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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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