Convergence of series in probability to 1 implies convergence of pointwise max over $n$ in probability to 0

If we have non-negative random variables $X_k \geq 0$ such that $\frac{1}{n}\sum_{i = 1}^n X_i \to 1$ in probability.

How do we show that $\frac{\max_{1 \leq k \leq n} X_k}{n} \to 0$ in probability?

So far I've tried toying with Borel Cantelli and strengthening to strong law, but haven't gotten much.

Solution 1:

Denote by $M_n$ the random variable $n^{-1}\max_{1\leqslant k\leqslant n}X_k$ and $S_n:=\sum_{i=1}^nX_i$. We will show that each subsequence of $\left(M_n\right)_{n\geqslant 1}$ admits a further subsequence which converges to $0$ in probability. We will use the following:

Lemma. Let $\left(n_k\right)_{k\geqslant 1}$ be an increasing sequence of positive integers. There exists an increasing sequence of integers $\left(k_\ell\right)_{\ell\geqslant 1}$ such that $$ \sup_{a\geqslant 1}\left\lvert \frac{S_{am_\ell}}{am_\ell}-1\right\rvert\to 0 \mbox{ almost surely as }\ell \to+\infty, $$ where $m_\ell=n_{k_\ell}$ and such that $m_\ell/m_{\ell-1}\to +\infty$.

If there was almost sure convergence, there would be nothing to do. Here we can not only extract an almost surely convergent sequence, but also such that the sequence $\left(S_{am_\ell}/(am_\ell)\right)_{n\geqslant 1}$ converge to $1$ uniformly with respect to $a$.

Proof of the Lemma. For all $a\geqslant 1$, let $n_{a,i}$ be integers such that the sequence $\left(n_{a,i}\right)_{i\geqslant 1}$ is increasing and for all $a$ and $i$, $$\tag{*} \mathbb P\left(\left\lvert \frac{S_{an_{a,i}}}{an_{a,i}}-1\right\rvert>2^{-i} \right)\leqslant 2^{-a-i}. $$ Such a sequence of integers can be constructed inductively, by using the fact that $\left(S_{an}/(an)\right)$ converges to $1$ in probability. We can also construct these subsequences such that for all $a$, $$ \left\{n_{a,i},i\geqslant 1\right\}\subset \left\{n_{a-1,i},i\geqslant 1\right\}\subset \{n_k,k\geqslant 1\}. $$ Then define $k_\ell$ in such a way that $$ m_\ell:=n_{k_\ell}=n_{\ell,\ell}. $$ Observe that $(*)$ implies that $$ \mathbb P\left(\sup_{a\geqslant 1}\sup_{i\geqslant j}\left\lvert \frac{S_{an_{a,i}}}{an_{a,i}}-1\right\rvert>2^{-j} \right)\leqslant \sum_{a\geqslant 1}\sum_{i\geqslant j}\leqslant 2^{-i-a}\leqslant 2^{1-j}. $$ Moreover, since $m_\ell > n_{a,\ell}$ for all $a$, it follows that $$ \mathbb P\left(\sup_{a\geqslant 1}\sup_{\ell\geqslant j}\left\lvert \frac{S_{am_\ell }}{am_\ell}-1\right\rvert>2^{-j} \right)\leqslant 2^{1-j} $$ and we conclude by the Borel-Cantelli lemma that $m_\ell$ does the job. [We extract a further subsequence if $m_\ell/m_{\ell-1}\to +\infty$ is not satisfied.]

Let us go back to $M_n$. We now show that $M_{m_\ell}\to 0$ almost surely. To ease the notations, let $r_\ell:=\left[m_\ell/m_{\ell-1}\right]$, where $[x]$ denotes the floor function of $x$. Then $$ M_{m_\ell}=\frac 1{m_\ell}\max_{1\leqslant k\leqslant m_\ell} X_k\leqslant \frac 1{m_\ell}\max_{1\leqslant k\leqslant m_{\ell-1}r_\ell} X_k +\frac 1{m_\ell}\max_{ m_{\ell-1}r_\ell+1 \leqslant k\leqslant m_\ell} X_k.$$ For the last term, using the assumption $X_k\geqslant 0$, we derive that $$ \frac 1{m_\ell}\max_{ m_{\ell-1}r_\ell+1 \leqslant k\leqslant m_\ell} X_k\leqslant \frac 1{m_\ell}\left(S_{m_\ell}-S_{m_{\ell-1}r_\ell}\right) $$ and using the fact that $S_n/n\to 1$ in probability and that $m_{\ell-1}r_\ell\sim m_\ell$, we get that $$ \frac 1{m_\ell}\max_{ m_{\ell-1}r_\ell+1 \leqslant k\leqslant m_\ell} X_k\to 0\mbox{ in probability}. $$ Now, for the first term, using again non-negativeness of the involved random variables, we derive that $$ \frac 1{m_\ell}\max_{1\leqslant k\leqslant m_{\ell-1}r_\ell} X_k\leqslant \frac 1{m_\ell}\max_{1\leqslant j\leqslant r_\ell} \max_{(j-1)m_{\ell-1}+1\leqslant i\leqslant jm_{\ell-1}}X_i\leqslant \frac 1{m_\ell}\max_{1\leqslant j\leqslant r_\ell} S_{jm_{\ell-1}}-S_{(j-1)m_{\ell-1}}. $$ Now, observe that $$ 0\leqslant\frac 1{m_\ell}\left( S_{jm_{\ell-1}}-S_{(j-1)m_{\ell-1}}\right) =\frac{m_{\ell-1}}{m_\ell}\left(\frac{ S_{jm_{\ell-1}}}{jm_{\ell-1}}j-\frac{ S_{(j-1)m_{\ell-1}}}{(j-1)m_{\ell-1}}(j-1)\right)= \frac{m_{\ell-1}}{m_\ell}\left(\frac{S_{jm_{\ell-1}}}{jm_{\ell-1}}-1\right)j -\frac{m_{\ell-1}}{m_\ell}\left(\frac{ S_{(j-1)m_{\ell-1}}}{(j-1)m_{\ell-1}}-1\right)(j-1)-\frac{m_{\ell-1}}{m_\ell} $$ hence $$ \frac 1{m_\ell}\max_{1\leqslant k\leqslant m_{\ell-1}r_\ell} X_k\leqslant \frac{m_{\ell-1}}{m_\ell}+ 2\sup_{a\geqslant 1}\left\lvert\frac{ S_{am_{\ell-1}}}{am_{\ell-1}}-1\right\rvert \frac{m_{\ell-1}r_\ell}{m_\ell}, $$ which goes to zero almost surely by the lemma and using again $m_{\ell-1}r_\ell\sim m_\ell$.