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Show that finitely many $A_n$ will occur

Probability of $limsup$ of a sequence of sets (Borel-Cantelli lemma)A Borel-Cantelli lemma related question$P(limsup A_n)=1 $ if $forall A in mathfrakF$ s.t. $sum_n=1^infty P(A cap A_n) = infty$how that if $P(lim sup A_n) = 1$ then, $P(bigcup_n=1^infty A_n)=1$Is this sigma-additivity or union bound?Show that $Y_n$ converges to Y almost surely.Proof of Borel-CantelliProof of the converse Borel-Cantelli lemmaShow that $mathbfP(cup_ngeq 1A_n)=1$ iff $mathbfP(A_ntext i.o.)=1$.Borel-Cantelli and “infinitely often”Show that $ P(A_n) = 1 forall n Leftrightarrow P( bigcap A_n) =1 $

2

$begingroup$

I tried this problem quite a bit but went nowhere. I wish to solve this using set-theoretic algebra.

Problem statement

Let $A_n$, $n geq 1$ be a sequence of events such that $ mathbb P(A_n) to 0$ as $n to infty$, and

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

Show that almost surely, only finitely many of the $A_n$s will occur.

While irrelevant, it might be worth mentioning this problem is from MIT OCW (Exercise 3).

My attempt

I did not get very far. While it is obvious that we have to use Borel-Cantelli lemma on the hypothesis, it did not lead me anywhere. Specifically, since we know that

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

we can conclude that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_n+1setminus A_n right ) = 0$$

Using the above and the hypothesis that $ mathbb P(A_n) to 0$ as $n to infty$, we need to somehow show that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_k right ) = 0$$

I tried a lot of set algebra but it did not lead me anywhere.

I did see a question which is similar to this one but the hypothesis is slightly different. It mentions

$$sum_n=1^inftymathbb P (A_nsetminus A_n+1) <infty$$

instead of

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

as given in this problem. More importantly, I wish to solve this problem using set-theoretic algebra rather than arguments on lim sup etc.

I would be grateful if you can provide some insight or a hint in order for me to prove this using set-theoretic methods (ie using infinite unions, intersections, De-Morgan's laws etc).

probability-theory measure-theory borel-cantelli-lemmas

share|cite|improve this question

edited Mar 22 at 0:03

TryingHardToBecomeAGoodPrSlvr

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

$endgroup$

add a comment | 

2

$begingroup$

I tried this problem quite a bit but went nowhere. I wish to solve this using set-theoretic algebra.

Problem statement

Let $A_n$, $n geq 1$ be a sequence of events such that $ mathbb P(A_n) to 0$ as $n to infty$, and

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

Show that almost surely, only finitely many of the $A_n$s will occur.

While irrelevant, it might be worth mentioning this problem is from MIT OCW (Exercise 3).

My attempt

I did not get very far. While it is obvious that we have to use Borel-Cantelli lemma on the hypothesis, it did not lead me anywhere. Specifically, since we know that

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

we can conclude that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_n+1setminus A_n right ) = 0$$

Using the above and the hypothesis that $ mathbb P(A_n) to 0$ as $n to infty$, we need to somehow show that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_k right ) = 0$$

I tried a lot of set algebra but it did not lead me anywhere.

I did see a question which is similar to this one but the hypothesis is slightly different. It mentions

$$sum_n=1^inftymathbb P (A_nsetminus A_n+1) <infty$$

instead of

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

as given in this problem. More importantly, I wish to solve this problem using set-theoretic algebra rather than arguments on lim sup etc.

I would be grateful if you can provide some insight or a hint in order for me to prove this using set-theoretic methods (ie using infinite unions, intersections, De-Morgan's laws etc).

probability-theory measure-theory borel-cantelli-lemmas

share|cite|improve this question

edited Mar 22 at 0:03

TryingHardToBecomeAGoodPrSlvr

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

$endgroup$

add a comment | 

$begingroup$

I tried this problem quite a bit but went nowhere. I wish to solve this using set-theoretic algebra.

Problem statement

Let $A_n$, $n geq 1$ be a sequence of events such that $ mathbb P(A_n) to 0$ as $n to infty$, and

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

Show that almost surely, only finitely many of the $A_n$s will occur.

While irrelevant, it might be worth mentioning this problem is from MIT OCW (Exercise 3).

My attempt

I did not get very far. While it is obvious that we have to use Borel-Cantelli lemma on the hypothesis, it did not lead me anywhere. Specifically, since we know that

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

we can conclude that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_n+1setminus A_n right ) = 0$$

Using the above and the hypothesis that $ mathbb P(A_n) to 0$ as $n to infty$, we need to somehow show that

$$mathbb P left ( bigcap_n = 1^inftybigcup_k = n^infty A_k right ) = 0$$

I tried a lot of set algebra but it did not lead me anywhere.

I did see a question which is similar to this one but the hypothesis is slightly different. It mentions

$$sum_n=1^inftymathbb P (A_nsetminus A_n+1) <infty$$

instead of

$$sum_n=1^inftymathbb P (A_n+1setminus A_n) <infty$$

as given in this problem. More importantly, I wish to solve this problem using set-theoretic algebra rather than arguments on lim sup etc.

