Spectral moments and integrable tails for wide Gaussian pseudoinverses

If $h=0$, define $v$ arbitrarily; this event is null under the density [F9]. If $k=1$, then $\mathrm{\mid }{v}_1\mathrm{\mid }=1$ and both displayed sphere-moment formulas equal $1$. Suppose $k\ge 2$. For a real standard coordinate $x$, [F13] on the two half-lines, with inverses $x=\pm \sqrt{u}$, gives square density $u^{-1\mathrm{/}2}{e}^{-u\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ on $u>0$, the Gamma$(1\mathrm{/}2,2)$ density by [F6]. For independent Gamma variables of shapes $a,b>0$ and common scale $\theta$, [F8] and [F13] give their sum density $e^{-s\mathrm{/}\theta }{\int }_0^s{u}^{a-1}(s-u{)}^{b-1}du\mathrm{/}({\theta }^{a+b}\mathrm{\Gamma }(a)\mathrm{\Gamma }(b))$. Substituting $u=st$ and using [F3] makes it Gamma$(a+b,\theta )$. A normalized complex coordinate has squared modulus $(x^2+y^2)\mathrm{/}2$ for independent standard real $x,y$, hence Gamma$(1,1)$ by this convolution and scaling. Iterating the convolution, the independent variables $\mathrm{\mid }{h}_1{\mathrm{\mid }}^2$ and ${\sum }_{j=2}^k\mathrm{\mid }{h}_j{\mathrm{\mid }}^2$ have Gamma shapes $(1\mathrm{/}2,(k-1)\mathrm{/}2)$ and common scale $2$ over $\mathbb{R}$, and shapes $(1,k-1)$ and common scale $1$ over $\mathbb{C}$. Apply [F13] on the open positive quadrant using the inverse diffeomorphism $(s,t)\mapsto (x,y)=(st,s(1-t))$, with $s>0$, $0<t<1$ and absolute Jacobian $s$. In these coordinates the density factors into a Gamma$(a+b,\theta )$ factor in $s$ and a Beta$(a,b)$ factor in $t$ after scaling by the common scale $\theta$. Thus $\mathbb{E}{t}^q=B(a+q,b)\mathrm{/}B(a,b)$; [F3] gives $c_R(q)=\mathrm{\Gamma }(k\mathrm{/}2)\mathrm{\Gamma }(q+1\mathrm{/}2)\mathrm{/}(\sqrt{\pi }\mathrm{\Gamma }(k\mathrm{/}2+q))$ over $\mathbb{R}$ and $c_C(q)=\mathrm{\Gamma }(k)\mathrm{\Gamma }(q+1)\mathrm{/}\mathrm{\Gamma }(k+q)$ over $\mathbb{C}$. Invariance makes the same moment hold for $\mathrm{\mid }⟨v,u⟩\mathrm{\mid }$ for every fixed unit $u$.

For $a,s>0$, $\mathrm{\Gamma }(a+s)\mathrm{/}\mathrm{\Gamma }(a)\le (a+s{)}^s$. If $U\sim \mathrm{\Gamma }(x,1)$, direct integration and [F10] give $\mathbb{E}U=\mathrm{\Gamma }(x+1)\mathrm{/}\mathrm{\Gamma }(x)=x$. The density makes $\log U$ integrable, since $u^{x-1}\mathrm{\mid }\log u\mathrm{\mid }$ is integrable near zero and the exponential factor controls infinity. Thus [F4] gives $\psi (x):={\mathrm{\Gamma }}^{\mathrm{\prime }}(x)\mathrm{/}\mathrm{\Gamma }(x)=\mathbb{E}\log U$. Jensen [F5] for the convex function $-\log$ yields $\mathbb{E}\log U\le \log \mathbb{E}U=\log x$. Integrating $\psi$ from $a$ to $a+s$ yields $\log \mathrm{\Gamma }(a+s)-\log \mathrm{\Gamma }(a)\le s\log (a+s)$ and hence the ratio bound.

