1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.