First Borel-Cantelli-Lemma - Two different proofs











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The Borel Cantelli Lemma states that for a sequence of events $(A_n)$ with finite sum $sum_{k=1}^{infty}P(A_k)< infty$ the probability of infinite many events happening is $0$, thus $P(limsup_{n to infty} A_n)=0.$



Before getting to my question, here are two different proofs that make sense to me:



First one:



Let $B_m=bigcup_{k geq m}A_k$, then $lim_{m to infty} B_m=limsup_{n toinfty}A_n$ from above. Hence, by the continuity, subadditivity of measures
$$P(limsup_{n to infty} A_n)=P(lim_{m to infty} B_m)=lim_{m to infty}P( bigcup_{k geq m}A_k)leq lim_{m to infty}sum_{k=m}^{infty}P(A_k)=0.$$



Second one:



$P(limsup_{n to infty} A_n)=0$ is equivalent to $sum_{k=1}^{infty}1_{A_k}(omega) < infty$. So we need to show that $P(sum_{k=1}^{infty}1_{A_k}(omega))geq infty)=0$.
First of all, by linearity of the expectation we have $$Ebigl (sum_{k=1}^{infty}1_{A_k}bigr )=sum_{k=1}^{infty}E(1_{A_k})=sum_{k=1}^{infty}P(A_k) < infty$$
Thus, by using Markov's inequality we obtain
$$P(sum_{k=1}^{infty}1_{A_k}(omega) geq epsilon) leq {1 over epsilon}sum_{k=1}^{infty}P(A_k) rightarrow0$$ for $epsilon to infty$ which completes the proof.



Question:



If both proofs are correct (please check), does any of those two proofs require something that is not needed in the other proof?






probability-theory proof-verification borel-cantelli-lemmas







share|cite|improve this question




















  • 1




    Both proofs are valid.
    – Kavi Rama Murthy
    Nov 22 at 12:13










  • @KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
    – drhab
    Nov 22 at 12:42












  • @drhab thanks for checking. Whats parts should be rewritten?
    – Tesla
    Nov 22 at 12:44










  • Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
    – drhab
    Nov 22 at 12:47












  • Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
    – drhab
    Nov 22 at 12:52

















up vote
1
down vote

favorite












The Borel Cantelli Lemma states that for a sequence of events $(A_n)$ with finite sum $sum_{k=1}^{infty}P(A_k)< infty$ the probability of infinite many events happening is $0$, thus $P(limsup_{n to infty} A_n)=0.$



Before getting to my question, here are two different proofs that make sense to me:



First one:



Let $B_m=bigcup_{k geq m}A_k$, then $lim_{m to infty} B_m=limsup_{n toinfty}A_n$ from above. Hence, by the continuity, subadditivity of measures
$$P(limsup_{n to infty} A_n)=P(lim_{m to infty} B_m)=lim_{m to infty}P( bigcup_{k geq m}A_k)leq lim_{m to infty}sum_{k=m}^{infty}P(A_k)=0.$$



Second one:



$P(limsup_{n to infty} A_n)=0$ is equivalent to $sum_{k=1}^{infty}1_{A_k}(omega) < infty$. So we need to show that $P(sum_{k=1}^{infty}1_{A_k}(omega))geq infty)=0$.
First of all, by linearity of the expectation we have $$Ebigl (sum_{k=1}^{infty}1_{A_k}bigr )=sum_{k=1}^{infty}E(1_{A_k})=sum_{k=1}^{infty}P(A_k) < infty$$
Thus, by using Markov's inequality we obtain
$$P(sum_{k=1}^{infty}1_{A_k}(omega) geq epsilon) leq {1 over epsilon}sum_{k=1}^{infty}P(A_k) rightarrow0$$ for $epsilon to infty$ which completes the proof.



Question:



If both proofs are correct (please check), does any of those two proofs require something that is not needed in the other proof?






probability-theory proof-verification borel-cantelli-lemmas







share|cite|improve this question




















  • 1




    Both proofs are valid.
    – Kavi Rama Murthy
    Nov 22 at 12:13










  • @KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
    – drhab
    Nov 22 at 12:42












  • @drhab thanks for checking. Whats parts should be rewritten?
    – Tesla
    Nov 22 at 12:44










  • Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
    – drhab
    Nov 22 at 12:47












  • Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
    – drhab
    Nov 22 at 12:52















up vote
1
down vote

favorite









up vote
1
down vote

favorite











The Borel Cantelli Lemma states that for a sequence of events $(A_n)$ with finite sum $sum_{k=1}^{infty}P(A_k)< infty$ the probability of infinite many events happening is $0$, thus $P(limsup_{n to infty} A_n)=0.$



Before getting to my question, here are two different proofs that make sense to me:



First one:



