$X_n/n^\alpha\to a$ almost surely if $EX_n\sim an^\alpha$ and $\mathrm{var}(X_n)\leq Bn^\beta$

Question

I am going through the exercises in the Borel-Cantelli section of Durrett, and stuck on this question. (Exercise 2.3.18)

Let $0\leq X_1\leq X_2\cdots$ be random variables with $EX_n\sim an^\alpha$ and $\mathrm{var}(X_n)\leq Bn^\beta$, where $a,\alpha>0$ and $\beta<2\alpha$. Show that $X_n/n^\alpha\to a$ almost surely.

My attempt: Since $(EX_n-an^\alpha)/n^\alpha\to0$ as $n\to\infty$, it suffices to show that $(X_n-EX_n)/n^\alpha\to0$ almost surely. For any $\epsilon>0$, by Chebyshev's inequality we have $$ P\left(\left|\frac{X_n-EX_n}{n^\alpha}\right|>\epsilon\right)\leq\epsilon^{-2}\cdot n^{-2\alpha}\cdot\mathrm{var}(X_n) \leq B\epsilon^{-2}\cdot n^{\beta-2\alpha}\to 0 $$ as $n\to\infty$, so $(X_n-EX_n)/n^\alpha\to0$ in probability. The condition $\beta<2\alpha$ is too weak, however, in a sense that the final estimate is not summable in $n$ and one cannot apply the first Borel-Cantelli lemma to ensure the almost sure convergence.

I think the non-decreasing condition $0\leq X_1\leq X_2\cdots$ will help us get to the destination, but I am not sure how to use this assumption.

Answer 1

The non-decreasing condition indeed helps. Since $$ \max_{2^N+1\leqslant n\leqslant 2^{N+1}}X_nn^{-\alpha}\leqslant X_{2^{N+1}}2^{-N\alpha}, $$ it actually suffices to prove that $\left(X_{2^N}2^{-N\alpha}\right)_{N\geqslant 1}$ goes to zero almost surely. Rewrite the estimates you got with $n=2^N$ to conclude.

Answer 2

To achieve summability the trick is to apply Chebyshev to a subsequence of $(X_n/n^\alpha)$: $$ P\left(\left|\frac{X_{n^p}-E(X_{n^p})}{(n^p)^\alpha}\right|>\epsilon\right)\le\epsilon^{-2}n^{-2p\alpha}\operatorname{Var}(X_{n^p})\le\epsilon^{-2}n^{-2p\alpha}Bn^{p\beta}=B\epsilon^{-2}n^{-p(2\alpha-\beta)}. $$ The RHS will be summable for $p$ sufficiently large. Hence for such a $p$ the subsequence $X_{n^p}/n^{p\alpha}$ converges almost surely as $n\to\infty$.

Now establish a.s. convergence for the overall sequence: Given $n$, find $N$ such that $N^p\le n\le (N+1)^p$. By monotonicity of $(X_n)$, establish the inequalities $$ \left(\frac N{N+1}\right)^{p\alpha}\frac{X_{N^p}}{N^{p\alpha}} =\frac{X_{N^p}}{(N+1)^{p\alpha}}\le\frac{X_n}{n^\alpha}\le \frac{X_{(N+1)^p}}{N^{p\alpha}}= \left(\frac{N+1}N\right)^{p\alpha}\frac{X_{(N+1)^p}}{(N+1)^{p\alpha}}, $$ and observe that almost surely the LHS and RHS tend to $a$ as $N\to\infty$.