I'm struggling at finishing an exercise and hope some of you can help me!
Let be $X_i$ a set of random variables being normal distributed $N(0, \sigma_i^2$ for $i = 1,...,n$. I need to prove, that $\mathbb{E}[\max_{1\leq i \leq n}|X_i|] \leq \sqrt{6 \log(n)} \max_{1\leq i \leq n}\sigma_i $.
I followed a hint and proved for a konvex $\psi: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ that, if there exist $c_i > 0, i=1, … ,n$ and $c>0$, such that $\mathbb{E}[\frac{|X_i|}{c_i}] < c, \forall i = 1,...,n $ it holds:
$\mathbb{E}[\max_{1\leq i \leq n}|X_i|] \leq \psi^{-1}(c~ n) \max_{1\leq i \leq n}c_i$
Now I tried to apply this little lemma and took $\psi(x) = e^{x^2}$. This one is convex and strictly positive and has the inverse $\psi^{-1}(y) = \sqrt{log(y)}$.
I guess now I should take into account, that the $X_i$ are normal distributed. Unfortunatelly I didn't get the expected result, so it seems as if I would do something wrong...
Can anyone help me?
Thank you very much! :)
I don't know why you didn't get the expected result. Take $c_i=\sigma_i$. Since $E|X_i| \leq \sqrt {EX_i^{2}}=\sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $E\max_i |X_i| \leq \sqrt {log\, cn} \max_i \sigma_i$ for any $c >1$. Letting $c \to 1$ we get a stronger result than what we are asked to prove.