How can a probability measure satisfy the finite super additivity property?Semialgebra logical errorfinite additivity&countable additivityShowing countable additivitiy of Lebesgue measureIs $P(bigcup_k=1^infty A_k)=lim_nrightarrowinftyP(bigcup_k=1^n A_k)$ equivalent to countable additivity?A Special Property of Finite Probability Space.Probability measure is countably additive over almost disjoint sets.Show that the function $mathbbQ(A) = sum _ i:omega _ i in A ^,p_i$ is a probability measure on $(Omega, mathcalF)$Proving Countable Additivity From Finite AdditivityFrom additivity of the measure, couldn't we prove the $sigma -$additivity?Countable additivity does not imply continuity at $emptyset$ for non-finite measure

How do ground effect vehicles perform turns?

Why is Arduino resetting while driving motors?

Is there a conventional notation or name for the slip angle?

If a character with the Alert feat rolls a crit fail on their Perception check, are they surprised?

MAXDOP Settings for SQL Server 2014

Can someone explain how this makes sense electrically?

When quoting, must I also copy hyphens used to divide words that continue on the next line?

How do I extrude a face to a single vertex

A Permanent Norse Presence in America

Freedom of speech and where it applies

Is it possible to have a strip of cold climate in the middle of a planet?

Do Legal Documents Require Signing In Standard Pen Colors?

Indicating multiple different modes of speech (fantasy language or telepathy)

Is camera lens focus an exact point or a range?

Can the Supreme Court overturn an impeachment?

Is it improper etiquette to ask your opponent what his/her rating is before the game?

A social experiment. What is the worst that can happen?

Should I install hardwood flooring or cabinets first?

Greco-Roman egalitarianism

How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?

How should I respond when I lied about my education and the company finds out through background check?

Query about absorption line spectra

How to color a curve

Do the concepts of IP address and network interface not belong to the same layer?



How can a probability measure satisfy the finite super additivity property?


Semialgebra logical errorfinite additivity&countable additivityShowing countable additivitiy of Lebesgue measureIs $P(bigcup_k=1^infty A_k)=lim_nrightarrowinftyP(bigcup_k=1^n A_k)$ equivalent to countable additivity?A Special Property of Finite Probability Space.Probability measure is countably additive over almost disjoint sets.Show that the function $mathbbQ(A) = sum _ i:omega _ i in A ^,p_i$ is a probability measure on $(Omega, mathcalF)$Proving Countable Additivity From Finite AdditivityFrom additivity of the measure, couldn't we prove the $sigma -$additivity?Countable additivity does not imply continuity at $emptyset$ for non-finite measure













1












$begingroup$


When introducing the Extention Theorem the book said:



"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.



P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "



I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?










share|cite|improve this question









$endgroup$











  • $begingroup$
    If only $=$ is satisfied still $geq$ is true.
    $endgroup$
    – user647486
    Mar 16 at 10:52










  • $begingroup$
    So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
    $endgroup$
    – GEdgar
    Mar 16 at 11:27
















1












$begingroup$


When introducing the Extention Theorem the book said:



"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.



P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "



I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?










share|cite|improve this question









$endgroup$











  • $begingroup$
    If only $=$ is satisfied still $geq$ is true.
    $endgroup$
    – user647486
    Mar 16 at 10:52










  • $begingroup$
    So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
    $endgroup$
    – GEdgar
    Mar 16 at 11:27














1












1








1





$begingroup$


When introducing the Extention Theorem the book said:



"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.



P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "



I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?










share|cite|improve this question









$endgroup$




When introducing the Extention Theorem the book said:



"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.



P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "



I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?







probability probability-theory measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 16 at 10:47









QwertfordQwertford

323212




323212











  • $begingroup$
    If only $=$ is satisfied still $geq$ is true.
    $endgroup$
    – user647486
    Mar 16 at 10:52










  • $begingroup$
    So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
    $endgroup$
    – GEdgar
    Mar 16 at 11:27

















  • $begingroup$
    If only $=$ is satisfied still $geq$ is true.
    $endgroup$
    – user647486
    Mar 16 at 10:52










  • $begingroup$
    So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
    $endgroup$
    – GEdgar
    Mar 16 at 11:27
















$begingroup$
If only $=$ is satisfied still $geq$ is true.
$endgroup$
– user647486
Mar 16 at 10:52




$begingroup$
If only $=$ is satisfied still $geq$ is true.
$endgroup$
– user647486
Mar 16 at 10:52












$begingroup$
So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
$endgroup$
– GEdgar
Mar 16 at 11:27





$begingroup$
So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
$endgroup$
– GEdgar
Mar 16 at 11:27











0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150280%2fhow-can-a-probability-measure-satisfy-the-finite-super-additivity-property%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150280%2fhow-can-a-probability-measure-satisfy-the-finite-super-additivity-property%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Solar Wings Breeze Design and development Specifications (Breeze) References Navigation menu1368-485X"Hang glider: Breeze (Solar Wings)"e

Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee

Method to test if a number is a perfect power? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Detecting perfect squares faster than by extracting square rooteffective way to get the integer sequence A181392 from oeisA rarely mentioned fact about perfect powersHow many numbers such $n$ are there that $n<100,lfloorsqrtn rfloor mid n$Check perfect squareness by modulo division against multiple basesFor what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.Do there exist any positive integers $n$ such that $lfloore^nrfloor$ is a perfect power? What is the probability that one exists?finding perfect power factors of an integerProve that the sequence contains a perfect square for any natural number $m $ in the domain of $f$ .Counting Perfect Powers