I would be grateful if you can provide some insight or a hint in order for me to prove this using set-theoretic methods (ie using infinite unions, intersections, De-Morgan's laws etc).

probability-theory measure-theory borel-cantelli-lemmas

share|cite|improve this question

edited Mar 22 at 0:03

TryingHardToBecomeAGoodPrSlvr

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

$endgroup$

share|cite|improve this question

edited Mar 22 at 0:03

TryingHardToBecomeAGoodPrSlvr

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

share|cite|improve this question

edited Mar 22 at 0:03

TryingHardToBecomeAGoodPrSlvr

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

asked Mar 21 at 23:36

TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr

13112

1 Answer
1

active

oldest

votes

1

$begingroup$

Note that
$$
beginalign
P(cup_k=n^infty A_k)
&=[P(A_n)+ P(A_n+1A_n^c)+P(A_n+2A_n+1^cA_n^c)+dotsb]\
&leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0
endalign
$$
as $nto infty$. The result follows from measure continuity.

share|cite|improve this answer

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So I gather that $A_2 cup A_1 = A_1 cup(A_2 cap A_1^c)$, $A_3 cup A_2 cup A_1 = A_1 cup(A_2 cap A_1^c) cup(A_3 cap A_2^c cap A_1^c)$ etc. This when extended to infinity gives the expression that you mentioned. Also, defining $B_n = cup_k=n^infty A_k$, we want to show that $$mathbb P left ( bigcap_n = 1^inftyB_n right ) = 0$$ or as $B_n+1 subset B_n$, we need to show that $$lim_n to inftyP(B_n) = 0 $$
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So you are saying that $$ beginalign P(B_n) &=[P(A_n)+ P(A_n+1 cap A_n^c)+P(A_n+2 cap A_n+1^c cap A_n^c)+dotsb]\ &leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0 endalign $$ as $nto infty$ which basically solves the problem. I suppose the above expression simply did the trick. I wonder how you thought of this expression. :) Thanks a lot for the help!
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:13
  

  
  
  
  

add a comment | 

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1

$begingroup$

Note that
$$
beginalign
P(cup_k=n^infty A_k)
&=[P(A_n)+ P(A_n+1A_n^c)+P(A_n+2A_n+1^cA_n^c)+dotsb]\
&leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0
endalign
$$
as $nto infty$. The result follows from measure continuity.

share|cite|improve this answer

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So I gather that $A_2 cup A_1 = A_1 cup(A_2 cap A_1^c)$, $A_3 cup A_2 cup A_1 = A_1 cup(A_2 cap A_1^c) cup(A_3 cap A_2^c cap A_1^c)$ etc. This when extended to infinity gives the expression that you mentioned. Also, defining $B_n = cup_k=n^infty A_k$, we want to show that $$mathbb P left ( bigcap_n = 1^inftyB_n right ) = 0$$ or as $B_n+1 subset B_n$, we need to show that $$lim_n to inftyP(B_n) = 0 $$
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So you are saying that $$ beginalign P(B_n) &=[P(A_n)+ P(A_n+1 cap A_n^c)+P(A_n+2 cap A_n+1^c cap A_n^c)+dotsb]\ &leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0 endalign $$ as $nto infty$ which basically solves the problem. I suppose the above expression simply did the trick. I wonder how you thought of this expression. :) Thanks a lot for the help!
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:13
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that
$$
beginalign
P(cup_k=n^infty A_k)
&=[P(A_n)+ P(A_n+1A_n^c)+P(A_n+2A_n+1^cA_n^c)+dotsb]\
&leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0
endalign
$$
as $nto infty$. The result follows from measure continuity.

share|cite|improve this answer

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So I gather that $A_2 cup A_1 = A_1 cup(A_2 cap A_1^c)$, $A_3 cup A_2 cup A_1 = A_1 cup(A_2 cap A_1^c) cup(A_3 cap A_2^c cap A_1^c)$ etc. This when extended to infinity gives the expression that you mentioned. Also, defining $B_n = cup_k=n^infty A_k$, we want to show that $$mathbb P left ( bigcap_n = 1^inftyB_n right ) = 0$$ or as $B_n+1 subset B_n$, we need to show that $$lim_n to inftyP(B_n) = 0 $$
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So you are saying that $$ beginalign P(B_n) &=[P(A_n)+ P(A_n+1 cap A_n^c)+P(A_n+2 cap A_n+1^c cap A_n^c)+dotsb]\ &leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0 endalign $$ as $nto infty$ which basically solves the problem. I suppose the above expression simply did the trick. I wonder how you thought of this expression. :) Thanks a lot for the help!
  $endgroup$
  – TryingHardToBecomeAGoodPrSlvr
  Mar 22 at 0:13
  

  
  
  
  

add a comment | 

$begingroup$

Note that
$$
beginalign
P(cup_k=n^infty A_k)
&=[P(A_n)+ P(A_n+1A_n^c)+P(A_n+2A_n+1^cA_n^c)+dotsb]\
&leq P(A_n)+sum_k=n^infty P(A_k+1setminus A_k)to 0
endalign
$$
as $nto infty$. The result follows from measure continuity.

share|cite|improve this answer

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

$endgroup$

share|cite|improve this answer

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552

answered Mar 21 at 23:44

Foobaz JohnFoobaz John

22.9k41552