Convergence in probability of running maximum Suppose we have a sequence of integrable random variables $(X_n)$ on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that $n^{-1}X_n\to 0$ in probability as $n\to\infty$. Suppose further that $\mathbb{E}[X_i]=\mathbb{E}[X_j]$ for all $i,j$.
Can we then deduce that $n^{-1}\max_{0\le k\le n}|X_k|\to 0$ in probability as $n\to\infty$?
At the moment, the best I can show is the final convergence holds on a subsequence (by passing from convergence in probability to almost sure convergence on a subsequence in the hypothesis, and then using that the conclusion holds for almost sure convergence, and finally that almost sure convergence implies convergence in probability).
If anyone has any ideas then I will be extremely grateful!
 A: It seems that in general, no. 
Let $\left(X_j\right)_{j\geqslant 1}$ be an independent sequence, where 
$$
\Pr(X_j=j)=\Pr(X_j=-j)=\frac 1{2j}; \Pr(X_j=0)=1-1/j.
$$
Then:


*

*For each $\varepsilon\in (0,1)$, $\Pr(\left\lvert X_n\right\rvert/n\gt \varepsilon)=n^{-1}$ hence $X_n/n\to 0$ in probability. 

*For each $j\geqslant 1$, $X_j$ is symmetric and integrable hence its expectation is zero. 

*We will show that the sequence $\left(n^{-1}\max_{1\leqslant j\leqslant n}\left\lvert X_j\right\rvert\right)_{n\geqslant 1}$ does not converge to $0$ in probability. 


*

*It suffices to show that $\left(2^{-n}\max_{2^n+1\leqslant j\leqslant 2^{n+1}}\left\lvert X_j\right\rvert\right)_{n\geqslant 1}$ does not converge to $0$ in probability.

*Observe that $$\left\{2^{-n}\max_{2^n+1\leqslant j\leqslant 2^{n+1}}\left\lvert X_j\right\rvert\geqslant 1\right\}\supset \bigcup_{j=2^n+1}^{2^{n+1}}A_j,$$
where $A_j:=\left\{\left\lvert X_j\right\rvert=j\right\}$.

*By Bonferroni's inequality, 
$$\Pr\left(\bigcup_{j=2^n+1}^{2^{n+1}}A_j\right)\geqslant \sum_{j=2^n+1}^{2^{n+1}}\Pr\left(A_j\right)-\sum_{2^n+1\leqslant i\lt j\leqslant 2^{n+1}}\Pr\left(A_i\cap A_j\right).
$$
We can use independence and the fact that $\sum_{j=2^n+1}^{2^{n+1}}\Pr\left(A_j\right)$ lies in the interval $(1/2,1)$ to conclude that $\Pr\left(\bigcup_{j=2^n+1}^{2^{n+1}}A_j\right)\geqslant 1/4$.


