Quotient of indicatorfunction and probability
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I am having trouble to proof following statement.
If $A_1,A_2,..$ are pairwise independent and $sum_{n=1}^infty mathbb{P}(A_n)=infty$ then as $nrightarrow infty$
$frac{sum_{m=1}^n mathbb{1}_{A_m}}{sum_{m=1}^n mathbb{P}(A_m) } rightarrow 1$ almost surely.
We already showed that this expression converges in probability to 1. Now i was told to choose a subsequence and show that this subsequences converges almost surely to 1.
probability probability-theory
asked Nov 14 at 15:45
Katakuri
214
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up vote
2
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favorite
I am having trouble to proof following statement.
If $A_1,A_2,..$ are pairwise independent and $sum_{n=1}^infty mathbb{P}(A_n)=infty$ then as $nrightarrow infty$
$frac{sum_{m=1}^n mathbb{1}_{A_m}}{sum_{m=1}^n mathbb{P}(A_m) } rightarrow 1$ almost surely.
We already showed that this expression converges in probability to 1. Now i was told to choose a subsequence and show that this subsequences converges almost surely to 1.
probability probability-theory
asked Nov 14 at 15:45
Katakuri
214
I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am having trouble to proof following statement.
If $A_1,A_2,..$ are pairwise independent and $sum_{n=1}^infty mathbb{P}(A_n)=infty$ then as $nrightarrow infty$
$frac{sum_{m=1}^n mathbb{1}_{A_m}}{sum_{m=1}^n mathbb{P}(A_m) } rightarrow 1$ almost surely.
We already showed that this expression converges in probability to 1. Now i was told to choose a subsequence and show that this subsequences converges almost surely to 1.
probability probability-theory
asked Nov 14 at 15:45
Katakuri
214
I am having trouble to proof following statement.
If $A_1,A_2,..$ are pairwise independent and $sum_{n=1}^infty mathbb{P}(A_n)=infty$ then as $nrightarrow infty$
$frac{sum_{m=1}^n mathbb{1}_{A_m}}{sum_{m=1}^n mathbb{P}(A_m) } rightarrow 1$ almost surely.
We already showed that this expression converges in probability to 1. Now i was told to choose a subsequence and show that this subsequences converges almost surely to 1.
probability probability-theory
probability probability-theory
asked Nov 14 at 15:45
Katakuri
214
asked Nov 14 at 15:45
Katakuri
214
asked Nov 14 at 15:45
Katakuri
214
asked Nov 14 at 15:45
Katakuri
214
asked Nov 14 at 15:45
Katakuri
214
214
I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29
add a comment |
I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29
I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29
I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29
add a comment |
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I do not see how to use the hint. Showing the almost sure convergence of a subsequence does not seem simpler than the original problem. We can use Chebyschev's inequality and we can conclude from the Borel-Cantelli lemma if the series $sum c_n^{-1}$ converges where $c_n=sum_{i=1}^nP(A_i)$.
– Davide Giraudo
Nov 16 at 16:29