Almost sure convergence

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?

• 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). – j.c. Dec 10 '17 at 20:42
• The formatting is indeed very strange, and most likely done by some robot (converting from another format?), maybe somebody will guess what exactly. – YCor Dec 10 '17 at 21:38
• Thank you @j.c.. I am very new to this. I will try that next time. – user0735 Dec 10 '17 at 21:44

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.
• 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... – Henry.L Dec 11 '17 at 15:34