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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.