How to prove that the centering inequality for the sub-gaussian norm does not hold Specifically, define the sub-gaussian norm for a r.v. X as
$$
\|X\|_{\psi_2}=\inf\{t>0:e^{X^2/t^2}\leq 2\}.
$$
How do we prove that the centering inequality with $C=1$ does not hold in general?
i.e.
$$
\|X-\mathbb{E}X\|_{\psi_2}\leq C\|X\|_{\psi_2}. 
$$
A counter example would work too.
Thanks in advance!
 A: The idea is to have a random variable with extreme values that are of opposite sign from the mean, so that shifting by the mean leads to the contributions of these terms to be augmented. To do this, those extreme values must correspondingly have low weight.
For an explicit example, let $X$ be $1$ with probability $.995$ and $-100$ with probability $.005$. Then $\mathbb{E}[X]=.495$. You can numerically check that
\begin{gather}
\mathbb{E}[\exp(X^2/43.5^2)]=.005\exp(100^2/43.5^2)+.995\exp(1^2/43.5^2)\approx 1.98\ldots\\
\mathbb{E}[\exp((X-\mathbb{E}[X])^2/43.5^2)]=.005\exp((100.495)^2/43.5^2)+.995\exp(.505^2/43.5^2)\approx 2.03.
\end{gather}
This means that
\begin{equation}
\|X\|_{\psi_2}<43.5<\|X-\mathbb{E}[X]\|_{\psi_2},
\end{equation}
so the inequality fails for $C=1$. There's likely more extreme examples, but this is just the first one I found from trial and error.
A: Let $X$ be a random variable taking the value $a$ with probability $p$ and $-a$ with probability $1-p$, with $a>0$ and $0<p<1$ to be specified later. If 
$$\tag{*}
\mathbb E\left[\psi_2\left(\lvert X-\mathbb EX\rvert\right)\right]>1\geqslant 
\mathbb E\left[\psi_2\left(\lvert X \rvert\right)\right], 
$$
then $\lVert X-\mathbb EX\rVert_{\psi_2}>\lVert X\rVert_{\psi_2}$. Let us compute the quantities involved in $(*)$. First, observe that $\lvert X \rvert=a$ hence 
$\mathbb E\left[\psi_2\left(\lvert X \rvert\right)\right]=\psi_2(a)$. Moreover, 
$$
\mathbb EX= pa+(1-p)(-a)=2pa-a
$$
hence 
$$
\mathbb E\left[\psi_2\left(\lvert X-\mathbb EX\rvert\right)\right]
=p\psi_2\left(\left\lvert a-(2pa-a) \right\rvert\right)+(1-p)\psi_2\left(\left\lvert -a-(2pa-a) \right\rvert\right)\\=
p\psi_2\left(  2a (1-p) \right)+(1-p)\psi_2\left(  2a p \right).
$$
In order to fulfill (*), we thus need 
$$
p\psi_2\left(  2a (1-p) \right)+(1-p)\psi_2\left(  2a p \right)>1\geqslant\psi_2(a).
$$
We have to choose $a$ as large as possible, hence the choice $a=\sqrt{\ln 2}$. Then we have to find $p$ such that 
$$
f(p):= p\exp\left(4(1-p)^2\ln 2\right)+(1-p)\exp\left(4 p^2\ln 2\right)>2.
$$
Letting $p=1/4$ gives $$f(1/4)=
\frac{7}{4} \times 2^{1/4}>2.$$
