# Approximating the asymptotic behavior of an improper integral

In my studies of approximation theory I have recently found this strange problem:

We examine the cumulative distribution function for the standard normal distribution, $\Phi(x) = \frac{1}{2}[1+erf(\frac{x}{\sqrt{2}})]$ and let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ be the probability density function of the normal distribution. I am looking at the following integral in terms of its asymptotic behavior $I(a,d) = \int_{-\infty}^{\infty}[\Phi(\frac{x}{a})]^{d}\phi(x)dx$ in the limit $a \to \infty$ and where $d = \alpha a^2$ for some constant $\alpha > 1$. Using some numerical analysis (not sure about its accuracy) I might have $$\log I(a,d) \leq - \kappa a^2 \log(d) + o(a^2\log(d))$$ for some positive constant $\kappa$

I tried using a linear approximation of the CDF but the saturation happens quickly and the integral is not so nice... I am mainly interested in asymptotic limits and a non-trivial upper bound.

I am writing here in the hopes of someone helping analyze this integral and find the true asymptotic behavior of it as it seems to me intractable and I need help finding a non-trivial upper bound on this, so I am writing this here. I do hope the question is clear, I thank all helpers.

• That should be $\Phi(x) = \frac{1}{2}\left( 1 + \text{erf}(x/\sqrt{2})\right)$. – Robert Israel Sep 30 '16 at 16:54
• @RobertIsrael : I fixed it, thanks for the correction – kroner Sep 30 '16 at 16:55

Change variables to $t = x/a$, so your integral (with $d = a^2 \alpha$) is

$$I(a, \alpha) = \frac{a}{\sqrt{2\pi}} \int_{-\infty}^\infty \left( \Phi(t)^\alpha e^{-t^2/2} \right)^{a^2}\; dt$$

The maximum of $\Phi(t)^\alpha e^{-t^2/2}$ is at $t = t_0$ where $$\sqrt{\pi} t_0 \left( \text{erf}(t_0/\sqrt{2}) + 1\right) - \sqrt{2} \alpha e^{-t_0^2/2} =0$$

You should be able to use Laplace's method to get the asymptotics.

EDIT: We write $\Phi(t)^\alpha e^{-t^2/2} = \exp(G(t))$ where $$G(t) = \alpha \log(\Phi(t)) - \frac{t^2}{2}$$ so that $$I(a,\alpha) = \dfrac{a}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(a^2 G(t))\; dt$$ Then according to Laplace's method we should have $$I(a,\alpha) \sim \dfrac{1}{\sqrt{|G''(t_0)|}} e^{a^2 G(t_0)}$$

That is, $$\log I(a,\alpha) = a^2 G(t_0) - \dfrac{1}{2} \log |G''(t_0)| + o(1)$$

• I cannot seem to apply Laplace's method as the function integrated is not of Laplace's form? At least not to me, could you please assist? – kroner Sep 30 '16 at 17:30
• could you please assist? – kroner Sep 30 '16 at 18:02