# Prove subgaussian norm of sugaurssian random variables is a norm

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.

$$\|X\|_{\psi_2}=\inf\{t>0:E e^{-\frac{X^2}{t^2}}\}$$

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:

$$\|X + Y\|_{\psi_2} \le \|X\|_{\psi_2}+\|Y\|_{\psi_2}.$$