I've been having some problems deriving large $ n $ asymptotics for the following integral:
$$
\int_{-\infty}^{\infty}\exp(n\log(\Phi(ax)))\phi(x)dx,
$$
where $ a > 0 $ and $ \Phi $ is the cdf and $ \phi $ is the pdf of a standard normal random variable. The issue is that the maximum of the concave function $ \log(\Phi(ax)) $ occurs at $ x = \infty $, at which point its derivative and $ \phi $ are both zero. Thus, Laplace's method is not directly applicable. Are there any results that would apply in this situation?
Best Answer
Not a solution but a direction.
In this problem the general Laplace principle that the lead asymptotic terms flow from places where $\log \Phi(ax)$ is maximized still holds, but the maximum is not attained at an interior calculus-type maximum but rather at the endpoint of the range of integration.
If $a=1$ the integral $I(n,a)=\int_{\mathbb R} \exp(n\log \Phi(ax) ) \phi(x)\,dx$ is, after the change of variable $u=\Phi(ax)$ the simpler integral $\int_0^1 u^n du = 1/(n+1)$.
If $a\ne 1$ the same change of variable $x=\Phi^{-1}(u)/a$ results in $$I(n,a)=\int_0^1 u^n \frac {\phi(x)}{a\phi(ax)}\,du=\int_0^1 u^n g(u,a) du,$$ where $g(\Phi(ax),a)=\phi(x)/(a\phi(ax))$. There is an asymptotic formula $1-\Phi(x)\sim \phi(x)(1+O(1/x^2))$ good for large $x$ that should allow you to work out the asymtotics of $g(u,a)$ close to $u=1$ and finish the job. (Disclaimer: I have not worked through this myself.)