I have recently been reading Vershynin's "High-Dimensional Probability" (which can be found for free here), and I would very much like to find an explicit representation of the absolute constants found in the equivalent properties of sub-gaussian random variables when restated in terms of the sub-gaussian norm.
Following the proof of Proposition 2.5.2, stated below, I think it should be possible to say that $C \geq 12\sqrt{e}$ for the absolute constant in the proposition, since we get the relations $$ K_2 \leq 3K_1, \quad K_3 = 2\sqrt{e}K_2, \quad K_1 = K_4 = 2K_5$$ and we can thus simply take the largest $C$ so that all relations are upheld, i.e. $C = 12\sqrt{e}$.
However, I would like to find the appropriate constants when restating these properties in terms of the sub-gaussian norm. From the Proposition it is apparent that we have $K_4 = \|X\|_{\psi_2}$, and so I would think we can simply take $K_1 = \|X\|_{\psi_2}$ as well, but in the book we have the following restatement of the properties:
\begin{align*} \mathbb{P}(|X|\geq t) &\leq 2 \exp(-ct^2/\|X\|_{\psi_2}^2) \quad \text{for all }t\geq 0 \\ \|X\|_{L^p} &\leq C \|X\|_{\psi_2}\sqrt{p} \\ \mathbb{E}\exp(X^2/\|X\|_{\psi_2}^2) &\leq 2 \\ \text{if } \mathbb{E}X=0 \text{ then } \mathbb{E}\exp(\lambda X) &\leq \exp(C\lambda^2\|X\|_{\psi_2}^2) \quad \text{for all }\lambda \in \mathbb{R}. \end{align*} So, basically I am wondering why we get a $c$ in the first property here... are these different constants? If so, why? And can we make them explicit?
For reference, this is the relevant proposition:
Proposition 2.5.2 Let $X$ be a random variable. Then, the following properties are equivalent; the parameters $K_i > 0$ appearing in these properties differ from each other by at most an absolute constant factor.
- The tails of $X$ satisfy $$ \mathbb{P}(|X| \geq t) \leq 2 \exp (-t^2/K_1^2) \quad \text{for all } t \geq 0. $$
- The moments of $X$ satisfy $$ \|X\|_{L^p} = (\mathbb{E}|X|^p)^{1/p} \leq K_2 \sqrt{p} \quad \text{for all } p \geq 1. $$
- The MGF of $X^2$ satisfies $$ \mathbb{E} \exp (\lambda^2 X^2) \leq \exp (K_3^2 \lambda^2) \quad \text{for all } \lambda \text{ such that } |\lambda| \leq \frac{1}{K_3}.$$
- The MGF of $X^2$ is bounded at some point, namely $$ \mathbb{E} \exp (X^2/K_4^2) \leq 2$$ Moreover, if $\mathbb{E}X = 0$ then properties 1-4 are also equivalent to the following one.
- The MGF of $X$ satisfies $$ \mathbb{E} \exp (\lambda X) \leq \exp (K_5^2 \lambda^2) \quad \text{for all } \lambda \in \mathbb{R}. $$
Edit
One of the constants was not too hard to get by using the implication $(3) \implies (4)$:
Since $K_4 = \|X\|$ we get $K_3 = \sqrt{\log2}K_4 = \sqrt{\log2}\|X\|$.