Suppose that we have a a tight/stochastically-bounded sequence $\{X_n\}_{n\in\mathbb{N}}$ (tightness results from zero expectation and uniformly bounded variance). Is it possible to derive that $$Y_n=\max_{i=1,\ldots n}X_i$$ is tight/stochastically-bounded?
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The following example can be generalised to sub-gaussian random variables and via truncation techniques to more general classes.
Let $\{X_n\}$ be a sequences of independent gaussian random variables with mean zero and uniformly bounded variance, then $$\max_{i=1,\ldots,n} X_i = O_p\left(\sqrt{\log(n)} \right).$$ Proof: Let $Var(X_i)\leq K$, then applying Markovs inequality with $\gamma$ determined in the sequel yield $$P(\max_{i=1,\ldots,n} X_i \geq Cb(n)) \leq P(\max_{i=1,\ldots,n} |X_i| \geq Cb(n)) \leq \sum_{i=1}^n P(|X_i|\geq Cb(n)) \\ \leq 2\sum_{i=1}^n e^{-C\gamma b(n)} E[e^{\gamma X_i}] \leq 2\sum_{i=1}^n e^{-C\gamma b(n)}e^{\gamma^2 K/2} = 2\sum_{i=1}^n e^{\log(n)(-C + K/2+1)} = 2n^{-C+K/2+1}=o(1)$$ with the choice $$b(n) = \gamma = \sqrt{\log(n)}$$ and $C$ larger than $K/2+1$.
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