Problem: Let $\phi(x)$ be the normal probability density function (pdf), and $\Phi(x)$ the normal cumulative distribution (cdf). I'm interested in the asymptotic behavior of the following integral


in the limit that $a \rightarrow \infty $, and $d =\alpha a^2$, with $\alpha>1$ some constant.

Background: Suppose we have an equivaraiant random Gaussian vector $\mathbf{x}$ in $\mathbb{R}^d$, i.e. $\mathbf{x}\sim\mathcal{N}(0,\boldsymbol{\Sigma})$, where $\Sigma_{ij} =\delta_{ij} + a^{-2}$. Then the integral above is the orthant probability: $ I(a,d)= P(\forall i:x_i >0). $

Conjecture: From numerical simulations and some intuition, I'm suspecting that in this limit $$ \log I(a,d) \leq - \kappa a^2 \log(d) + o(a^2\log(d)) \quad(\ast) $$

for some positive constant $\kappa $. The numerical results might be wrong, since I'm getting warnings on precision accuracy. The intuition behind this bound is that $\left[\Phi\left(x/a\right)\right]^{d}$ is "approximately" a step function. In other words, we can choose some constant $y$ so that $\left[\Phi\left(x/a\right)\right]^{d}$ is very small for $x<x_0\triangleq a\Phi^{-1}({y^{1/q})}$ ($\Phi^{-1}$ is the inverse CDF), and upper bounded by $1$ if $x>x_0$. Integrating over this bound when $x>x_0$ and using standard bounds we get $(\ast)$, as long as $y$ is not too small. However, so far, I haven't found a good way to upper bound $\left[\Phi\left(x/a\right)\right]^{d}$ when $x<x_0$ so that the integral in this range is smaller than the integral in the range $x>x_0$ (while keeping $y$ sufficiently large).

Goal: A valid answer to this problem can either

(1) Prove this bound and find $\kappa $.

(2) Disprove this bound (show it is too low) and find a different (non trivial) upper bound.

(3) Find a better (lower) upper bound.

Any help would be appreciated, and thanks in advance!

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    $\begingroup$ My tex is proving inadequate for this, but I think you can calculate asymptotics just using $\Phi(z) \approx \frac 1 2 + \frac z {\sqrt{2 \pi}}$ $\endgroup$
    – user83457
    Sep 27, 2016 at 12:06
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    $\begingroup$ Thanks. I tried using a linear approximation. I found this approximation becomes saturated (larger then 1) at some point, so I had to divide the integral into cases (e.g., below and above saturation), or the integral becomes too large. Then, what do we do above saturation? If we just bound $\Phi(z)$ with $1$, then our bound will not depend on $d$. $\endgroup$ Sep 27, 2016 at 12:29

1 Answer 1


I have asymptotics sketched out like this: first break the integral into $|x| > a^{\frac 2 3 }$ and the complement. Bound the first one by $\int_{|x| > a^{\frac 2 3 }} \phi(x) dx \approx. e^{-a^{1.2}}$ which will be much smaller than the other term. For the other term use $\Phi(\frac x a ) \approx \frac 1 2 + \phi(0) \frac x a$ and therefore $\Phi(\frac x a )^d \approx (\frac 12 )^d e^{2x\alpha \phi(0)}$, and the x part just integrates to a constant, giving $(\frac 12 )^d$ in all cases. I believe positively correlated guassians have an fkg type inequality, and that the probability of the same for uncorrelated gaussians is an actual lower bound.

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    $\begingroup$ Many apologies, I have noticed a typo in my question: I accidentally wrote $a$ instead of $a^2$ in the bound $(\ast)$. My typo made the question too easy... After the correction, your bound is insufficiently tight. Thanks for helping me find this mistake! $\endgroup$ Sep 27, 2016 at 13:40
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    $\begingroup$ Maple confirms that in some sense, outputting $${{\rm e}^{-d\ln \left( 2 \right) }}+{\frac {{{\rm e}^{-d\ln \left( 2 \right) }}d \left( d-1 \right) }{\pi\,{a}^{2}}}.$$ See the code here dropbox.com/s/29jxbwlf9nac48l/asymptotics.pdf?dl=0 . $\endgroup$
    – user64494
    Sep 27, 2016 at 19:31
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    $\begingroup$ Note your code check the limit that $a$ goes to infinity, but $d$ stays constant. I'm interested in the limit in which both goes to infinity, and $d=\alpha a^2$ with $\alpha>1$, so $d$ is much larger then $a$. Thanks for checking! It helped me find another $a^2$ to $a$ typo I missed. $\endgroup$ Sep 28, 2016 at 4:49
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    $\begingroup$ Here dropbox.com/s/r0rt6pg1qelv85j/asympt.pdf?dl=0 it is. $\endgroup$
    – user64494
    Sep 28, 2016 at 7:57
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    $\begingroup$ Thanks for checking, but in your code the CDF should be raised to the power of $ka^2$ and not $ka$. $\endgroup$ Sep 28, 2016 at 9:22

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