Lower bound on $P(X>\lambda)$ where $X$ is Gaussian. Suppose X is a 0 mean Gaussian random variable with variance 1.   I'm trying to find a lower bound on $P(X>\lambda)$.   Specifically I'd like to derive a lower bound of the form
$c e^{-C\lambda^2}$ for positive constants $c,C$.  I know there exists several upper bounds of this form.   But I am looking for a lower bound. 
 A: Using the change of variable $x=\lambda+t/\lambda$, one gets, for every positive $\lambda$,
$$
\mathbb P(X\gt\lambda)=\frac1{\sqrt{2\pi}}\int_\lambda^{+\infty}\mathrm e^{-x^2/2}\mathrm dx=\frac1{\lambda\sqrt{2\pi}}\mathrm e^{-\lambda^2/2}\int_0^{+\infty}\mathrm e^{-t}\mathrm e^{-t^2/(2\lambda^2)}\mathrm dt.
$$
When $\lambda\to\infty$, the very last exponential goes to $1$ hence the last integral converges to $1$ and 
$$
\lim_{\lambda\to+\infty}\sqrt{2\pi}\lambda\mathrm e^{\lambda^2/2} \mathbb P(X\gt\lambda)=1.
$$
To get nonasymptotic bounds, one can use the fact that $\mathrm e^{-t^2/(2\lambda^2)}\geqslant\mathrm e^{-1/2}$ for every $t\leqslant\lambda$ and $\mathrm e^{-t^2/(2\lambda^2)}\leqslant1$ everywhere. This yields, for every positive $\lambda$,
$$
\mathrm e^{-1/2}(1-\mathrm e^{-\lambda})\leqslant\sqrt{2\pi}\lambda\mathrm e^{\lambda^2/2} \mathbb P(X\gt\lambda)\leqslant1.
$$
A: Let $Q(\alpha)$ denote $P\{X > \alpha\}$, the probability that
a standard normal random variable exceeds $\alpha$.  As shown
in this answer,
for $\alpha > 0$,
$$ \frac{\exp(-\alpha^2/2)}{\sqrt{2\pi}}
\left (\frac{1}{\alpha} - \frac{1}{\alpha^3}\right )
< Q(\alpha) < \frac{\exp(-\alpha^2/2)}{\alpha\sqrt{2\pi}}
$$
which is not quite a bound of the desired form $c\exp(-C\alpha^2)$
where $c$ and $C$ are constants. But the result does indicate
that no lower bound on $Q(\alpha)$ can have both
$c$ a constant and $C = \frac{1}{2}$; 
$C$ must be larger than $\frac{1}{2}$.
(Note that the answers provided by @did do have $C = \frac{1}{2}$
but are of the form $g(\alpha)\exp(-\alpha^2/2)$ where
the $g(\alpha)$ are decreasing functions of $\alpha$.)  
A weak bound of the desired form, with $c$ and $C$ both constants
(but unfortunately with $C = 1$) can be obtained as follows.
Suppose that $X$ and $Y$ are independent standard normal random variables. 
Then, for $\alpha \geq 0$, $P\{|X| \leq \alpha,|Y| \leq \alpha\} = [1-2Q(\alpha)]^2$ where $Q(\alpha)$ is the
complementary cumulative probability distribution function of the standard normal
random variable.  But this probability is the integral of
the joint density of $X$ and $Y$ over the square region of side $2\alpha$ centered at the origin, and is bounded above by the integral over the
circumscribed circle of radius $\sqrt{2}\alpha$.
Thus, we have
$$\begin{align*}
P\{|X| \leq \alpha, |Y| \leq \alpha\} 
&= \int_{-\alpha}^{\alpha}\int_{-\alpha}^{\alpha}
\frac{1}{2\pi}\exp[(-x^2-y^2)/2]\,\mathrm dx\,\mathrm dy\\
&\leq \int_{0}^{\sqrt{2}\alpha} \int_{0}^{2\pi}\frac{1}{2\pi}\exp(-r^2/2)
\,r\mathrm d\theta \,\mathrm dr\\
&= 1 - \exp(-\alpha^2) ~~ \text{for} ~\alpha \geq 0.
\end{align*}$$
giving 
$[1-2Q(\alpha)]^2 
\leq 1 - \exp(-\alpha^2)$ for $\alpha \geq 0$, or, equivalently, 
$\exp(-\alpha^2) \leq 4Q(\alpha) - 4Q^2(\alpha)$. But, since
$4Q^2(\alpha) > 0$ for all $\alpha$, we get that
$$Q(\alpha) > \frac{1}{4}\exp(-\alpha^2) ~ \text{for}~ \alpha \geq 0$$
which is a lower bound of the form desired.
But, as mentioned earlier, this is a very loose bound.
