Suppose $A$ is a diagonal matrix with eigenvalues $1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n}$ and $x$ is drawn from standard Gaussian in $n$ dimensions. In numerical simulations, the following quantity seems to converge to $2$ as $n\rightarrow \infty$

$$z_n=E_{x\sim \mathcal{N}\left(0, I_n\right)}\left[\frac{x^T A^2 x}{x^T A^3 x}\right]$$

Can this be proven or disproven?

$z_n$ can also be written as the following sum

$$z_n=\sum_{i=1}^n i E_{y\sim \mathcal{N}\left(0,A^3\right)}\left[\frac{y_i^2}{\|y\|^2}\right]$$

This quantity can be viewed as the average ratio of quadratic forms $A^2$ and $A^3$ on the surface of $n$-dimensional sphere. enter image description here

Here's what the distribution looks like for a few values of $n$, means are tending towards $2$ enter image description here


3 Answers 3


We can attempt to simplify this multidimensional integral as follows. The expression for the expectation value reads

$$E[\frac{x^TA^2 x}{x^TA^3 x}]=\int d^nx \frac{e^{-|x|^2/2}}{(2\pi)^{n/2}}\frac{\sum_{k=1}^n x^2_k/k^2}{\sum_{k=1}^n x^2_k/k^3}=:I_n$$

We insert the identity $\frac{1}{\Delta}=\int_0^{\infty}e^{-s\Delta}ds$, with $\Delta=\sum_{k=1}^n x^2_k/k^3$, to make the denominator disappear. The integrals over the random vector coordinates become Gaussian and we obtain the much simpler one-dimensional integral

$$I_n=\int_0^{\infty}ds F_n(s)G_n(s)~,~ F_n(s)=\prod_{k=1}^{n}\frac{1}{\sqrt{1+2s/k^3}}~,~ G_n(s)=\sum_{k=1}^n\frac{k}{k^3+2s}$$

The explicit forms for $F_{\infty}(s), G_{\infty}(s)$ are not very illuminating, but performing some quick integral asymptotics reveals that when $s\to\infty$

$$F_{\infty}(s)\sim \exp\left(-\frac{\pi}{\sqrt{3}}(2s)^{1/3}\right)$$ $$G_{\infty}(s)\sim \frac{2\pi}{3\sqrt{3}(2s)^{1/3}}$$

and since the sequences $F_n, G_n$ are monotonic and upper bounded by their respective infinite sum and product limits and those limits exist and the integral converges, we are allowed to use dominated convergence to take the limit inside the integral. Numerical integration for these functions for various values of $n$ indicates that the limit is most likely only approximating the value $2$ in the following manner:

enter image description here

  • $\begingroup$ btw if I substitute your $F_\infty$ and $G_\infty$, the integral can be solved exactly, but the result is $\approx 0.5513$ which seems wrong.... wolframcloud.com/obj/yaroslavvb/newton/… $\endgroup$ Aug 27, 2021 at 13:27
  • 1
    $\begingroup$ This is an asymptotic estimate meant to demonstrate the convergence of the integral as $s\to\infty$- it is by no means an exact answer. $\endgroup$ Aug 27, 2021 at 18:27
  • $\begingroup$ follow-up here which computes the value of integral to be $1.99218056...$ $\endgroup$ Sep 1, 2021 at 14:22

I think this is covered, but not very accessible, in this paper from 1953 by Gurland on "Distribution of Quadratic Forms and Ratios of Quadratic Forms". It considers ratios for positive symmetric $Q$ and $P$ for: $$\frac{x^TQx}{x^TPx}$$


Also this reference may be of interest: https://www.researchgate.net/publication/224817325_Quadratic_Forms_in_Random_Variables_Theory_and_Applications


(I didn't know of that other posting. My references above were redundant.)

I think @DinosaurEgg is right. I verified that integral by the one given in this reference (for integer exponents): https://www-2.rotman.utoronto.ca/~kan/papers/mupq.pdf

I also numerically integrated up to n=200000 and got the value 1.99217 It seems to stabilize to a value below 2.0 so most likely the value is not 2.0. I also provided some graphs below to build some level of confidence but certainly no proof.

Convergence (expectation vs. n): enter image description here

Integrand (s<=1000, n=200000): enter image description here

Log of Integrand (s<=1000, n=200000): enter image description here

  • $\begingroup$ So.....is the limit equal to 2 or not? ;) There's a pretty large literature on ratios of quadratic forms, another post links to 4 more papers here -- stats.stackexchange.com/a/540580/511 $\endgroup$ Aug 28, 2021 at 11:33
  • $\begingroup$ @YaroslavBulatov Sorry. I missed that other posting. There seems to be little hope to derive a closed form solution. I provided some numerical results above. $\endgroup$
    – Jap88
    Aug 31, 2021 at 3:19
  • $\begingroup$ Thanks for double checking. These numerical results rule out fast convergence to 2, but I think convergence to 2 at the rate of $1/\log(n)$ is still possible $\endgroup$ Aug 31, 2021 at 8:55
  • $\begingroup$ Yes, you might be right. $\endgroup$
    – Jap88
    Aug 31, 2021 at 10:13

Not an answer, but wanted to point out an interesting result leading to an identity with the zeta functions. Let $A = \sum_{k\ge 1} X_k^2/k^2$ and $B = \sum_{k \ge 1}X_k^2/k^3$. Then a Taylor series expansion of $f(A, B)$ around $(\mu_A, \mu_B)=\mathbb{E}[(A, B)] = (\zeta(2), \zeta(3))$ is \begin{align*} \frac{A}{B} = \frac{\mu_A}{\mu_B} + \begin{pmatrix} 1/\mu_B \\ -\mu_A/\mu_B^2\end{pmatrix}^\intercal \begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \frac{1}{2}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix}^\intercal\begin{pmatrix} 0 & -1/\mu_B^2 \\ -1/\mu_B^2 & 2 \mu_A/\mu_B^3 \end{pmatrix}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \cdots \end{align*} The first few terms of this expansion are \begin{align*} \mathbb{E}\left[\frac{A}{B}\right] &= \frac{\mu_A}{\mu_B} - \frac{\mathbb{E}(A - \mu_A)(B-\mu_B)}{\mu_B^2} + \frac{\mu_A\mathbb{E}(B-\mu_B)^2}{\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)^2(B-\mu_B)}{6\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)(B-\mu_B)^2}{3\mu_B^2} - \frac{\mathbb{E}(B-\mu_B)^3\mu_A}{2\mu_B^4} + \cdots \\ &= \frac{\zeta(2)}{\zeta(3)} - 2 \frac{\zeta(5)}{\zeta(3)^2} + 2 \frac{\zeta(6)\zeta(2)}{\zeta(3)^3} + \cdots \end{align*}

  • $\begingroup$ Interesting....I tried computing a few partial sums but they seem to grow exponentially...may need another identity to compute this series (maybe from projecteuclid.org/journals/…) $\endgroup$ Aug 25, 2021 at 14:04
  • $\begingroup$ Interesting as well... I'm not too familiar with Wolfram notation, but the linear term in the expansion should have expected value 0, and therefore the output should also be 1.36843. $\endgroup$
    – Tom Chen
    Aug 25, 2021 at 14:51
  • $\begingroup$ good catch, there was a bug in my implementation, updated notebook. But my CAS approach wasn't too enlightening because I had to do this for $n=2$ to be tractable $\endgroup$ Aug 26, 2021 at 8:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .