bound of standard normal random variable Let $X$ be a standard normal random variable, $h>1$ is a constant. How can I get an estimation of $M$, if $$\mathbb{P}[-M\leq X\leq M] =1-\frac{1}{h^2}.$$ For example, an upper bound and a lower bound of $M$ in terms of $h$, if a closed form solution for $M$ is not possible.
 A: $$M = \sqrt{2} \,\mathrm{erf}^{-1}\left(1 - \frac{1}{h^2}\right)  \text{,}  $$
where $\mathrm{erf}^{-1}$ is the inverse error function.  Depending on your definition of "closed form", we are done.  (For instance, see the Python implementation scipy.special.erfinv().)
A series expansion centered at $\infty$ simplifies to
$$  M \approx \sqrt{-\log \left(\pi  \log \left(\frac{2 h^4}{\pi }\right)\right) + 4 \log(h) + \log(2)}  \text{.}  $$
It is an upper bound for $h$ up to about $3.8$.
It is egregiously terrible for $h \in [1,3/2]$, but the error is less than $0.066$ for $h \geq 2$ and less than $0.011$ for $h \geq 3$.
Using the Taylor series for the inverse error function centered at $0$ (and the argument $1-1/h^2$), we can get a lower bound
$$ M \geq \sqrt{\frac{\pi }{2}} \left(
\left(1-\frac{1}{h^2}\right) +
    \frac{\pi}{12}  
   \left(1-\frac{1}{h^2}\right)^3 + 
    \frac{7\pi^2}{480}
   \left(1-\frac{1}{h^2}\right)^5 + 
    \frac{127 \pi ^3}{40320} 
   \left(1-\frac{1}{h^2}\right)^7
\right)  \text{,}  $$
which can be improved by using more terms of the series.  The error is less than $0.13$ on $[1,2]$.
So a set of bounds is...  Let
\begin{align*}
a_1 &= \sqrt{-\log \left(\pi  \log \left(\frac{2 h^4}{\pi }\right)\right) + 4 \log(h) + \log(2)}  \\
a_2 &= \sqrt{\frac{\pi }{2}} \left(
\left(1-\frac{1}{h^2}\right) +
    \frac{\pi}{12}  
   \left(1-\frac{1}{h^2}\right)^3 + 
    \frac{7\pi^2}{480}
   \left(1-\frac{1}{h^2}\right)^5 + 
    \frac{127 \pi ^3}{40320} 
   \left(1-\frac{1}{h^2}\right)^7
\right)  \text{.}
\end{align*}
Then
$$  \min\{ a_1 - 0.066, a_2\} \leq M \leq a_1 + 0.066  \text{.}  $$
If you want tighter bounds than this, I recommend using bisection to polish (since you know the forward function, erf, and you know that erf is monotonic).
