0
$\begingroup$

I'm trying to check a bound on Stieltjes transform of a probability measure, that's given in equation (2.92) on P. 170 in Terence Tao's notes "Topics in Random Matrix Theory". Denote the Stieltjes transform of the probability measure $\mu$ by $s_{\mu}(z)$. Then the bound mentioned in his notes is:

$zs_{\mu}(z)= 1 + o_{\mu}(z)$ as $z= x+iy \to \infty$ so that $|\frac{x}{y}|$ is bounded. N.B. here "$o_{\mu}(z)$" is a notation used to denote $o(z)$ but with highlighting the fact that $z=x+iy\to \infty$ with $|x/y|$ bounded, and that the convergence rate depends on $\mu$.

But all I'm getting, at least under a special case, is: under the same condition of convergence, mentioned just now, $zs_{\mu}(z)= -1 + o_{\mu}(z)$, which I demonstrate below.

For the special case that I'll treat, just assume that: $\frac{x}{y}=K$. But if you follow my computation below, you'll see that the end result wouldn't change in the limit if you assume $|\frac{x}{y}|\leq K$.

$$zs_{\mu}(z) = \int_{\mathbb{R}} \frac{z}{t-z}d\mu(t)=\int_{\mathbb{R}} \frac{z(t-\bar{z})}{|t-z|^2}d\mu(t)= \int_{\mathbb{R}}\frac{(tx - x^2 - y^2)+i(ty)} {(t-x)^2 + y^2 }d\mu(t)= \int_{\mathbb{R}}\frac{(\frac{x}{y}.\frac{t}{y}- (\frac{x}{y})^2-1) + i(\frac{t}{y})}{(\frac{x}{y})^2+1}d\mu(t)= -1 + \int_{\mathbb{R}}\frac{K+1}{K^2 + 1}\frac{t}{y}d\mu(t)= -1 + \frac{K+1}{K^2 + 1}.\frac{\mathbb{E}[\mathbb{I}]}{y}$$, where $\mathbb{E}[\mathbb{I}]$ is really the expectation of the identity function $\mathbb{I}(t):=t$ w.r.t. $\mu$. Assume it exists for now!

Note that, above, since $z=x+iy \to \infty$ but $|x/y|$ is bounded (actually I assumed that $|x/y|$ is constant to make things bit easy), we must have $y \to \infty$, yielding:

$zs_{\mu}(z)= -1 + o_{\mu}(z)$, disproving $zs_{\mu}(z)= 1 + o_{\mu}(z)$. Did I do something wrong in my calculation?

Also note that: if you take: $\mu$ to be the Dirac measure at $0$, i.e. $\mu = \delta_0$, then $s_{\mu}(z)= -1/z$, which does satisfy: $zs_{\mu}(z)= -1 + o_{\mu}(z)$, but not $zs_{\mu}(z)= 1 + o_{\mu}(z)$.

Thanks for taking a look!

$\endgroup$
  • $\begingroup$ Tao defines $X = o(Y)$ by the requirement $|X|≤c(n)Y$. Note the modulus on the left side. I think the equation $s_\mu = \frac{1 + o_\mu (z)}{z}$ on P. 170 should accordingly be interpreted as $\vert z s_\mu (z) - 1 \vert = o_\mu (z)$. $\endgroup$ – Bruno Krams Dec 5 '19 at 18:47
  • $\begingroup$ @BrunoKrams Thanks, but Tao writes on P.170: "...where $o_{\mu}(z)$ is an expression that, for any fixed $\mu$, goes to zero as $z\to\infty$ non-tangentially in the sense that $|Re(z)/Im(z)|$ is kept bounded, where the rate of convergence is allowed to depend on $\mu$". Did I miss something? Where did you find that definition of "o" in Tao's book? Also, does your comment contradict $zs_{\mu}(z) + 1 =o_{\mu}(z) ?$ Following Tao's notation of $o_{\mu}(z) $, I did interpret it exactly as you wrote: $|zs_{\mu}(z) - 1| =o_{\mu}(z) $. Still, the counterexamples I gave above are valid. $\endgroup$ – Mathmath Dec 5 '19 at 18:54
  • $\begingroup$ The definition can be found on page 6. If the o-Notation is interpreted as I wrote, than what you gave is not a counterexample but in perfect agreement with the equation (2.92) from Tao's book. Also your calculation is - with slight modifications - a proof of (2.92) :) $\endgroup$ – Bruno Krams Dec 5 '19 at 19:06
  • $\begingroup$ By the way on page 169 Tao himself gives the series expansion $$s_n(z) = - \frac{1}{z} - \frac{1}{z^2} \frac{1}{n} \operatorname{tr} M_n - ...$$ so it would be a very surprising mistake if he missed a sign in (2.92) on the next page. $\endgroup$ – Bruno Krams Dec 5 '19 at 19:12
  • $\begingroup$ Sorry, I just looked at Page 6. He defined there something entirely different: he's talking about the dimension $n$ of random matrices there, which, in my question itself, is not relevant at all. My question concerns not random matrix, so you can ignore that definition of $o$, and consider that of $o_{\mu}$ that he writes on P.170, specially related to my questions. Also, could you please be so kind to point out why my calculation and counterexamples aligns with $zs_{\mu}(z)= 1 + o_{\mu}(z)$? In fact, I totally agree with what he wrote on P.169, because it validates my calculation. $\endgroup$ – Mathmath Dec 5 '19 at 19:19
0
$\begingroup$

