Tail probabilities of Gaussians

There are many many tail probabilities for the Gaussian, and most use
some form of integration by parts. Duembgen’s paper on bounding
standard tail probabilities has
several, for example.

Let
$$\Prob( X \geq t) = T(t) = \int_{t}^{\infty} \exp(-x^2/2) \frac{1}{\sqrt{2\pi}} dx$$
Two simple inequalities obtained by integration by parts are:
$$\begin{aligned}
\frac{1}{\sqrt{2\pi}} \exp(-t^2/2) \left(\frac{1}{t} - \frac{1}{t^3}\right) \leq T(t) \leq \frac{1}{t} \exp(-t^2/2) \end{aligned}$$
Fleshed out versions of these calculations can be found anywhere; see
https://mikespivey.wordpress.com/2011/10/21/normaltails/.
Essentially one simply introduces $x/t$ into the integrand to obtain an
upper bound. The integrand now has a simple anti-derivative.

Here is a better bound, using a slightly cooler trick of introducing a
new independent variable.

Introduce an independent Gaussian variable $Y$ and consider
$T(t) = \Prob( Y \in \R, X \geq t)$ again. This is bounded above by the
probability of the exterior of the circle: $$\begin{aligned}
\Prob( Y \in \R, X \geq t)
& \leq \Prob( X^2 + Y^2 \geq t^2)\
& = \frac1{2 \pi} \int_t^{\infty} r e^{-r^2/2} dr \
& = \frac1{2 \pi} e^{-t^2/2} \end{aligned}$$ Certainly
$\frac1{2\pi} \leq \frac12$; this is the bound that appears frequently
in papers: $$
\frac{1}{L(1 + t)} \exp\left( -\frac{t^2}{2}\right) \leq T(t) \leq \frac{1}{2} \exp\left( -\frac{t^2}{2} \right).
$$ I haven’t worked out the details of the lower bound, but I assume
it’s quite similar.