For an integer $p\ge 1$, $\mathrm{\Gamma }((p+1)\mathrm{/}2)\ge \sqrt{\pi }(p\mathrm{/}(2e){)}^{p\mathrm{/}2}$. If $p=2r$, use [F10] and [F6] to write $\mathrm{\Gamma }(r+1\mathrm{/}2)=\sqrt{\pi }{\prod }_{j=1}^r(j-1\mathrm{/}2)$; [F7] gives $(\log x{)}^{\mathrm{\prime }\mathrm{\prime }}=-1\mathrm{/}{x}^2<0$. For $j\ge 2$, midpoint concavity yields $\log (j-1\mathrm{/}2)\ge {\int }_{j-1}^j\log xdx$; for $j=1$, $\log (1\mathrm{/}2)\ge {\int }_0^1\log xdx=-1$. The product is at least $\sqrt{\pi }(r\mathrm{/}e{)}^r$. If $p=2r+1$, [F10] gives $\mathrm{\Gamma }((p+1)\mathrm{/}2)=r!$ and midpoint concavity gives $\log (r!)\ge {\int }_{1\mathrm{/}2}^{r+1\mathrm{/}2}\log xdx$; hence $r!\ge \sqrt{2e}((r+1\mathrm{/}2)\mathrm{/}e{)}^{r+1\mathrm{/}2}\ge \sqrt{\pi }((r+1\mathrm{/}2)\mathrm{/}e{)}^{r+1\mathrm{/}2}$, since $e>2$ from its series and $\pi <4$ from the unit disk lying in a square of area $4$. Also $p!\ge (p\mathrm{/}e{)}^p$ follows from ${\sum }_{j=1}^p\log j\ge {\int }_1^p\log xdx$.

Put $M=(G{G}^{\ast }{)}^{-1}$. By the pseudoinverse identity in [F1], $X^2={\lambda }_{\max }(M)$ by [F11]. For each fixed positive $M$, choose a unit eigenvector $u$ for its largest eigenvalue. Then $v^{\ast }Mv\ge {\lambda }_{\max }(M)\mathrm{\mid }⟨v,u⟩{\mathrm{\mid }}^2$, so integrating over $v$ and using the sphere moment $c_F(q)$ from step 1.1 gives ${\mathbb{E}}_v(v^{\ast }Mv{)}^q\ge {\lambda }_{\max }(M{)}^q{c}_F(q)$. Tonelli [F8] interchanges the nonnegative integrals over $G$ and $v$. For each fixed $v$, invariance [F2] makes $v^{\ast }Mv$ distributed as $M_{11}$. Using the reciprocal Gamma law [F1] and dividing by $c_F(q)$ gives exactly $C_R(q)$ or $C_C(q)$ in the statement.

Use the real moment bound from step 2.1 with $q=p\mathrm{/}2$, the Gamma-ratio estimate in step 1.2, and the lower bound in step 1.3 to obtain $\mathbb{E}{X}^p\le (e\sqrt{k+p}\mathrm{/}p{)}^p$. For the complex case, use the $C_C$ moment bound from step 2.1 with $q=p$, the ratio estimate from step 1.2, and the factorial bound in step 1.3 twice to obtain $\mathbb{E}{X}^{2p}\le (e\sqrt{k+p}\mathrm{/}p{)}^{2p}$. Jensen for the convex powers gives $\mathbb{E}X\le (\mathbb{E}{X}^p{)}^{1\mathrm{/}p}$ in the real case and $\mathbb{E}X\le (\mathbb{E}{X}^{2p}{)}^{1\mathrm{/}(2p)}$ in the complex case. For either exponent $r=p$ or $r=2p$, the pointwise inequality $t^r{\mathbf{1}}_{\{X>t\}}\le X^r$ implies $\mathbb{P}\{X>t\}\le \mathbb{E}{X}^r\mathrm{/}{t}^r$; using the moment bounds and also $\mathbb{P}\{X>t\}\le 1$ gives the displayed tails. Since $r>1$ for $p\ge 2$, $t\mapsto (a\mathrm{/}t{)}^r$ is integrable on $[1,\mathrm{\infty })$, so these tails are integrable at infinity.