Mixing
First Borel-Cantelli-Lemma - Two different proofs
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
asked Nov 22 '18 at 12:02
TeslaTesla
940426
|
show 1 more comment
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
asked Nov 22 '18 at 12:02
TeslaTesla
940426
1
Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 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 '18 at 12:42
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 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 '18 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 '18 at 12:52
|
show 1 more comment
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
asked Nov 22 '18 at 12:02
TeslaTesla
940426
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
asked Nov 22 '18 at 12:02
TeslaTesla
940426
asked Nov 22 '18 at 12:02
TeslaTesla
940426
edited Nov 26 '18 at 11:00
Tesla
asked Nov 22 '18 at 12:02
TeslaTesla
940426
asked Nov 22 '18 at 12:02
TeslaTesla
940426
asked Nov 22 '18 at 12:02
TeslaTesla
940426
940426
1
Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 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 '18 at 12:42
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 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 '18 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 '18 at 12:52
|
show 1 more comment
1
Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 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 '18 at 12:42
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 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 '18 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 '18 at 12:52
1
1
Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 at 12:13
Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 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 '18 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 '18 at 12:42
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 at 12:44
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 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 '18 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 '18 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 '18 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 '18 at 12:52
|
show 1 more comment
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Both proofs are valid.
– Kavi Rama Murthy
Nov 22 '18 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 '18 at 12:42
@drhab thanks for checking. Whats parts should be rewritten?
– Tesla
Nov 22 '18 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 '18 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 '18 at 12:52