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Name for this condition: $forallepsilon > 0:quad P(exists mge n: |X_m - X|>epsilon)to 0,quad ntoinfty$
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A sequence of random variables $X_n$ converges in probability to a random variable $X$ if
$$forallepsilon > 0:quad P(|X_n - X|>epsilon)to 0,quad ntoinfty. tag1$$
Note that the following condition is stronger:
$$forallepsilon > 0:quad P(exists mge n: |X_m - X|>epsilon)to 0,quad ntoinfty. tag2$$
Question: Is there a name for condition $(2)$?
probability analysis probability-theory convergence terminology
asked Aug 3 at 17:10
Epiousios
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A sequence of random variables $X_n$ converges in probability to a random variable $X$ if
$$forallepsilon > 0:quad P(|X_n - X|>epsilon)to 0,quad ntoinfty. tag1$$
Note that the following condition is stronger:
$$forallepsilon > 0:quad P(exists mge n: |X_m - X|>epsilon)to 0,quad ntoinfty. tag2$$
Question: Is there a name for condition $(2)$?
probability analysis probability-theory convergence terminology
asked Aug 3 at 17:10
Epiousios
1,481422
Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39
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up vote
1
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up vote
1
down vote
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A sequence of random variables $X_n$ converges in probability to a random variable $X$ if
$$forallepsilon > 0:quad P(|X_n - X|>epsilon)to 0,quad ntoinfty. tag1$$
Note that the following condition is stronger:
$$forallepsilon > 0:quad P(exists mge n: |X_m - X|>epsilon)to 0,quad ntoinfty. tag2$$
Question: Is there a name for condition $(2)$?
probability analysis probability-theory convergence terminology
asked Aug 3 at 17:10
Epiousios
1,481422
A sequence of random variables $X_n$ converges in probability to a random variable $X$ if
$$forallepsilon > 0:quad P(|X_n - X|>epsilon)to 0,quad ntoinfty. tag1$$
Note that the following condition is stronger:
$$forallepsilon > 0:quad P(exists mge n: |X_m - X|>epsilon)to 0,quad ntoinfty. tag2$$
Question: Is there a name for condition $(2)$?
probability analysis probability-theory convergence terminology
asked Aug 3 at 17:10
Epiousios
1,481422
asked Aug 3 at 17:10
Epiousios
1,481422
asked Aug 3 at 17:10
Epiousios
1,481422
asked Aug 3 at 17:10
Epiousios
1,481422
1,481422
Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39
add a comment |Â
Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39
Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39
Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39
add a comment |Â
2 Answers
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accepted
Condition (2) is equivalent to $$tag*S_n:=sup_mgeqslant nleftlvert X_m-Xrightrvert to 0mbox in probability,$$ which is equivalent to $X_nto X$ almost surely. Indeed, if $X_nto X$ almost surely, then $sup_mgeqslant nleftlvert X_m-Xrightrvert to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $left(S_nright)_ngeqslant 1$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
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up vote
0
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I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.
If we denote by $A_n(varepsilon)$ the event $$, then the event $X_n-X$ is just the
union of events
$$B_n(varepsilon)=bigcup_mgeq nA_m(varepsilon).$$
Obviously the sequence $B_n$ decreases, in the sense that $B_n+1subset B_n$ for all $n$. Elementary measure theory tells us that
$$lim_ntoinftyP(B_n(varepsilon))=Pleft(bigcap_n=1^inftyB_n(varepsilon)right)$$
but
$$bigcap_n=1^inftyB_n(varepsilon)=bigcap_n=1^inftybigcup_mgeq nA_m(varepsilon)=overlinelim_ntoinftyA_n(varepsilon)$$
where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation:
$$lim_ntoinftyP(B_n(varepsilon))=P(overlinelim_ntoinftyA_n(varepsilon))$$
which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have
$$P(overlinelim_ntoinftyA_n(varepsilon))geqoverlinelim_ntoinftyP(A_n(varepsilon)$$
answered Aug 3 at 18:36
uniquesolution
7,493721
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Condition (2) is equivalent to $$tag*S_n:=sup_mgeqslant nleftlvert X_m-Xrightrvert to 0mbox in probability,$$ which is equivalent to $X_nto X$ almost surely. Indeed, if $X_nto X$ almost surely, then $sup_mgeqslant nleftlvert X_m-Xrightrvert to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $left(S_nright)_ngeqslant 1$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
add a comment |Â
up vote
1
down vote
accepted
Condition (2) is equivalent to $$tag*S_n:=sup_mgeqslant nleftlvert X_m-Xrightrvert to 0mbox in probability,$$ which is equivalent to $X_nto X$ almost surely. Indeed, if $X_nto X$ almost surely, then $sup_mgeqslant nleftlvert X_m-Xrightrvert to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $left(S_nright)_ngeqslant 1$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Condition (2) is equivalent to $$tag*S_n:=sup_mgeqslant nleftlvert X_m-Xrightrvert to 0mbox in probability,$$ which is equivalent to $X_nto X$ almost surely. Indeed, if $X_nto X$ almost surely, then $sup_mgeqslant nleftlvert X_m-Xrightrvert to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $left(S_nright)_ngeqslant 1$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
Condition (2) is equivalent to $$tag*S_n:=sup_mgeqslant nleftlvert X_m-Xrightrvert to 0mbox in probability,$$ which is equivalent to $X_nto X$ almost surely. Indeed, if $X_nto X$ almost surely, then $sup_mgeqslant nleftlvert X_m-Xrightrvert to 0$ almost surely hence in probability. Conversely, the sequence of non-negative random variables $left(S_nright)_ngeqslant 1$ is non-increasing and has a subsequence which converges to $0$ almost surely hence it converges to $0$ almost surely.
