Infinitely often and almost always - Revision history

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:37:35Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:37, 31 July 2019
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	! perspective !! infinitely often !! almost always		! perspective !! infinitely often !! almost always
	\|-		\|-
	\| \|\| <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent \|\| <math>(a_n)_{n=1}^\infty</math> is eventually <math>\epsilon</math>-close to <math>x</math>		\| Tao's terminology (see his ''Analysis'') \|\| <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent \|\| <math>(a_n)_{n=1}^\infty</math> is eventually <math>\epsilon</math>-close to <math>x</math>
			\|-
			\| first-order quantifier \|\| for every <math>N \geq 1</math> there exists some <math>n \geq N</math> such that <math>\|a_n - x\| \leq \epsilon</math> \|\| there exists some <math>N \geq 1</math> such that for all <math>n \geq N</math> we have <math>\|a_n - x\| \leq \epsilon</math>
	\|-		\|-
	\| \|\| <math>\|a_n - x\| \leq \epsilon</math> for infinitely many <math>n</math> \|\| <math>\|a_n - x\| \leq \epsilon</math> for all but finitely many <math>n</math>		\| \|\| <math>\|a_n - x\| \leq \epsilon</math> for infinitely many <math>n</math> \|\| <math>\|a_n - x\| \leq \epsilon</math> for all but finitely many <math>n</math>

IssaRice at 22:35, 31 July 2019

2019-07-31T22:35:05Z

← Older revision		Revision as of 22:35, 31 July 2019
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	\|-		\|-
	\| \|\| <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> \|\| <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math>		\| \|\| <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> \|\| <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math>
			\|-
			\| \|\| <math>x \in \limsup_{n\to\infty} A_n</math> \|\| <math>x \in \liminf_{n\to\infty} A_n</math>
	\|}		\|}

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:34:27Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:34, 31 July 2019
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	but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).		but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).

			{\| class="wikitable"
			\|-
			! perspective !! infinitely often !! almost always
			\|-
			\| \|\| <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent \|\| <math>(a_n)_{n=1}^\infty</math> is eventually <math>\epsilon</math>-close to <math>x</math>
			\|-
			\| \|\| <math>\|a_n - x\| \leq \epsilon</math> for infinitely many <math>n</math> \|\| <math>\|a_n - x\| \leq \epsilon</math> for all but finitely many <math>n</math>
			\|-
			\| \|\| <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> \|\| <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math>
			\|}

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:31:23Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:31, 31 July 2019
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	<math>\liminf_{n\to\infty} \|a_n-x\| = \sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.		<math>\liminf_{n\to\infty} \|a_n-x\| = \sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.

	but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n~~)_{n=1}^\infty~~</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n~~)_{n=1}^\infty~~</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).		but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>a_n</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:23:55Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:23, 31 July 2019
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	I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.		I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.

	<math>\inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>		<math>\limsup_{n\to\infty} \|a_n-x\| = \inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>

	<math>\sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.		<math>\liminf_{n\to\infty} \|a_n-x\| = \sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.

			but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n)_{n=1}^\infty</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n)_{n=1}^\infty</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:10:48Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:10, 31 July 2019
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	<math>\inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>		<math>\inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>

	<math>\sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math>		<math>\sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T22:06:44Z

Analogy with sequences of real numbers

← Older revision		Revision as of 22:06, 31 July 2019
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	I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.		I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.

			<math>\inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>

			<math>\sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math>

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T21:54:33Z

Analogy with sequences of real numbers

← Older revision		Revision as of 21:54, 31 July 2019
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	==Analogy with sequences of real numbers==		==Analogy with sequences of real numbers==

	Let <math>(a_n)_{n=1}^\infty</math> be a sequence of real numbers, ~~and~~ let <math>\epsilon > 0</math> be a real number.		Let <math>(a_n)_{n=1}^\infty</math> be a sequence of real numbers, let <math>\epsilon > 0</math> be a real number, and let <math>x</math> be a real number.

