \ge 1$ and $d \ge 1$,

$$\begin{aligned} \frac{(d^{1/2}/2)_k}{(d/2)_k} \le \gamma _1^{k-1} ((k-1)!)^{1/2} d^{-k/2}. \end{aligned}$$

(A.32)

Proof

Denote by $r_k$ the left-hand side of (A.32). Then $r_1 = d^{-1/2}$ and, for $j \ge 1$,

$$\begin{aligned} \frac{r_{j+1}}{r_j} = \frac{(d^{1/2}/2)_{j+1}}{(d/2)_{j+1}} \cdot \frac{(d/2)_j}{(d^{1/2}/2)_j} = \frac{d^{1/2}+2j}{d+2j}. \end{aligned}$$

We claim that

$$\begin{aligned} \frac{d^{1/2}+2j}{d+2j} \le \gamma _1 j^{1/2} d^{-1/2}, \end{aligned}$$

(A.33)

$j \ge 1$, and we prove this as follows. Define $y \equiv y(t) = (t^2+2jt)/(t^2+2j)$, $t \ge 1$. It is simple to verify that y(t) attains its maximum at $t_j:= 1 + (2j+1)^{1/2}$ and that

$$\begin{aligned} y(t_j) = \frac{1}{2} \big (1 + (2j+1)^{1/2}\big ). \end{aligned}$$

For $j \ge 1$, it is also straightforward to show that

$$\begin{aligned} \frac{1}{2} \big (1 + (2j+1)^{1/2}\big ) \le \gamma _1 j^{1/2}. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{d+2jd^{1/2}}{d+2j} = y(d^{1/2}) \le y(t_j) \le \gamma _1 j^{1/2}. \end{aligned}$$

Multiplying the left- and right-hand sides of the above inequality by $d^{-1/2}$, we obtain (A.33). $\square$