Asymptotic Linear Spectral Statistics for Spiked Hermitian Random Matrices

Abstract

Using the Coulomb Fluid method, this paper derives central limit theorems (CLTs) for linear spectral statistics of three “spiked” Hermitian random matrix ensembles. These include Johnstone’s spiked model (i.e., central Wishart with spiked correlation), non-central Wishart with rank-one non-centrality, and a related class of non-central $F$ matrices. For a generic linear statistic, we derive simple and explicit CLT expressions as the matrix dimensions grow large. For all three ensembles under consideration, we find that the primary effect of the spike is to introduce an $O(1)$ correction term to the asymptotic mean of the linear spectral statistic, which we characterize with simple formulas. The utility of our proposed framework is demonstrated through application to three different linear statistics problems: the classical likelihood ratio test for a population covariance, the capacity analysis of multi-antenna wireless communication systems with a line-of-sight transmission path, and a classical multiple sample significance testing problem.

Appendices

Appendix 1: Statement and Proof of Lemma 2

Lemma 2

Let $u>0$ and $v>0$ such that $nu+1>0$ and $n(v-u)\notin \mathbb {N}$. Assume that $\mathfrak {R}z>1$ and $\gamma \ge 0$. As $n\rightarrow \infty$, we have

$$\begin{aligned}&\,_1F_1(nu+1,nv+1,n\gamma z) \nonumber \\& \approx \frac{1}{\sqrt{2\pi n}}\frac{\Gamma (n(u-v))\Gamma (nv+1)}{\Gamma (nu+1)}\frac{\mathrm {e}^{n\gamma z t(z)}t(z)^{nu+1}(t(z)-1)^{n(u-v)}}{\sqrt{-v(t(z)-1)^2+(u-v)(2t(z)-1)}}, \end{aligned}$$

(69)

where $t(z)= \frac{\gamma z -v+\sqrt{(v-\gamma z)^2+4\gamma z u}}{2\gamma z}$.

Proof

Under the prescribed conditions on $u$ and $v$, we may use the following integral representation [94]

$$\begin{aligned} \,_1F_1(nu+1,nv+1,n\gamma z)=\frac{\Gamma (n(u-v))\Gamma (nv+1)}{2\imath \pi \Gamma (nu+1)}\oint _C\frac{t^{nu}e^{n\gamma z t}}{(t-1)^{n(u-v)+1}}\,\mathrm {d}t, \end{aligned}$$

(70)

where the contour $C$ starts at 0, traverses anti-clockwise around 1 and returns to 0. For large $n$, the Laplace approximation yields

$$\begin{aligned} \oint _C\frac{t^{nu}e^{n\gamma z t}}{(t-1)^{n(u-v)+1}}\,\mathrm {d}t\approx \sqrt{\frac{2\pi }{n}}\frac{q(t_0)\mathrm {e}^{-np(t_0)}}{\sqrt{p''(t_0)}}, \end{aligned}$$

(71)

for which

$$\begin{aligned} p(t)&=-\left( \gamma z t + u\text {Log}(t)+(v-u)\text {Log}(t-1)\right) \nonumber \\ q(t)&=(t-1)^{-1} \end{aligned}$$

(72)

and where $t_0$ is the saddlepoint, which is the solution to

$$\begin{aligned} p'(t_0) = 0 . \end{aligned}$$

Thus, $t_0$ must satisfy

$$\begin{aligned} \gamma z t_0^2+t_0(v-\gamma z)-u=0, \end{aligned}$$

with $t_0 \notin \{0,1$}. There is one solution which lies outside the contour for $\mathfrak {R}z > 1$:

$$\begin{aligned} t_0(z)=\frac{\gamma z-v+\sqrt{(v-\gamma z)^2+4u\gamma z}}{2\gamma z}. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} p''(t_0(z))=-\frac{-v(t_0(z)-1)^2+(u-v)(2t_0(z)-1)}{t_0(z)^2(t_0(z)-1)^2} \end{aligned}$$

so that, taking the root with the correct phase factor (see [89, Chap. 4] or [90, Chap. 7] for more details), we get

$$\begin{aligned} \frac{1}{\sqrt{p''(t_0(z))}}=\imath \frac{t_0(z)(t_0(z)-1)}{\sqrt{-v(t_0(z)-1)^2+(u-v)(2t_0(z)-1)}} . \end{aligned}$$

Substituting this quantity with (72) into (71) together with (70), we find the desired result (69). $\square$

