Exponential bound on the tail of a gaussian Let $Z$ be a centered normal variable of variance $\sigma^2$, I am trying to prove that,
$$\sup_{t>0} \left( \mathbb{P}(Z \geq t) \exp\left( \frac{t^2}{2 \sigma^2} \right) \right) = \frac{1}{2} $$
I have proven the quantity is at least $\frac{1}{2}$, by looking at the limit for $t \to 0$, I'm trying to upper bound the quantity inside the $\sup$ by $\frac{1}{2}$.
I've obviously tried re-writing the probability as an integral, and with a change of variable I'm able to write:
$$ \mathbb{P}(Z \geq t) \exp\left( \frac{t^2}{2 \sigma^2} \right) = \int_0^{\infty} \exp \left( \frac{-u^2 - 2tu}{2 \sigma^2} \right) du $$
If I'm not mistaken we don't know how to explicitly calculate these types of integrals, and I see no obvious upper bounds. I've tried re-writing the exponential as a series (both in this expression and the original integral). I've tried integrating by parts the product of exponentials in this last expression, but it simply leads to a difference between two terms making it even harder to upper bound.
I've also tried a Cauchy-Schwarz upper bound on the product of exponentials but it yields something proportional to $\frac{1}{t}$, thus insufficient (which makes sense because only one of the functions depends on $t$).
 A: Let $Q(x)$ denote the complementary standard normal CDF given by
$$Q(x) = \int_x^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{t^2}{2}\right) \, \mathrm dt.$$ Suppose that $t > x > 0$. Since $t+ x > t - x > 0$, we have that $(t + x)(t - x) = t^2 - x^2 > (t - x)^2 > 0$.  Hence,
\begin{align}\exp\left(\frac{x^2}{2}\right)Q(x) &= \int_x^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left (- \frac{t^2 - x^2}{2} \right ) \, \mathrm dt\\
&< \int_x^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left (- \frac{(t - x)^2}{2} \right )\, dt\\
&= \frac 12.
\end{align}
That last integral displayed above is just the area to the right of $x$ under the pdf of a $N(x,1)$ random variable.
A: Substitute $z = t/\sigma$ to simplify notation. You need to show that the function $$f(z) := \dfrac{1}{\sqrt{2\pi}}e^{z^2/2}\int_{z}^{\infty}e^{-u^2/2}\,du$$ is decreasing over $z > 0$ so that you can conclude that $\sup_{z > 0} f(z) = f(0) = \dfrac{1}{2}$.
Differentiate to get:
\begin{align*}
f'(z) &= \dfrac{1}{\sqrt{2\pi}}ze^{z^2/2} \cdot \int_{z}^{\infty}e^{-u^2/2}\,du + \dfrac{1}{\sqrt{2\pi}}e^{z^2/2} \cdot -e^{-z^2/2}
\\
&= zf(z)-\dfrac{1}{\sqrt{2\pi}}
\end{align*}
So to show that $f'(z) \le 0$ for all $z > 0$, you just need to show that $f(z) \le \dfrac{1}{z\sqrt{2\pi}}$, i.e. $$\int_{z}^{\infty}e^{-u^2/2}\,du \le \dfrac{1}{z}e^{-z^2/2}.$$ Based on the last three lines of your question, I suspect you have already done something like this. But if not, bound the integral above by multiplying the integrand by $\tfrac{u}{z}$ and see what that gets you.
