**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 $x<x_0\triangleq a\Phi^{-1}({y^{1/q})}$ ($\Phi^{-1}$ 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!