# Asymptotics of a 1D integral, or the orthant probability of an equicorrelated random Gaussian vector

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

$$I(a,d)=\int_{-\infty}^{\infty}dx\left[\Phi\left(x/a\right)\right]^{d}\phi(x)$$

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 $$x ($$\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 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!

• My tex is proving inadequate for this, but I think you can calculate asymptotics just using $\Phi(z) \approx \frac 1 2 + \frac z {\sqrt{2 \pi}}$
– user83457
Sep 27, 2016 at 12:06
• 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 $d$. Sep 27, 2016 at 12:29

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.
• 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! Sep 27, 2016 at 13:40
• 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 . Sep 27, 2016 at 19:31
• 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. Sep 28, 2016 at 4:49
• Thanks for checking, but in your code the CDF should be raised to the power of $ka^2$ and not $ka$. Sep 28, 2016 at 9:22