Let $B_m=bigcup_{k geq m}A_k$, then $lim_{m to infty} B_m=limsup_{n toinfty}A_n$ from above. Hence, by the continuity, subadditivity of measures
$$P(limsup_{n to infty} A_n)=P(lim_{m to infty} B_m)=lim_{m to infty}P( bigcup_{k geq m}A_k)leq lim_{m to infty}sum_{k=m}^{infty}P(A_k)=0.$$



Second one:



$P(limsup_{n to infty} A_n)=0$ is equivalent to $sum_{k=1}^{infty}1_{A_k}(omega) < infty$. So we need to show that $P(sum_{k=1}^{infty}1_{A_k}(omega))geq infty)=0$.
First of all, by linearity of the expectation we have $$Ebigl (sum_{k=1}^{infty}1_{A_k}bigr )=sum_{k=1}^{infty}E(1_{A_k})=sum_{k=1}^{infty}P(A_k) < infty$$
Thus, by using Markov's inequality we obtain
$$P(sum_{k=1}^{infty}1_{A_k}(omega) geq epsilon) leq {1 over epsilon}sum_{k=1}^{infty}P(A_k) rightarrow0$$ for $epsilon to infty$ which completes the proof.



Question:



If both proofs are correct (please check), does any of those two proofs require something that is not needed in the other proof?






probability-theory proof-verification borel-cantelli-lemmas







share|cite|improve this question















The Borel Cantelli Lemma states that for a sequence of events $(A_n)$ with finite sum $sum_{k=1}^{infty}P(A_k)< infty$ the probability of infinite many events happening is $0$, thus $P(limsup_{n to infty} A_n)=0.$



Before getting to my question, here are two different proofs that make sense to me:



First one:



Let $B_m=bigcup_{k geq m}A_k$, then $lim_{m to infty} B_m=limsup_{n toinfty}A_n$ from above. Hence, by the continuity, subadditivity of measures
$$P(limsup_{n to infty} A_n)=P(lim_{m to infty} B_m)=lim_{m to infty}P( bigcup_{k geq m}A_k)leq lim_{m to infty}sum_{k=m}^{infty}P(A_k)=0.$$



Second one:



$P(limsup_{n to infty} A_n)=0$ is equivalent to $sum_{k=1}^{infty}1_{A_k}(omega) < infty$. So we need to show that $P(sum_{k=1}^{infty}1_{A_k}(omega))geq infty)=0$.
First of all, by linearity of the expectation we have $$Ebigl (sum_{k=1}^{infty}1_{A_k}bigr )=sum_{k=1}^{infty}E(1_{A_k})=sum_{k=1}^{infty}P(A_k) < infty$$
Thus, by using Markov's inequality we obtain
$$P(sum_{k=1}^{infty}1_{A_k}(omega) geq epsilon) leq {1 over epsilon}sum_{k=1}^{infty}P(A_k) rightarrow0$$ for $epsilon to infty$ which completes the proof.



Question:



If both proofs are correct (please check), does any of those two proofs require something that is not needed in the other proof?






probability-theory proof-verification borel-cantelli-lemmas




probability-theory proof-verification borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 11:00

























asked Nov 22 at 12:02









Tesla

928426




928426








  • 1




    Both proofs are valid.
    – Kavi Rama Murthy
    Nov 22 at 12:13










  • @KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
    – drhab
    Nov 22 at 12:42












  • @drhab thanks for checking. Whats parts should be rewritten?
    – Tesla
    Nov 22 at 12:44










  • Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
    – drhab
    Nov 22 at 12:47












  • Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
    – drhab
    Nov 22 at 12:52
















  • 1




    Both proofs are valid.
    – Kavi Rama Murthy
    Nov 22 at 12:13










  • @KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
    – drhab
    Nov 22 at 12:42












  • @drhab thanks for checking. Whats parts should be rewritten?
    – Tesla
    Nov 22 at 12:44










  • Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
    – drhab
    Nov 22 at 12:47












  • Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
    – drhab
    Nov 22 at 12:52










1




1




Both proofs are valid.
– Kavi Rama Murthy
Nov 22 at 12:13




Both proofs are valid.
– Kavi Rama Murthy
Nov 22 at 12:13












@KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
– drhab
Nov 22 at 12:42






@KaviRamaMurthy IMO the second proof should be rewritten before it can be judged. The idea might be okay, but very weird things are said in that proof.
– drhab
Nov 22 at 12:42














@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 at 12:44




@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 at 12:44












Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
– drhab
Nov 22 at 12:47






Start with looking at spots where your write an $omega$ (so just one arbitrary outcome) and wonder whether it is on its place. E.g. we can talk about $mathbb E1_A$ but not about $mathbb E1_A(omega)$.
– drhab
Nov 22 at 12:47














Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
– drhab
Nov 22 at 12:52






Also note that $P(limsup_{ntoinfty}A_n=0)iffsum_{k=1}^{infty}1_{A_k}(omega)<infty$ must become something like $omeganotinlimsup_{ntoinfty}A_niffsum_{k=1}^{infty}1_{A_k}(omega)<infty$.
– drhab
Nov 22 at 12:52

















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