From the definition of sub-Gaussian distribution $X$ w.r.t. $\sigma$ i.e.


$$\mathbb{P}(|X-\mathbb{E}(X)|\geq t) \leq 2 \exp(-\frac{t^2}{2\sigma^2}).$$


It's natural that when $X \sim \mathcal{N}(\mu, \sigma^2)$,
$$\mathbb{P}(X-\mu\geq t) \leq \exp(-\frac{t^2}{2\sigma^2}).$$


But this bound is too loose since when $t=0$, $\mathbb{P}(X-\mu\geq t) = \frac{1}{2} << 1$.


Thus I guessed somehow


$$\mathbb{P}(X-\mu\geq t) \leq \frac{1}{2}\exp(-\frac{t^2}{2\sigma^2}) \tag{1}$$


when $t\geq0$.


It can be verfied using the computer that $(1)$ is true. But how to prove it?




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Mill's inequality $$\mathbb{P}(X-\mu\geq t) \leq \sqrt{\frac{2}{\pi}}\frac{\exp(-\frac{t^2}{2\sigma^2})}{\frac{t}{\sigma}}$$ helps solve my guess when $t$ is big. But when $t$ is near zero, how to solve it?




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I have previously mistyped $\mathbb{P}(X-\mu\geq t)$ by $\mathbb{P}(X-\mu\leq t)$. Great thanks to [Brendan McKay](https://mathoverflow.net/users/9025/brendan-mckay) to point this out!



Upper bound about Gaussian tail bound
Letting
$Z:=\frac{X-\mu}\sigma$ and $z:=\frac t\sigma$, rewrite the inequality in question as
$$d(z):=P(Z\ge z)-e^{-z^2/2}/2\le0 \tag{1}\label{1}$$
for real $z\ge0$.


Note that
$$d'(z)=(z-\sqrt{2/\pi})e^{-z^2/2}/2.$$
So, $d$ is decreasing on the interval $[0,\sqrt{2/\pi}]$ and increasing on the interval $[\sqrt{2/\pi},\infty)$. Also,$d(0)=0$ and $d(z)\to0$ as $z\to\infty$. So, inequality \eqref{1} follows. $\quad\Box$


