Asymptotics of inverse of normal CDF

Let $\Phi(x)$ denote the cdf of the standard normal distribution. What are the asymptotics of $\Phi^{-1}(p)$, as $p \to 1$? In particular, is there an asymptotic expression for $\Phi^{-1}(1-x)$, as $x \to 0$? A first-order approximation would be fine.

We have $$\Phi(x) = \frac 1 2 + \frac 1 2 \operatorname{erf} \frac x {\sqrt 2} \sim 1 - \frac 1 {x \sqrt {2 \pi}} e^{-x^2/2}, \\ \ln (1 - \Phi(x)) \sim -\frac {x^2} 2 - \ln (x \sqrt {2 \pi}) \sim -\frac {x^2} 2, \quad x \to \infty, \\ \Phi^{-1}(1 - y) \sim \sqrt {-2 \ln y}, \quad y \to 0^+.$$ More terms are given in Blair, Edwards, Johnson, Rational Chebyshev Approximations for the Inverse of the Error Function.

One way to get estimates of $\Phi(x)$ for large $x$ is to repeatedly integrate by parts. First in integrating $e^{-y^2/2} dy$, you write $dv=ye^{-y^2/2} dy,u=1/y$, resulting in

$$\int_x^\infty e^{-y^2/2} dy = \frac{e^{-x^2/2}}{x} - \int_x^\infty \frac{e^{-y^2/2}}{y^2} dy.$$

You can now repeat this same technique arbitrarily many times (although for any fixed $x$, the bounds obtained will eventually become less tight). One more step is what is required for a first asymptotic:

$$\int_x^\infty e^{-y^2/2} dy = \frac{e^{-x^2/2}}{x} - \frac{e^{-x^2/2}}{x^3} + 3 \int_x^\infty \frac{e^{-y^2/2}}{y^4} dy.$$

Hence

$$\frac{e^{-x^2/2}}{x} - \frac{e^{-x^2/2}}{x^3} \leq \int_x^\infty e^{-y^2/2} dy \leq \frac{e^{-x^2/2}}{x}.$$

Now write the inside as $\sqrt{2\pi}-\sqrt{2\pi} \Phi(x)$ and do some algebra:

$$\frac{\sqrt{2\pi}-\frac{e^{-x^2/2}}{x} + \frac{e^{-x^2/2}}{x^3}}{\sqrt{2\pi}} \geq \Phi(x) \geq \frac{\sqrt{2\pi}-\frac{e^{-x^2/2}}{x}}{\sqrt{2\pi}}.$$

This gives the bounds

$$\Phi^{-1} \left ( 1-\frac{\frac{e^{-x^2/2}}{x}}{\sqrt{2\pi}} \right ) \leq x \\ \Phi^{-1} \left ( 1-\frac{\frac{e^{-x^2/2}}{x} - \frac{e^{-x^2/2}}{x^3}}{\sqrt{2\pi}} \right ) \geq x.$$

This reduces your question to finding estimates for local inverses of $e^{-x^2/2}(1/x)$ and $e^{-x^2/2}(1/x-1/x^3)$ for large $x$ (so that you can replace the argument by $1-x$ and the right side by the appropriate local inverse).

The former may be readily made explicit:

$$\Phi^{-1}(1-x) \leq (W(x^{-2}))^{1/2}$$

where $W$ is the Lambert W function. The latter isn't as easy.