# Subgaussian Absolute Constant

I am looking at the proofs of equivalent properties for subgaussian random variables. Two listed below:

$$P(|X|\ge t)\le 2\text{exp}(\frac{-t^2}{K_1^2})$$

$$(\mathbb{E}|X|^{p})^{\frac{1}{p}}\le K_2\sqrt{p}$$

I am able to follow the proofs but the constants $$K_i$$ are stated to differ only by an absolute constant. According to the definition they propose that if $$i$$ implies $$j$$ then $$K_j\le C K_i$$ where $$C$$ is an absolute constant. Does this means that $$C$$ is the same for any $$(i,j)$$ pairing? Looking at the proofs that doesn't seem to be true. Can someone help clarify what they mean by absolute constant?

• Which proof are you looking at? Commented Dec 14, 2020 at 3:55
• The ones in High-Dimensional Probability: An Introduction with Applications in Data Science
– nvm
Commented Dec 14, 2020 at 4:01

There are only finitely many pairs, so $$C$$ can be taken to be a number larger than any of the values that appear in each part of the proof of Prop. 2.5.2.
• It means that instead of the $K_2=3$ obtained in the proof, the second constant will be $K_2=3K_1$. Basically, this is saying "apply the proof to $X'=X/K_1$ instead of $X$" (you can check that all conditions are homogeneous, i.e., that using $aX$ instead of $X$ in the proofs just changes the constants obtained by a factor $a$). Commented Dec 14, 2020 at 4:42