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
answered Aug 3 at 20:34
Davide Giraudo
121k15146249
121k15146249
add a comment |Â
add a comment |Â
up vote
0
down vote
I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.
If we denote by $A_n(varepsilon)$ the event $$, then the event $X_n-X$ is just the
union of events
$$B_n(varepsilon)=bigcup_mgeq nA_m(varepsilon).$$
Obviously the sequence $B_n$ decreases, in the sense that $B_n+1subset B_n$ for all $n$. Elementary measure theory tells us that
$$lim_ntoinftyP(B_n(varepsilon))=Pleft(bigcap_n=1^inftyB_n(varepsilon)right)$$
but
$$bigcap_n=1^inftyB_n(varepsilon)=bigcap_n=1^inftybigcup_mgeq nA_m(varepsilon)=overlinelim_ntoinftyA_n(varepsilon)$$
where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation:
$$lim_ntoinftyP(B_n(varepsilon))=P(overlinelim_ntoinftyA_n(varepsilon))$$
which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have
$$P(overlinelim_ntoinftyA_n(varepsilon))geqoverlinelim_ntoinftyP(A_n(varepsilon)$$
answered Aug 3 at 18:36
uniquesolution
7,493721
add a comment |Â
up vote
0
down vote
I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.
If we denote by $A_n(varepsilon)$ the event $$, then the event $X_n-X$ is just the
union of events
$$B_n(varepsilon)=bigcup_mgeq nA_m(varepsilon).$$
Obviously the sequence $B_n$ decreases, in the sense that $B_n+1subset B_n$ for all $n$. Elementary measure theory tells us that
$$lim_ntoinftyP(B_n(varepsilon))=Pleft(bigcap_n=1^inftyB_n(varepsilon)right)$$
but
$$bigcap_n=1^inftyB_n(varepsilon)=bigcap_n=1^inftybigcup_mgeq nA_m(varepsilon)=overlinelim_ntoinftyA_n(varepsilon)$$
where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation:
$$lim_ntoinftyP(B_n(varepsilon))=P(overlinelim_ntoinftyA_n(varepsilon))$$
which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have
$$P(overlinelim_ntoinftyA_n(varepsilon))geqoverlinelim_ntoinftyP(A_n(varepsilon)$$
answered Aug 3 at 18:36
uniquesolution
7,493721
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.
If we denote by $A_n(varepsilon)$ the event $$, then the event $X_n-X$ is just the
union of events
$$B_n(varepsilon)=bigcup_mgeq nA_m(varepsilon).$$
Obviously the sequence $B_n$ decreases, in the sense that $B_n+1subset B_n$ for all $n$. Elementary measure theory tells us that
$$lim_ntoinftyP(B_n(varepsilon))=Pleft(bigcap_n=1^inftyB_n(varepsilon)right)$$
but
$$bigcap_n=1^inftyB_n(varepsilon)=bigcap_n=1^inftybigcup_mgeq nA_m(varepsilon)=overlinelim_ntoinftyA_n(varepsilon)$$
where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation:
$$lim_ntoinftyP(B_n(varepsilon))=P(overlinelim_ntoinftyA_n(varepsilon))$$
which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have
$$P(overlinelim_ntoinftyA_n(varepsilon))geqoverlinelim_ntoinftyP(A_n(varepsilon)$$
answered Aug 3 at 18:36
uniquesolution
7,493721
I don't know if there is a special name for the set you mentioned, but it can however be put in context within a measure-theoretic framework, as follows.
If we denote by $A_n(varepsilon)$ the event $$, then the event $X_n-X$ is just the
union of events
$$B_n(varepsilon)=bigcup_mgeq nA_m(varepsilon).$$
Obviously the sequence $B_n$ decreases, in the sense that $B_n+1subset B_n$ for all $n$. Elementary measure theory tells us that
$$lim_ntoinftyP(B_n(varepsilon))=Pleft(bigcap_n=1^inftyB_n(varepsilon)right)$$
but
$$bigcap_n=1^inftyB_n(varepsilon)=bigcap_n=1^inftybigcup_mgeq nA_m(varepsilon)=overlinelim_ntoinftyA_n(varepsilon)$$
where here we use directly the definition of an upper limit of a sequence of measurable sets. So we obtain the relation:
$$lim_ntoinftyP(B_n(varepsilon))=P(overlinelim_ntoinftyA_n(varepsilon))$$
which quantifies precisely the connection between your sets and the original ones, and shows why the condition is indeed stronger in general, because we always have
$$P(overlinelim_ntoinftyA_n(varepsilon))geqoverlinelim_ntoinftyP(A_n(varepsilon)$$
answered Aug 3 at 18:36
uniquesolution
7,493721
answered Aug 3 at 18:36
uniquesolution
7,493721
answered Aug 3 at 18:36
uniquesolution
7,493721
answered Aug 3 at 18:36
uniquesolution
7,493721
7,493721
add a comment |Â
add a comment |Â
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Maybe you mean something like $forallepsilon > 0:quad Pleft(suplimits_ m> n |X_m - X|geq epsilonright)to 0,quad ntoinfty. tag2$
– callculus
Aug 3 at 17:39