Probability – Value of Specific Limit Involving Inverse Functions

Question

Closed. This question does not meet Mathematics Stack Exchange guidelines. It is not currently accepting answers.

---

Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.

Closed yesterday.

Improve this question

Denote by $\Phi^{-1}$ the inverse normal cdf. I want to calculate the following limit:
$$\lim_{x \to 0} \Phi^{-1}(x) – \Phi^{-1}\left(\frac{1}{2}x\right).$$
I think it should be $0$ but I am not able to prove it. Can someone help me?

Answer 1

Around $x=0$ $$\Phi^{-1}(x)=-\sqrt{a-\log (a)}\qquad \text{with} \qquad a= -\log \left(2 \pi x^2\right)$$

The required difference seems to be difficult (not to say more) to express formally.

If we let $x=10^{-n}$, it seems that a decent approximation could be $$ \Phi^{-1}(x) - \Phi^{-1}\left(\frac{x}{2}\right)\sim \frac 1 {3 \sqrt n}\qquad \text{with} \qquad x=10^{-n}$$

Edit

Notice that , still with $x=10^{-n}$, asymptotically

$$ \Bigg(\Phi^{-1}(x)\Bigg)^2 - \Bigg(\Phi^{-1}\left(\frac{x}{2}\right) \Bigg)^2=\log \left(\frac{4 \log \left(2 \pi 10^{-2 n}\right)}{\log \left(\frac{1}{2} \pi 10^{-2 n}\right)}\right)$$ and $$ \Bigg(\frac{\Phi^{-1}(x) }{ \Phi^{-1}\left(\frac{x}{2}\right)} \Bigg)^2 \sim \frac{1579}{1139}-\frac{33}{109\, n}$$ These two relations would allow several uses.