Appendix 2: Equivalence Between $\int _a^b f(x) \rho _2 (x, z)\mathrm {d}x$ and $\int _a^b \text {Log}(z-x) \rho _1(x) \mathrm {d}x$ in (39)

We want to show that, for $z \notin [a,b]$,

$$\begin{aligned} \int _a^b f(x) \rho _2 (x, z)\mathrm {d}x = \int _a^b \text {Log}(z-x) \rho _1(x) \mathrm {d}x, \end{aligned}$$

where

$$\begin{aligned} \int _a^b f(x) \rho _2 (x, z)\mathrm {d}x&=\frac{1}{2\pi } \int _a^b \frac{f(x)}{\sqrt{(b-x)(x-a)}}\left( \frac{\sqrt{(z-a)(z-b)}}{z-x}-1 \right) \, \mathrm {d}x \end{aligned}$$

(73)

and

$$\begin{aligned} \int _a^b \text {Log}(z-x) \rho _1(x)\, \mathrm {d}x=\frac{1}{2\pi ^2} \int _a^b \frac{\text {Log}(z-x)}{\sqrt{(b-x)(x-a)}} \left[ \mathcal {P} \int _a^b \frac{f'(y)\sqrt{(b-y)(y-a)}}{y-x} \,\mathrm {d}y \right] \, \mathrm {d}x . \end{aligned}$$

(74)

This identity is not straightforward and appears difficult to show directly using the above expressions. Thus, here we adopt an approach based on first showing that the derivative with respect to $z$ of (73) and (74) are equal. First considering (73), we have

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}z}\left[ \int _a^b f(x) \rho _2 (x, z)\,\mathrm {d}x\right]&=\frac{1}{2\pi \sqrt{(z-a)(z-b)}} \int _a^b \frac{f(x)\left( (a+b)(x+z)-2ab \right) }{2(z-x)^2\sqrt{(b-x)(x-a)}} \, \mathrm {d}x. \end{aligned}$$

(75)

Now taking the derivative of (74), we get

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}z}\left[ \int _a^b \text {Log}(z-x) \rho _1(x)\,\mathrm {d}x \right] \\& = \int _a^b \frac{\sqrt{(b-y)(y-a)}}{2\pi ^2}f'(y) \left[ \mathcal {P} \int _a^b \frac{\mathrm {d}x}{(x-y)(z-x)\sqrt{(b-x)(x-a)}} \right] \, \mathrm {d}y \\& =\int _a^b \frac{\sqrt{(b-y)(y-a)}}{2\pi ^2(z-y)}f'(y) \left[ \mathcal {P} \int _a^b \frac{(z-x)^{-1}+(x-y)^{-1}}{\sqrt{(b-x)(x-a)}} \mathrm {d}x\right] \, \mathrm {d}y, \end{aligned}$$

which, after applying the identities (86) and (89), yields

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}z}\left[ \int _a^b \text {Log}(z-x) \rho _1(x)\,\mathrm {d}x \right]&=\frac{1}{2\pi \sqrt{(z-a)(z-b)}}\int _a^b \frac{\sqrt{(b-y)(y-a)}}{y-z}f'(y)\,\mathrm {d}y. \end{aligned}$$

Application of integration by parts to the last integral gives

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}z}\left[ \int _a^b \text {Log}(z-x) \rho _1(x)\, \mathrm {d}x \right]&=\frac{1}{2\pi \sqrt{(z-a)(z-b)}}\ \int _a^b \frac{f(y)\left( (a+b)(y+z)-2ab \right) }{2(z-y)^2\sqrt{(b-y)(y-a)}} \, \mathrm {d}y, \end{aligned}$$

which is the same as (75). So we have proved that

$$\begin{aligned} \int _a^b \text {Log}(z-x) \rho _1(x)\, \mathrm {d}x = \int _a^b f(x) \rho _2 (x, z)\mathrm {d}x + \text {Constant} . \end{aligned}$$

(76)

Now, note that with $z$ such that $\mathfrak {I}z=0$, as $\mathfrak {R}z \rightarrow \infty$,

$$\begin{aligned} \frac{\sqrt{(z-a)(z-b)}}{z-x}-1 = \frac{x-\frac{a+b}{2}}{z}+ O\left( \frac{1}{z^2} \right) . \end{aligned}$$

