I am trying to show that if $X$ is a mean zero ($\mathbb{E}[X]=0$) sub-Gaussian random variable, then $$\mathbb{E}[\exp(\lambda X)]\leq \exp(C\lambda^2||X||_{\psi_2}^2)$$ for all $\lambda \in \mathbb{R}$, where $||X||_{\psi_2}^2 = \inf\{t>0: \mathbb{E}(|X|/t)] \leq2\}$ Any hints on the first step as I am kind of stuck. I know from the definition, $\mathbb{E}[\exp(\lambda X)]\leq\exp(C^2\lambda^2)$ but not sure how to use this.
1 Answer
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I highly recommend checking out Vershynin's High-Dimensional Probability for this one (you can find it for free online). If you understand the proof of the equivalent characterizations of sub-gaussian variables (Proposition 2.5.2 in the book), it becomes rather trivial.
It is proven that if $X$ is mean zero, the following two properties are equivalent:
- $\mathbb{E}\exp(X^2/K_4^2) \leq 2$
- $\mathbb{E}\exp(\lambda X) \leq \exp(K_5^2 \lambda^2) \quad \forall \lambda \in \mathbb{R}$
where $K_4$ and $K_5$ are constants which differ at most by an absolute constant $C >0$. Knowing this, you can note that $\|X\|_{\psi_2}$ is precisely the smallest possible $K_4$ for which this property holds, and thus the $K_5$ is simply $C\|X\|_{\psi_2}$.
