Gaussian estimate

Question

I am trying to find some $a\gt 0$ such that $$\int_{-2}^{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt\leq e^{-a}$$ Any good method to do so? I'm looking for an analytical argument rather than a numerical solution. Edit: there was a typo in the integral.

Answer 1

with standard procedure you can see that

$$\int_{-\infty}^2 \phi(x)dx\approx 0.97725$$

thus you can find an approximated value for $a$, that is

$$a\approx0.023$$

and it is valid with $=$ instead of $\leq$

Answer 2

To make the problem general, you search for $a$ such that, for a given $x$, $$\int_{-x}^{+x}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt\leq e^{-a}$$ that is to say $$\text{erf}\left(\frac{x}{\sqrt{2}}\right)\leq e^{-a}$$ First, we can solve the equation using some approximation. For example $$\text{erf}(t)\sim \sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha\, t^2}{1+\beta \,t^2}\,t^2 \Big)}$$ where $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi } \qquad \text{and} \qquad \beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$ so $$a > -\frac 12 \log \Bigg[1-\exp\Bigg(-\frac{2 x^2 \left(\alpha x^2+2\right)}{\pi \left(\beta x^2+2\right)} \Bigg)\Bigg]$$

Using $x=2$, this approximation gives $a >0.04673$ while the exact solution would be $a >0.04657$.

We can have much better using the second approximation I gave here.