Plugging this expression into (73) gives

$$\begin{aligned} \int _a^b f(x) \rho _2 (x, z)\mathrm {d}x = \frac{1}{z} \int _a^b \frac{f(x)}{\sqrt{(b-x)(x-a)}}\left( x-\frac{a+b}{2} \right) \, \mathrm {d}x + O\left( \frac{1}{z^2} \right) \end{aligned}$$

which tends to zero as $\mathfrak {R}z \rightarrow \infty$. Furthermore, with

$$\begin{aligned} \text {Log}(z-x) = \text {Log}(z)-\sum _{k=1}^\infty \frac{x^k}{z^k} , \end{aligned}$$

the expression in (74) becomes, upon interchanging the integrals,

$$\begin{aligned} \int _a^b \text {Log}(z-x) \rho _1(x)\, \mathrm {d}x&=\frac{1}{2\pi ^2} \int _a^b f'(y)\sqrt{(b-y)(y-a)} \left[ \mathcal {P} \int _a^b \frac{\text {Log}(z)\,\mathrm {d}x}{(y-x)\sqrt{(b-x)(x-a)}}\right. \\& -\left. \sum _{k=1}^\infty \frac{\mathcal {P}}{z^k}\int _a^b\frac{x^k\,\mathrm {d}x}{(y-x)\sqrt{(b-x)(x-a)}} \right] \, \mathrm {d}y . \end{aligned}$$

Here, the first principal value integral is zero by (86), whilst the remaining terms tends to zero when $z$ is such that $\mathfrak {R}z \rightarrow \infty$ and $\mathfrak {I}z=0$.

Consequently, taking $z$ such that $\mathfrak {R}z \rightarrow \infty$ and $\mathfrak {I}z=0$ in (76), we find that the constant term is zero, thus proving the result.

Appendix 3: Useful Formulas

For the derivations of our results, we will require numerous integrals; these are summarized in (77–93). Note that for all definite integrals involving the variable $t$, these are valid for $\mathfrak {R}t >b$, while in all cases we assume $0<a<b$.

$$\begin{aligned}&\int _a^b \frac{\text {Log}(x+t)}{\sqrt{(b-x)(x-a)}}\, \mathrm {d}x=2\pi \text {Log}\left( \frac{\sqrt{t+a}+\sqrt{t+b}}{2} \right) \end{aligned}$$

(77)

$$\begin{aligned}&\int _a^b \frac{\mathrm {d}x}{(x+t)\sqrt{(b-x)(x-a)}} =\frac{\pi }{\sqrt{(t+a)(t+b)}} \end{aligned}$$

(78)

$$\begin{aligned}&\int _a^b \frac{\mathrm {d}x}{\sqrt{(b-x)(x-a)}}=\pi \end{aligned}$$

(79)

$$\begin{aligned}&\int _a^b \frac{x\,\mathrm {d}x}{\sqrt{(b-x)(x-a)}}=\pi \frac{a+b}{2} \end{aligned}$$

(80)

$$\begin{aligned}&\int _a^b \frac{\text {Log}(x+t)}{\sqrt{(b-x)(x-a)}x}\,\mathrm {d}x=\frac{\pi }{\sqrt{ab}}\text {Log}\left( \frac{(\sqrt{ab}+\sqrt{(t+a)(t+b)})^2-t^2}{(\sqrt{a}+\sqrt{b})^2}\right) \end{aligned}$$

(81)

$$\begin{aligned}&\int _a^b \frac{x\text {Log}(x+t)}{\sqrt{(b-x)(x-a)}}\,\mathrm {d}x=\pi \frac{(\sqrt{a+t}-\sqrt{b+t})^2}{2}\nonumber \\&\qquad \quad +\pi \frac{a+b}{2}\text {Log}\left( \frac{2(t+(a+t)(b+t))+a+b}{4}\right) \end{aligned}$$

(82)

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)}{\sqrt{(b-x)(x-a)}}\, \mathrm {d}x=2\pi \text {Log}\left( \frac{\sqrt{1-a}+\sqrt{1-b}}{2} \right) \end{aligned}$$

(83)

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)}{\sqrt{(b-x)(x-a)}x}\,\mathrm {d}x=\frac{\pi }{\sqrt{ab}}\text {Log}\left( \frac{1-(\sqrt{ab}-\sqrt{(1-a)(1-b)})^2}{(\sqrt{a}+\sqrt{b})^2}\right) \end{aligned}$$

(84)

