# Constructing sub-gaussian random variable with arbitrary norm and variance

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.

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 $$1$$.