I know how to prove the zero and scaling property of norm. However I'm stuck on proving triangle inequality. The definition of norm of sub-Gaussian random variable is. Sub-Gaussian random variable is such norm exists.
$$\|X\|_{\psi_2}=\inf\{t>0:E e^{-\frac{X^2}{t^2}}\}$$
Best Answer
I think you mistake the definition.
sub-gaussian norm is
$$ \|X\|_{\psi_2} = \inf\left\{t>0:\mathbb{E}\exp\left( X^2/t^2\right)\le2\right\}. $$
What you want to show that is
$$ \|X+Y\|_{\psi_2} \le \|X\|_{\psi_2} + \|Y\|_{\psi_2}. $$
To show this, Let $f(x) = e^{x^2} $ which is increasing and convex.
Then, we have
$$ f\left(\frac{|X+Y|}{a+b} \right) \le f\left(\frac{|X|+|Y|}{a+b} \right)\\ \le \frac{a}{a+b}f\left(\frac{|X|}{a}\right) + \frac{b}{a+b}f\left(\frac{|Y|}{b}\right) $$ by Jensen's Inequality. After that, taking expectations, $$ \mathbb{E}f\left(\frac{|X+Y|}{a+b} \right) \le \frac{a}{a+b}\mathbb{E}f\left(\frac{|X|}{a}\right) + \frac{b}{a+b}\mathbb{E}f\left(\frac{|Y|}{b}\right). $$
If we insert $a=\|X\|_{\psi_2}, b=\|Y\|_{\psi_2}$, then by definition, we have
$$ \mathbb{E}f\left(\frac{|X+Y|}{\|X\|_{\psi_2}+\|Y\|_{\psi_2}} \right) \le \frac{a}{a+b}\times2 + \frac{b}{a+b}\times2 = 2. $$
So, $\|X\|_{\psi_2}+\|Y\|_{\psi_2}$ is in the set $\left\{t>0:\mathbb{E}\exp\left( X^2/t^2\right)\le2\right\}$, and it completes the proof as follows:
$$ \|X + Y\|_{\psi_2} \le \|X\|_{\psi_2}+\|Y\|_{\psi_2}. $$