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)}{\sqrt{(b-x)(x-a)}(x-1)}\,\mathrm {d}x=\frac{2\pi }{\sqrt{(1-a)(1-b)}}\text {Log}\left( \frac{1}{2\sqrt{1-a}}+\frac{1}{2\sqrt{1-b}}\right) \end{aligned}$$

(85)

$$\begin{aligned}&\mathcal {P}\int _a^b \frac{1}{(y-x)\sqrt{(b-x)(x-a)}}\,\mathrm {d}x=0\end{aligned}$$

(86)

$$\begin{aligned}&\mathcal {P} \int _a^b \frac{\sqrt{(b-x)(x-a)}}{x(y-x)} \, \mathrm {d}x =\pi \left( 1-\frac{\sqrt{ab}}{y} \right) \end{aligned}$$

(87)

$$\begin{aligned}&\mathcal {P} \int _a^b \frac{\sqrt{(b-y)(y-a)}}{(1-y)(x-y)} \,\mathrm {d}y = \pi \left( \frac{\sqrt{(1-a)(1-b)}}{1-x}-1\right) \end{aligned}$$

(88)

$$\begin{aligned}&\int _a^b \frac{1}{(x-t)\sqrt{(b-x)(x-a)}}\,\mathrm {d}x= - \frac{\pi }{\sqrt{(t-a)(t-b)}} \end{aligned}$$

(89)

$$\begin{aligned}&\int _a^b \frac{1}{\sqrt{(b-x)(x-a)}}\left( \frac{\sqrt{(z-a)(z-b)}}{t-x}-1 \right) \, \mathrm {d}x=0\end{aligned}$$

(90)

$$\begin{aligned}&\mathcal {P} \int _a^b \frac{\sqrt{(b-x)(x-a)}}{y-x} \,\mathrm {d}x = \pi \left( y-\frac{a+b}{2} \right) \end{aligned}$$

(91)

Moreover, for $z \in \mathbb {C}$ and $b<1$,

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)}{(x-z)\sqrt{(b-x)(x-a)}} \,\mathrm {d}x\nonumber \\&\qquad \quad =\frac{\pi }{A} \text {Log}\left( \frac{2A+2z-a-b}{2A\sqrt{(1-a)(1-b)}+z(2-a-b)-a-b+2ab}\right) \end{aligned}$$

(92)

and for $z \in \mathbb {C}$,

$$\begin{aligned}&\int _a^b \frac{\text {Log}(x+t) \, \mathrm {d}x}{(x-z)\sqrt{(b-x)(x-a)}}\nonumber \\& =\frac{\pi }{A}\text {Log}\left( \frac{2A+(a+b-2z)(t+z)+2A\sqrt{A^2+(a+b-2z)(t+z)+(t+z)^2}}{(t+z)^2(a+b-2z+2A)}\right) \end{aligned}$$

(93)

with $A=\sqrt{(z-a)(z-b)}$.

Equations (77–88) are given in [18], whilst (89) is a slight modification of (78), and (90) follows using (78) and (79). The expression (91) follows upon multiplying the numerator and the denominator of the integrand by $\sqrt{(b-x)(x-a)}$, applying a partial fraction decomposition, then invoking (79), (80) and (86). For (92) and (93), the derivations are more involved, and we describe each in turn.

For (92), we use the parametrization

$$\begin{aligned} \text {Log}(1-x) = \int _0^1 \frac{x}{\lambda x-1} \,\mathrm {d}\lambda \end{aligned}$$

and the partial fraction decomposition

$$\begin{aligned} \frac{x}{(x-1/\lambda )(x-z)}=\frac{1/\lambda }{(x-1/\lambda )(1/\lambda -z)}-\frac{z}{(x-z)(1/\lambda -z)} \end{aligned}$$

to arrive at

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)\,\mathrm {d}x}{(x-z)\sqrt{(b-x)(x-a)}} =\int _0^1 \frac{\mathrm {d}\lambda }{1-\lambda z} \int _a^b \left( \frac{1/\lambda }{x-1/\lambda } + \frac{z}{z-x} \right) \frac{\mathrm {d}x}{\sqrt{(b-x)(x-a)}}\\&\quad =\, \pi \int _0^1 \left( \frac{-1}{\sqrt{(1-\lambda a)(1-\lambda b)}} + \frac{z}{\sqrt{(z-a)(z-b)}}\right) \frac{\mathrm {d}\lambda }{ 1-\lambda z}. \end{aligned}$$

