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Consider $$D=|D|=\frac{\sum_{i=1}^n \sum_{j=1}^n I[|{\sqrt{T}}\hat\theta_{ij,T}|>f(n)]}{g(n)}, $$ where $i \ne j$, $I$ is the indicator function, $ |{\hat\theta_{ij,T}}|<1$, $f(n)=O(\ln(n))$, $g(n)=O(n^2)$, and $f(n)/T=o(1)$.

The aim is to show that $D$ converges almost surely to $0$.

Taking the expectation, I know the following:

$$ E|D|=\frac{\sum_{i=1}^n \sum_{j=1}^n Pr[|{\sqrt{T}\hat\theta_{ij,T}}|>f(n)]}{g(n)}\, {\le}\,\,{\sup_{ij}}\,Pr[|{\sqrt{T}\hat\theta_{ij,T}}|>f(n)] \,\, {\le} \,\, {e^{-\frac{f(n)}{\rho}}} $$

where ${\rho}={\sup_{ij}}\, {\rho_{ij}}$ is positive and bounded.

From the Markov inequality: $$ Pr(|D|>{\epsilon})\,\,{\le}\,\,\frac{E|D|}{\epsilon} \,\,{\le}\,\, \frac{{e^{-\frac{f(n)}{\rho}}}}{\epsilon} $$ for some positive $\epsilon$ and hence $\lim_{n,T\to\infty} Pr(|D|>{\epsilon})=0.$

Is this result sufficient for $D$ to converge almost surely to 0? If not, how can I show that $D$ converges almost surely to 0?

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    $\begingroup$ This is just a suggestion but I recommend restricting the use of latex to the equations in your question, since mathjax takes extra time to render in the browser (especially on mobile devices). $\endgroup$ – j.c. Dec 10 '17 at 20:42
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    $\begingroup$ The formatting is indeed very strange, and most likely done by some robot (converting from another format?), maybe somebody will guess what exactly. $\endgroup$ – YCor Dec 10 '17 at 21:38
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    $\begingroup$ Thank you @j.c.. I am very new to this. I will try that next time. $\endgroup$ – user0735 Dec 10 '17 at 21:44
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What you have proved is that the $D$ converges in probability as $n\rightarrow\infty$, and I do not see why $T\rightarrow\infty$ should involve in the limit in your conclusion since you neither assume decay speed of $\hat{\theta}_{i,j,T}$ nor restrict the increasing speed of $T$. So I think you missed some assumption here.

Moreover, if $I[|{\hat\theta_{ij,T}}|>f(n)]$(or the double sum itself is monotonic) is monotonic as $T,n\rightarrow\infty$, which is very likely to be true since $|{\hat\theta_{i,j,T}}|<1$. Then using Skorohod Theorem you know there is a subsequence $Z_{k_n}$ of $$Z_n:=\sum_{i=1}^n \sum_{j=1}^n I[|{\hat\theta_{ij,T}}|>f(n)]$$ which converges to $0$ almost surely $P$, and since such a sequence $Z_{k_n}$ is selected from a monotonic sequence $Z_n$, the original sequence must also converge almost surely by comparison lemma.

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  • $\begingroup$ Thank you @Henry.L for your response. Sorry I forgot T, which I have now added. Without the use of monotonicity, would it make sense to show that the sum of n from 1 to infinity of the exponential term is bounded in order to get almost sure convergence? (thinking along the lines of the Borel-Cantelli lemma). $\endgroup$ – user0735 Dec 10 '17 at 21:55
  • $\begingroup$ I think you need monotonicity with $T$ taken into consideration now. I think a bit into BC lemma but do not quite see how it comes into play in this scenario... $\endgroup$ – Henry.L Dec 11 '17 at 15:34
  • $\begingroup$ I don't quite see how with an indicator function I can have monotonicity. $\endgroup$ – user0735 Dec 13 '17 at 12:23

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