# Finding Chernoff bounds maximum estimators

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?