The last equation was obtained by invoking (78) and (89). From a further change of variable $x=1- \lambda z$, we have

$$\begin{aligned}&\int _a^b \frac{\text {Log}(1-x)}{(x-z)\sqrt{(b-x)(x-a)}} \,\mathrm {d}x = \pi \int _{1-z}^1 \left( \frac{-1}{\sqrt{(z-a+ax)(z-b+bx)}} + \frac{1}{A}\right) \frac{\mathrm {d}x}{x}\\&\quad =\frac{\pi }{A} \Bigg ( -\text {Log}(1-z) \\&\qquad + \left. \left[ \text {Log}\left( \frac{2A\sqrt{(z-a+ax)(z-b+bx)}+2A^2+x(z(a+b)-2ab)}{x}\right) \right] _{1-z}^1\right) \\&\quad =\frac{\pi }{A} \text {Log}\left( \frac{2A+2z-a-b}{2A\sqrt{(1-a)(1-b)}+z(2-a-b)-a-b+2ab}\right) . \end{aligned}$$

Now consider (93). In this case, we use the relation

$$\begin{aligned} \text {Log}(x+t) = \text {Log}t + \int _0^1 \frac{x}{t+ yx} \mathrm {d}y \end{aligned}$$

to give

$$\begin{aligned}&\int _a^b \frac{\text {Log}(x+t) \, \mathrm {d}x}{(x-z)\sqrt{(b-x)(x-a)}}\\& =\int _a^b \frac{\text {Log}t \, \mathrm {d}x}{(x-z)\sqrt{(b-x)(x-a)}}+\int _0^1 \left[ \int _a^b \frac{1}{\sqrt{(b-x)(x-a)}}\frac{x}{t+yx}\frac{\mathrm {d}x}{x-z}\right] \mathrm {d}y{.} \end{aligned}$$

The first integral is given by (89), whereas the double integral is

$$\begin{aligned}&\int _0^1 \left[ \int _a^b \frac{1}{\sqrt{(b-x)(x-a)}} \frac{x}{x+\frac{t}{y}}\frac{\mathrm {d}x}{x-z} \right] \frac{\mathrm {d}y}{y} \\& = \int _0^1\left[ \int _a^b \frac{1}{\sqrt{(b-x)(x-a)}} \left( \frac{z}{x-z}+ \frac{\frac{t}{y}}{x+\frac{t}{y}}\right) \mathrm {d}x \right] \frac{\mathrm {d}y}{y \left( \frac{t}{y}+z\right) }\\& = \int _0^1 \frac{z}{y\left( \frac{t}{y}+z\right) }\left[ \int _a^b \frac{ \, \mathrm {d}x}{(x-z)\sqrt{(b-x)(x-a)}}\right] \mathrm {d}y \\&\quad +\int _0^1 \frac{\frac{t}{y}}{y \left( \frac{t}{y}+z\right) } \left[ \int _a^b \frac{1}{\sqrt{(b-x)(x-a)}} \frac{\mathrm {d}x}{x+\frac{t}{y}} \right] \mathrm {d}y . \end{aligned}$$

The first term on the right-hand side above is obtained using (89), whereas the second is

$$\begin{aligned}&\int _0^1 \frac{1}{y}\left[ \frac{\frac{t}{y}}{\frac{t}{y}+z} \int _a^b \frac{1}{\sqrt{(b-x)(x-a)}} \frac{\mathrm {d}x}{x+\frac{t}{y}}\right] \mathrm {d}y\\& = \pi \int _0^1 \frac{1}{y}\frac{\frac{t}{y}}{\frac{t}{y}+z} \frac{ \mathrm {d}y}{\sqrt{\left( b+\frac{t}{y}\right) \left( a+\frac{t}{y}\right) }} \quad \quad \text {[using (78)]}\\& = \pi \int _{t+z}^{\infty } \frac{\, \mathrm {d}x}{x\sqrt{(x-z+a)(x-z+b)}} \quad \quad [\text {setting} x= t/y +z]\\& = \pi \int _{t+z}^{\infty } \frac{\, \mathrm {d}x}{x\sqrt{-(b-z)(z-a)+(b+a-2z)x+x^2}}\\& = \frac{\pi }{A} \text {Log}\left( \frac{2A+(a+b-2z)(t+z)+2A\sqrt{A^2+(a+b-2z)(t+z)+(t+z)^2}}{(t+z)(a+b-2z+2A)}\right) , \end{aligned}$$

using the first identity in [61, Eq. 2.266]. Combining the previous calculations, we get the result.