I believe there is a minor typo in the definition of the Stieltjes transform on page 169 of those notes, namely instead of $$ s_\mu(z):=\int_{\mathbb R}\frac{1}{x-z}d\mu(x),\qquad \Im z>0 $$ the intended definition is $$ s_\mu(z):=\int_{\mathbb R}\frac{1}{z-x}d\mu(x),\qquad \Im z>0. $$ After fixing the sign mistake caused by swapping $x$ and $z$, this matches the usual definition and also allows a straightforward justification of $(2.92)$ on page 170 as follows. Let $z_n$ be any sequence of complex numbers with $\Im(z_n)>0$ such that $z_n\to\infty$ non-tangentially, and let $f_n(x)=(z_n-x)^{-1}$. Then for all $x\in\mathbb R$ $$ \lim_{n\to\infty}z_nf_n(x)=1, $$ thus by the dominated convergence theorem it follows that $$ \lim_{n\to\infty}z_ns_{\mu}(z_n)=\lim_{n\to\infty}\int_{\mathbb R}z_nf_n(x)d\mu(x)=\int_{\mathbb R}\lim_{n\to \infty}z_nf_n(x)d\mu(x)=\int_{\mathbb R}d\mu(x)=1, $$ which in Tao's notation is equivalent to stating that $zs_{\mu}(z)=1+o_{\mu}(1)$, as claimed.

Note that the applicability of the dominated convergence theorem is justified in the exact same manner as in the observation $(2.91)$ on page 170, since the integrand satisfies $$ |z_nf_n(x)|\leq \frac{|z_n|}{|\Im(z_n)|}\leq 1. $$ In other words, we are actually using the special case of the dominated convergence theorem known as the bounded convergence theorem.

$\endgroup$
  • $\begingroup$ thanks for your answer. But the definitions of Stieltjes transform seems to vary quite bit, without affecting the main results upto a sign change it seems: see this very early paper - jstor.org/stable/1989901?seq=1 (look at the preview), the wiki definition (en.wikipedia.org/wiki/Stieltjes_transformation) that you already pointed out, Silverstein's definition (ims.nus.edu.sg/Programs/randommatrix/files/jsilverstein_ln.pdf) and so on. But the other results on P. 169 of Tao's notes are soncistent with (contd.) $\endgroup$ – Mathmath Dec 9 '19 at 9:31
  • $\begingroup$ (contd.) $\int_{\mathbb{R}}\frac{d\mu(t)}{t-z}$, because look at the equation right after "Whereas the moment method started from the identity (2.83), the Stieltjes transform method proceeds from the identity" on P.169: it'd be $\frac{1}{n}tr((zI - \frac{M_n}{\sqrt(n)})^{-1})$ had he defined the transform same as wiki's defintion (the one you used). Also in the analytic expression, see that the leading term is $\frac{-1}{z}$ in Tao's notes, which'd be $\frac{1}{z}$ , had he defined using the one you mentioned. But I guess we both agree that at least either way, there's a typo. $\endgroup$ – Mathmath Dec 9 '19 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.