0
$\begingroup$

Write $X \in sG(\sigma^2)$ if $X$ is sub-Gaussian of parameter $\sigma^2$, that is $\mathbb{E}(e^{\lambda X}) \le e^{\lambda^2 \sigma^2 / 2}$.

I'm interested in showing that, given $\epsilon > 0$, there exists a random variable $X$ with the property that $X \in sG(1)$ but $X \notin sG(\tau^2)$ for any $\tau < 1$ and $E(X^2) = \epsilon$.

Now, if I just take the standard normal, this satisfies the sub-Gaussian part of the condition. However, I can't adapt it to reduce the variance without altering the sub-Gaussian part. (There might be a way, but I can't see it!) I can't come up with any other random variables satisfying the sub-Gaussian conditions though!

I then want to use this to show that it's not possible to obtain a bound that is valid for all $X \in sG(1)$ and all $t > 0$ of the form $$\mathbb{P}(X > t) \le \exp\left(-\frac{t^2}{\mathbb{E}(X^2)}\right).$$

This seemed easy enough: if it were the case, then I'd be able to take a sequence $(X_n)$ with $E(X_n) = 1/n$ and $X \in sG(1)$ but $X \notin sG(\tau^2)$ for any $\tau < 1$. However,choosing $X_n$ so that $X_n \to X$, we must have that $X$ is almost surely constant, since its variance is $0$, and thus almost surely $0$ since its expectation is $0$. Then $X \notin sG(\sigma^2)$ for any $\sigma$ whatsoever. However, all this does is show that $sG(\cdot)$ is not closed.

Any help on these issues would be most appreciated! Thanks! :)

Plus, if anyone can think of any better tags, please add away! :)

$\endgroup$
  • $\begingroup$ you introduce epsilon but then do nothing with it. $\endgroup$ – Thoth Jan 30 '16 at 23:02
  • $\begingroup$ oops, corrected! $\endgroup$ – Sam T Jan 30 '16 at 23:03
  • $\begingroup$ Just let $X$ be a mixture of constant $0$ (probability $1-\epsilon$) and standard normal (probability $\epsilon$). $\endgroup$ – A.S. Jan 31 '16 at 0:55
  • $\begingroup$ Oh of course! This just multiplies the mgf by $\epsilon^2$, which is the same order for $\lambda$. (I was trying to use some sort of scaling.) $\endgroup$ – Sam T Jan 31 '16 at 11:30
  • $\begingroup$ Any advice fur the second part? (Not the solution, of course, just some advice :).) $\endgroup$ – Sam T Jan 31 '16 at 11:30

Your Answer

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

Browse other questions tagged or ask your own question.