	We say <math>(a_n)_{n=1}^\infty</math> is eventually <math>\epsilon</math>-close to <math>x</math> iff there exists some <math>N \geq 1</math> such that for all <math>n \geq N</math> we have <math>\|a_n - x\| \leq \epsilon</math>.		We say <math>(a_n)_{n=1}^\infty</math> is eventually <math>\epsilon</math>-close to <math>x</math> iff there exists some <math>N \geq 1</math> such that for all <math>n \geq N</math> we have <math>\|a_n - x\| \leq \epsilon</math>.
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	We say that <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent iff for every <math>N \geq 1</math> there exists some <math>n \geq N</math> such that <math>\|a_n - x\| \leq \epsilon</math>.		We say that <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent iff for every <math>N \geq 1</math> there exists some <math>n \geq N</math> such that <math>\|a_n - x\| \leq \epsilon</math>.

	I think we can even define <math>A_n := \{x : \|a_n-x\| \leq \epsilon\}</math>.		I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.

	[[Category:Probability]]		[[Category:Probability]]

IssaRice: /* Analogy with sequences of real numbers */

2019-07-31T21:53:20Z

Analogy with sequences of real numbers

← Older revision		Revision as of 21:53, 31 July 2019
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	We say that <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent iff for every <math>N \geq 1</math> there exists some <math>n \geq N</math> such that <math>\|a_n - x\| \leq \epsilon</math>.		We say that <math>(a_n)_{n=1}^\infty</math> is continually <math>\epsilon</math>-adherent iff for every <math>N \geq 1</math> there exists some <math>n \geq N</math> such that <math>\|a_n - x\| \leq \epsilon</math>.

			I think we can even define <math>A_n := \{x : \|a_n-x\| \leq \epsilon\}</math>.

	[[Category:Probability]]		[[Category:Probability]]

IssaRice at 21:49, 31 July 2019

2019-07-31T21:49:04Z

← Older revision		Revision as of 21:49, 31 July 2019
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	\| unions and intersections \|\| <math>\omega \in \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty A_n</math> \|\| <math>\omega \in \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty A_n</math>		\| unions and intersections \|\| <math>\omega \in \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty A_n</math> \|\| <math>\omega \in \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty A_n</math>
	\|-		\|-
	\| first-order quantifiers \|\| <math>\forall N\geq 1\ \exists n \geq N\colon \omega \in A_n</math> \|\| <math>\exists N \~~leq~~ 1\ \forall n \geq N\colon \omega \in A_n</math>		\| first-order quantifiers \|\| <math>\forall N\geq 1\ \exists n \geq N\colon \omega \in A_n</math> \|\| <math>\exists N \geq 1\ \forall n \geq N\colon \omega \in A_n</math>
	\|-		\|-
	\| verbal expression \|\| <math>\omega \in A_n</math> for infinitely many <math>n\geq 1</math> \|\| <math>\omega \in A_n</math> for almost all <math>n\geq 1</math>, i.e. <math>\omega \in A_n</math> for all but finitely many <math>n \geq 1</math>, i.e. <math>\omega \notin A_n</math> for finitely many <math>n \geq 1</math>		\| verbal expression \|\| <math>\omega \in A_n</math> for infinitely many <math>n\geq 1</math> \|\| <math>\omega \in A_n</math> for almost all <math>n\geq 1</math>, i.e. <math>\omega \in A_n</math> for all but finitely many <math>n \geq 1</math>, i.e. <math>\omega \notin A_n</math> for finitely many <math>n \geq 1</math>

← Older revision		Revision as of 22:23, 31 July 2019
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	I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.		I think we can even define <math>A_n := \{x \in \mathbf R : \|a_n-x\| \leq \epsilon\}</math>.

	<math>\inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>		<math>\limsup_{n\to\infty} \|a_n-x\| = \inf_{N \geq 1} \sup_{n \geq N} \|a_n - x\| \leq \epsilon</math>

	<math>\sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.		<math>\liminf_{n\to\infty} \|a_n-x\| = \sup_{N \geq 1} \inf_{n \geq N} \|a_n - x\| \leq \epsilon</math> -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.

			but this makes sense in terms of strength of "infinitely often" vs "almost always". We have <math>\liminf_{n\to\infty} \|a_n-x\| \leq \limsup_{n\to\infty} \|a_n-x\|</math>, so if <math>\limsup_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n)_{n=1}^\infty</math> is <math>\epsilon</math>-close to <math>x</math> almost always) then <math>\liminf_{n\to\infty} \|a_n-x\| \leq \epsilon</math> (i.e. <math>(a_n)_{n=1}^\infty</math> is <math>\epsilon</math>-close to <math>x</math> infinitely often).

	[[Category:Probability]]		[[Category:Probability]]