Mixing
Lim sup and Borel-Cantelli solution
Clash Royale CLAN TAG#URR8PPP
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I was seeking a solution to a textbook's problem to check if mine is correct and found one that got me a doubt.
We'd like to show that:
beginalignsum_n=1^inftymathbbPleft( fracX_nn>1 right) < infty implies lim sup_nfracX_nnleq 1endalign
My first thought was applying Borel-Cantelli lemma, so we show that it will be higher than 1 infinitely often with probability $0$.
The author did the above and proceeded with the following step:
$mathbbPleft(lim sup left[ fracX_nn>1 right]right) = 0 implies mathbbPleft(lim sup left[ fracX_nn leq 1right]right) = 1$
Is this because we know the following from the series?
$mathbbPleft( fracX_nn>1 right) downarrow 0 impliesmathbbPleft( left[fracX_nn>1right]^c right) = mathbbPleft( fracX_nnleq1 right) uparrow 1$
And thus a series based upon the complement sequence won't converge atributing probability $1$ to $lim sup_nfracX_nnleq 1$.
probability convergence limsup-and-liminf
asked Jul 22 at 16:44
Sergio Andrade
143112
add a comment |Â
up vote
0
down vote
favorite
I was seeking a solution to a textbook's problem to check if mine is correct and found one that got me a doubt.
We'd like to show that:
beginalignsum_n=1^inftymathbbPleft( fracX_nn>1 right) < infty implies lim sup_nfracX_nnleq 1endalign
My first thought was applying Borel-Cantelli lemma, so we show that it will be higher than 1 infinitely often with probability $0$.
The author did the above and proceeded with the following step:
$mathbbPleft(lim sup left[ fracX_nn>1 right]right) = 0 implies mathbbPleft(lim sup left[ fracX_nn leq 1right]right) = 1$
Is this because we know the following from the series?
$mathbbPleft( fracX_nn>1 right) downarrow 0 impliesmathbbPleft( left[fracX_nn>1right]^c right) = mathbbPleft( fracX_nnleq1 right) uparrow 1$
And thus a series based upon the complement sequence won't converge atributing probability $1$ to $lim sup_nfracX_nnleq 1$.
probability convergence limsup-and-liminf
asked Jul 22 at 16:44
Sergio Andrade
143112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was seeking a solution to a textbook's problem to check if mine is correct and found one that got me a doubt.
We'd like to show that:
beginalignsum_n=1^inftymathbbPleft( fracX_nn>1 right) < infty implies lim sup_nfracX_nnleq 1endalign
My first thought was applying Borel-Cantelli lemma, so we show that it will be higher than 1 infinitely often with probability $0$.
The author did the above and proceeded with the following step:
$mathbbPleft(lim sup left[ fracX_nn>1 right]right) = 0 implies mathbbPleft(lim sup left[ fracX_nn leq 1right]right) = 1$
Is this because we know the following from the series?
$mathbbPleft( fracX_nn>1 right) downarrow 0 impliesmathbbPleft( left[fracX_nn>1right]^c right) = mathbbPleft( fracX_nnleq1 right) uparrow 1$
And thus a series based upon the complement sequence won't converge atributing probability $1$ to $lim sup_nfracX_nnleq 1$.
probability convergence limsup-and-liminf
asked Jul 22 at 16:44
Sergio Andrade
143112
I was seeking a solution to a textbook's problem to check if mine is correct and found one that got me a doubt.
We'd like to show that:
beginalignsum_n=1^inftymathbbPleft( fracX_nn>1 right) < infty implies lim sup_nfracX_nnleq 1endalign
My first thought was applying Borel-Cantelli lemma, so we show that it will be higher than 1 infinitely often with probability $0$.
The author did the above and proceeded with the following step:
$mathbbPleft(lim sup left[ fracX_nn>1 right]right) = 0 implies mathbbPleft(lim sup left[ fracX_nn leq 1right]right) = 1$
Is this because we know the following from the series?
$mathbbPleft( fracX_nn>1 right) downarrow 0 impliesmathbbPleft( left[fracX_nn>1right]^c right) = mathbbPleft( fracX_nnleq1 right) uparrow 1$
And thus a series based upon the complement sequence won't converge atributing probability $1$ to $lim sup_nfracX_nnleq 1$.
probability convergence limsup-and-liminf
asked Jul 22 at 16:44
Sergio Andrade
143112
edited Jul 22 at 19:25
asked Jul 22 at 16:44
Sergio Andrade
143112
asked Jul 22 at 16:44
Sergio Andrade
143112
asked Jul 22 at 16:44
Sergio Andrade
143112
143112
add a comment |Â
add a comment |Â
1 Answer
1
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accepted
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so
beginalign*
Pleft(bigcap_n X_nright) = 1
endalign*
Hence $limsup |X_n|/n leq 1$ with probability $1$.
answered Jul 22 at 18:45
Daniel Xiang
1,823414
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so
beginalign*
Pleft(bigcap_n X_nright) = 1
endalign*
Hence $limsup |X_n|/n leq 1$ with probability $1$.
answered Jul 22 at 18:45
Daniel Xiang
1,823414
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
add a comment |Â
up vote
2
down vote
accepted
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so
beginalign*
Pleft(bigcap_n X_nright) = 1
endalign*
Hence $limsup |X_n|/n leq 1$ with probability $1$.
answered Jul 22 at 18:45
Daniel Xiang
1,823414
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so
beginalign*
Pleft(bigcap_n X_nright) = 1
endalign*
Hence $limsup |X_n|/n leq 1$ with probability $1$.
answered Jul 22 at 18:45
Daniel Xiang
1,823414
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so
beginalign*
Pleft(bigcap_n X_nright) = 1
endalign*
Hence $limsup |X_n|/n leq 1$ with probability $1$.
answered Jul 22 at 18:45
Daniel Xiang
1,823414
answered Jul 22 at 18:45
Daniel Xiang
1,823414
answered Jul 22 at 18:45
Daniel Xiang
1,823414
answered Jul 22 at 18:45
Daniel Xiang
1,823414
1,823414
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
add a comment |Â
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
I'm really sorry, sum = 0 was a typo.
– Sergio Andrade
Jul 22 at 19:26
1
1
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
Then $P(limsup X_n/n > 1) leq P(X/n > 1 text i.o.) = 0$ by the first Borel Cantelli Lemma
– Daniel Xiang
Jul 23 at 0:13
add a comment |Â
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