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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?

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?

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:

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

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?

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?

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

$ \text {The aim is to show that D converges almost surely to 0.} $

$ \text {Taking the expectation, I know the following:} $

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

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

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

$ \text {Is this result sufficient for D to converge almost surely to 0?} $

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:

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

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?