1
$\begingroup$

I am currently trying to resolve the following exercise about Chernoff bounds:

  1. Let $X_{1}, X_{2}, \dots, X_{n}$ be independent, identically distributed (i.i.d) random variables with distribution $N(0,\sigma^{2})$. Show that for every $\varepsilon > 0$: $$P(\overline{X_{n}} > \varepsilon) \le e^{\frac{-n\varepsilon^{2}}{2\sigma^{2}}}$$

  2. A random variable $X$ with $E(X) = 0$ is said to be Subgaussian for the parameter $\sigma > 0$ if its moment generating function $M_{x}(t)$ is such that $M_{x}(t) \le e^{\frac{t^{2}\sigma^{2}}{2}}$ for all $t \in \mathbb{R} $. Show that the inequality in the previous point holds if $X_{1}, X_{2}, \dots, X_{n}$ are i.i.d random variables, and Subgaussian for the parameter $\sigma > 0$

In the first point I know I am working with Chernoff bounds since the exercise already gives its upper tail, with $\mu = 0$, so I have something like:

$$P(\overline{X_{n}} > \varepsilon + 0) \le e^{-ng(t)}$$

Where $$g(t) = \varepsilon t - \log M_{X_{1}}(t)$$ Recalling that, in this case, $$M_{X_{1}}(t) = e^{\frac{t^{2}\sigma^{2}}{2}}$$ We have $$g(t) = \varepsilon t - \frac{t^{2}\sigma^{2}}{2}$$

Now here comes the problem. In order to discover my upper tail, I have to find a maximum estimator $t^{\star}$ so I can compute $g(t^{*})$. The solution of the exercise says $t^{\star} = \frac{\varepsilon}{\sigma^{2}}$, but I don't know why. So, what I ask is how I can find maximum estimators when it comes to this type of exercises.

About the second point, since I already have $g(t^{\star})$, how can this result will be useful to prove the inequality for all the Subgaussian. Wasn't this implicitly proved in the previous point?

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.