It that true to conclude that if a random $f(z)$ is a sub-Gaussian random variable for a constant value of z, its derivative $f'(z)|_{z=k}$ with respect to variable $z$ is also sub-Gaussian?

In particular, my random function is $f(z) = cos wz$ where $w$ is drawn from a normal distribution. Since $f(z)$ is a bounded random variable Hoeffding's inequality shows an exponential concentration bound for it. However, I need to prove the same bound for $f'(z) = -w sin wz$ which seems to be more complicated since it is not easy to show that $f'(z)$ is a sub-Gaussian random variable anymore.

  • 2
    $\begingroup$ Why on Earth would you expect this for general $f$, given that the derivative may not even exist? $\endgroup$ – Alexander Shamov Sep 25 '14 at 14:26

No, $f(z)$ being sub-Gaussian does not imply that $f'(z)$ is sub-Gaussian.

An easy way to see this is to use the fact that a sub-Gaussian must have zero mean. Thus in your example, $z$ is such that $\mathbb E(\cos(zW))=0$. For instance it could be that $z=1$ and $W\sim\mathcal N(\frac\pi{2},1)$. Then $f'(1)=-W\sin(W)$. But $$ \mathbb E(W\sin(W))\ne 0 $$ in this case. So $f'(1)$ is not sub-Gaussian.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. However, my question is whether $W\sin(W) - \mathbb E(W\sin(W))$ is subgaussian. $\endgroup$ – Amirreza Shaban May 16 '14 at 14:52

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.