# Question about relating variance to existing tailbound. Apparently “straightforward”.

I'm trying to make sense of a claim made in this paper (in particular justifying the equation in between equations 25 and 26).

Suppose $X$ is a random variable with mean zero, and we know that $$Pr(|\vec v^TX| > t)\le \exp(-e^{-t^2/2}).$$

How does this imply that $$\sup_{\vec v:\|\vec v\| = 1}\text{Var}(\vec{v}^T X)\le \int_0^\infty \exp(e^{-t/2}),$$ which reduces to $2$. Since this is offered without proof in the above paper, I take it this should be relatively easy.

Am I wrong?

Getting a bound on moments coming from bounds on the tails are usually derived using the layer cake formula, i.e.

$$\int |Y|^p \, d \Bbb{P} = p \cdot \int_0^\infty \lambda^{p-1} \cdot \mathbb{P}(|Y|\geq \lambda) \, d\lambda.$$

The proof of this is an application of Fubini's theorem after you have written $\Bbb{P}(|Y| \geq \lambda) = \int \Bbb{1}_{|Y| \geq \lambda} \, d\Bbb{P}$.

If you plug this in for $Y = |v^T X|$, you get (since $Y$ has mean zero, because $X$ has mean zero)

$${\rm Var}(Y) = \int |Y|^2 \, d\Bbb{P} = 2 \cdot \int_0^\infty \lambda \cdot \Bbb{P}(|Y| \geq \lambda) \, d \lambda \leq 2 \cdot \int_0^\infty \lambda e^{-\lambda^2 /2 }\, d\lambda.$$

From here you should be able to derive the desired conclusion by elementary calculations.