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?


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