If $X \sim Normal(0,1)$, then we have the tail bound:
 $$ (\*) \qquad\Pr[X > t] \leq \mathcal{O}\left(\frac{e^{-t^2/2}}{t}\right) .$$


Now for general variables $X$, a nice condition is that $X$ be *subgaussian*, meaning that $\mathbb{E}[e^{tX}] \leq e^{t^2/2}$. In this case we traditionally get the tail bound
 $$ \Pr[X > t] \leq e^{-t^2/2} . $$


My question is, can we actually get the tail bound of the form $(\*)$, or is there a counterexample? I ask because it seems intuitively plausible, but I have never seen such a result in references such as [Boucheron, Lugosi, and Massart](http://rads.stackoverflow.com/amzn/click/0199535256).


(**Edit**) After some helpful answers I want to clarify: first, let's only consider the regime, say, $t \geq 1$. Second, I'm not worried about the constant in the big-O, but the constant of $1/2$ in the exponent should stay fixed -- note that $e^{-t^2/2}/t = e^{-t^2/2 - \ln(t)} \leq e^{-O(t^2)}$. So this is really a very fine distinction I am asking about (I'm tempted to say "it's entirely academic").


One stronger property that is not true is that a subgaussian variable's tail is dominated by $\Phi$. For example, a Rademacher variable $X$ (in $\{\pm 1\}$ with probability $0.5$ each) has $\Pr[X \geq 1] = 0.5$, which is larger than $\Pr[N \geq 1]$ for a standard normal $N$. But this seems like the "worst case", so it still seems hopeful to me that $(\*)$ could hold for some constant.


(P.S. I was initially assuming that any conclusions here would translate smoothly to $\sigma$-subgaussian variables, where $\mathbb{E}[e^{tX}] \leq e^{\sigma^2 t^2/2}$. But maybe that's not true....)



Do subgaussian variables obey the slightly-stronger-than-Chernoff tail bound?
$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb{R}}\newcommand{\ch}{\operatorname{ch}}$Let $X$ be a $\sigma$-sub-Gaussian random variable (r.v.) for some real $\sigma>0$, that is,
\begin{equation\*}
 M\_X(t):=Ee^{tX}\le e^{\sigma^2t^2/2}\quad\text{for all real $t$}. \tag{1}
\end{equation\*}
Then the standard upper bound on $P(X\ge u)$ is $e^{-u^2/(2\si^2)}$,
for each real $u\ge0$.


Note that (1) implies that the left and right derivatives of the moment generating function $M\_X$ are $0$, and hence $EX=0$. So, the standard upper bound $e^{-u^2/(2\si^2)}$ on $P(X\ge u)$ can be indeed improved -- see [Section 3.2](https://arxiv.org/abs/2103.15740).


However, any such improvement must be asymptotically negligible for large $u$. Specifically, for any upper bound $B(u)$ on $P(X\ge u)$ valid for all $\si$-sub-Gaussian random r.v.'s $X$ we have $B(u)\ge e^{-u^2/(2\si^2)}(1-o(1))$; here and in what follows, all asymptotic relations are given for $u\to\infty$.


Indeed, by simple rescaling, without loss of generality (wlog) $\si=1$. So, we have to show that
\begin{equation\*}
 P\_u:=\sup\_{X\in S}P(X\ge u)\ge e^{-u^2/2}(1-o(1)), \tag{2}
\end{equation\*}
where $S$ is the set of all $1$-sub-Gaussian random r.v.'s.


Here is a proof of (2). For $p\in(0,1)$ and real $u>0$, let $X\_{p,u}$ be any r.v. such that
\begin{equation\*}
 P(X\_{p,u}=u)=P(X\_{p,u}=-u)=p/2,\quad P(X\_{p,u}=0)=1-p. \tag{2.5}
\end{equation\*}
We will have $X\_{p,u}\in S$ iff $1-p+p\ch tu\le e^{t^2/2}$ for all real $t$, that is, iff
\begin{equation\*}
 p\le p\_u:=\inf\_{t\in\R}r\_u(t)=\inf\_{t\ge0}r\_u(t) \tag{3} 
\end{equation\*}
where $\ch:=\cosh$ and
\begin{equation\*}
 r\_u(t):=\frac{e^{t^2/2}-1}{\ch tu-1} 
\end{equation\*}
for real $t\ne0$, with $r\_u(0):=r\_u(0+)=r\_u(0-)=1/u^2$. Since $r\_u(\infty-)=\infty$ and the function $r\_u$ is even and continuous, the infimum in (3) is attained at some $t\_u\in[0,\infty)$, so that
\begin{equation\*}
 p\_u=r\_u(t\_u). \tag{4} 
\end{equation\*}


Since the set $[0,\infty]$ is compact, wlog one of the following cases must occur:


*Case 1: $t\_u=0$.* Then
\begin{equation}
 p\_u=r\_u(0)=\frac1{u^2}>>e^{-u^2/2}. \tag{5}
\end{equation}
Here and in the sequel, we write $a>>b$ for $a/b\to\infty$.


*Case 2: $t\_u\downarrow0$ and even $t\_u u\to0$.* Then
\begin{equation}
 p\_u=\frac{e^{t\_u^2/2}-1}{\ch t\_u u-1}\sim\frac1{u^2}>>e^{-u^2/2}. \tag{6}
\end{equation}


*Case 3: $t\_u\downarrow0$ and $t\_u u\to c$ for some $c\in(0,\infty)$.* Then $t\_u\sim c/u$ and hence
\begin{equation}
 p\_u=\frac{e^{t\_u^2/2}-1}{\ch t\_u u-1}\sim\frac{t\_u^2/2}{\ch c-1}
 \sim\frac{c^2/2}{\ch c-1}\frac1{u^2}>>e^{-u^2/2}. \tag{7}
\end{equation}


*Case 4: $t\_u\downarrow0$ and $v:=t\_u u\to\infty$.* Then $v=o(u)$ and hence
\begin{equation}
 p\_u=\frac{e^{t\_u^2/2}-1}{\ch t\_u u-1}\sim\frac{v^2/2}{e^v/2}\frac1{u^2}
 >>\frac{e^{-v}}{u^2}>>e^{-u^2/2}. \tag{8}
\end{equation}


*Case 5: $t\_u\to c$ for some $c\in(0,\infty)$.* Then
\begin{equation}
 p\_u=\frac{e^{t\_u^2/2}-1}{\ch t\_u u-1}\sim\frac{e^{c^2/2}-1}{e^{cu}/2}>>e^{-u^2/2}. \tag{9}
\end{equation}


*Case 6: $t\_u\to\infty$.* Then
\begin{equation}
 p\_u=\frac{e^{t\_u^2/2}-1}{\ch t\_u u-1}\sim\frac{e^{t\_u^2/2}}{e^{t\_u u}/2}\ge2e^{-u^2/2}. \tag{10}
\end{equation}


Thus, in all cases
\begin{equation}
 p\_u\ge(2-o(1))e^{-u^2/2}. \tag{11}
\end{equation}
Also, if we choose $p=p\_u$, then clearly the non-strict inequality in (3) will hold. So, $X\_{p\_u,u}\in S$ and hence, by (2) and (2.5),
\begin{equation\*}
 P\_u\ge P(X\_{p\_u,u}\ge u)=p\_u/2. 
\end{equation\*}
Now (2) follows by (11). $\Box$


