Let $\Phi$ (resp. $\Phi^{-1}$) denote the standard Gaussian cumulative density (resp. quantile) function. Given $p \in [0, 1/2]$, what is a reasonable lower bound for $\Phi^{-1}(1-p)$ only involving elementary functions like logs, eponentials polynomials, etc. ?
Important note: The bound $\Phi^{-1}(1-p) \ge 0$ is trivial and not interesting. I am interested in a positive expression (in terms of elementary functions) that can serve as the bound.
Observations
- Some simulations suggest $\Phi^{-1}(1-p) \ge \sqrt{2(\log(C/p)}$ with $C=1/10$. The bound is only "accurate" for very very small $p$.
- More simulations show that $\Phi^{-1}(1-p) \ge \sqrt{\log\left(\frac{1/p^2}{10\log(1/p)}\right)}$ is an even better approximation for $ p \le .25$ (even though I haven't succeeded in proving it formally). I've also plotted Ian's proposed approximation for $q(p) = \sqrt{2\log(\sqrt{2\pi}/p)}$
One way goes like this. Derive an upper bound:
$$\int_x^\infty e^{-y^2/2} dy = \int_x^\infty \frac{y}{y} e^{-y^2/2} dy \leq \frac{1}{x} \int_x^\infty y e^{-y^2/2} dy = \frac{e^{-x^2/2}}{x}.$$ For a lower bound:
$$\int_x^\infty e^{-y^2/2} dy = \int_x^\infty \frac{y}{y} e^{-y^2/2} dy = \frac{e^{-x^2/2}}{x} - \int_x^\infty \frac{1}{y^2} e^{-y^2/2} dy \\ \int_x^\infty \frac{1}{y^2} e^{-y^2/2} dy = \int_x^\infty \frac{y}{y^3} e^{-y^2/2} dy \leq \frac{e^{-x^2/2}}{x^3}$$
and combining these gives
$$\int_x^\infty e^{-y^2/2} dy \geq \left ( \frac{1}{x} - \frac{1}{x^3} \right ) e^{-x^2/2}.$$
Thus
$$\frac{1}{\sqrt{2\pi}}\left ( \frac{1}{x} - \frac{1}{x^3} \right ) e^{-x^2/2} \leq 1-\Phi(x) \leq \frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-x^2/2}.$$
Thus
$$\frac{1}{\sqrt{2\pi}}\left ( \frac{1}{\Phi^{-1}(1-p)} - \frac{1}{\Phi^{-1}(1-p)^3} \right ) e^{-\Phi^{-1}(1-p)^2/2} \leq p \leq \frac{1}{\sqrt{2\pi}} \frac{1}{\Phi^{-1}(1-p)} e^{-\Phi^{-1}(1-p)^2/2}.$$
You now get an elementary lower bound in the small $p$ regime by getting a further lower bound for the LHS that only involves $\Phi^{-1}(1-p)$ through the exponential. Doing that requires substituting an upper bound for $\Phi^{-1}(1-p)$. One crude upper bound can be obtained by using the upper bound we just derived when $\Phi^{-1}(1-p) \geq 1$ (so it suffices to have $p \leq 0.15$). This bound is $q(p):=\sqrt{2\log(\sqrt{2\pi}/p)}$. In terms of this function:
$$\frac{1}{\sqrt{2\pi}} \left ( \frac{1}{q(p)} - \frac{1}{q(p)^3} \right ) e^{-\Phi^{-1}(1-p)^2/2} \leq p.$$
Now when $q(p) \geq 1$ you have:
$$e^{-\Phi^{-1}(1-p)^2/2} \leq \frac{\sqrt{2\pi} p q(p)^3}{q(p)^2-1}$$
and
$$\Phi^{-1}(1-p) \geq \sqrt{2 \log \left ( \frac{q(p)^2-1}{\sqrt{2 \pi} p q(p)^3} \right )}.$$
There is one additional requirement on $p$ which is simply that the argument of the log should be $\geq 1$ i.e. $q(p)^2-1 \geq \sqrt{2 \pi} p q(p)^3$, which constrains $p$ slightly more; numerically I find you're restricted to a slightly smaller range than $p \leq 0.137$. In the little range where this inequality fails but $\Phi^{-1}(1-p) \geq 1$, the previous inequality is still true but it is not telling you anything meaningful anymore.
This bound can be improved at the cost of further restricting its domain by restricting attention to $\Phi^{-1}(1-p) \geq C$ i.e. $p \leq 1-\Phi(C)$ for a larger $C$. In this case you get instead $q(p)=\sqrt{2 \log(\sqrt{2 \pi} C/p)}$ and a bound of the same form.