bound of standard normal random variable

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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.

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$$M = \sqrt{2} \,\mathrm{erf}^{-1}\left(1 - \frac{1}{h^2}\right) \text{,} $$ 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).