Constructing sub-gaussian random variable with arbitrary norm and variance

Question

Consider the following definition for $\sigma$-sub-gaussian random variable:

$$\mathbb{E}\left[\exp\left(\lambda\left(X-\mathbb{E}\left[X\right]\right)\right)\right]\leqslant\exp\left(\frac{\lambda^{2}\sigma^{2}}{2}\right)$$

and the sub-gaussian norm $||\cdot||$ is the minimal value of $\sigma$ for which $X$ is sub-gaussian.

I need to show that for any $C\geqslant1$ there exists some random variable $X$ such that var$(X)=1$ and $||X||=C$, with respect to the sub-gaussian norm.

However, I cannot see how can I scale the sub-gaussian norm without scailng the variance as well; For example, since $||X||=\sqrt{Var(X)}$ I thought about starting with $X$ with $Var=C^2$ , but when scailing its variance to $1$ the norm scales as well.

Answer 1

Let $X$ be a Bernoulli random variable with mean $p \in (0,1)$. The variance of $X$ is given as $\sigma_{\text{var}}^2 = p(1-p)$ and the optimal sub-Gaussian proxy is given as $\sigma_{\text{subG}}^2 = \dfrac{0.5 - p}{\ln\left(\dfrac{1}{p} - 1\right)}$ (reference).

It is straightforward to verify that the ratio $\dfrac{\sigma_{\text{subG}}^2}{\sigma_{\text{var}}^2}$ diverges as $p \to 0$. Thus, for any $C \geq 1$ you can obtain a value of $p \in $ such that $\dfrac{\sigma_{\text{subG}}^2}{\sigma_{\text{var}}^2} = C$. If you want the variance to be $1$, then simply scale the random variable to ensure variance is $1$.