What is an uncountable union of events? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Sequence of events probability proofWhat are the chances of getting $K$ identical outcomes in a row in a series of $Y$ independent events?Probability of a UnionAbout event in sample spaceShow that the collection of all subsets of a finite sample space satisfies the 3 conditions to be called a collection of events.How to find union and intersection of events?Number of possible events is $2^N$ ( $N$ is the number of outcomes )?Populations and Sample SpacesEvents and sample spaceHow can we define the Ordered Union of events?
ListPlot join points by nearest neighbor rather than order
Why constant symbols in a language?
How to deal with a team lead who never gives me credit?
Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?
Is the Standard Deduction better than Itemized when both are the same amount?
Why did the IBM 650 use bi-quinary?
If Jon Snow became King of the Seven Kingdoms what would his regnal number be?
Gastric acid as a weapon
Did Kevin spill real chili?
Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?
Proof involving the spectral radius and the Jordan canonical form
Stars Make Stars
Is there a documented rationale why the House Ways and Means chairman can demand tax info?
How to recreate this effect in Photoshop?
Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?
How can players work together to take actions that are otherwise impossible?
What is the correct way to use the pinch test for dehydration?
Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?
What is a Meta algorithm?
3 doors, three guards, one stone
Dominant seventh chord in the major scale contains diminished triad of the seventh?
What are the motives behind Cersei's orders given to Bronn?
Right-skewed distribution with mean equals to mode?
Do I really need recursive chmod to restrict access to a folder?
What is an uncountable union of events?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Sequence of events probability proofWhat are the chances of getting $K$ identical outcomes in a row in a series of $Y$ independent events?Probability of a UnionAbout event in sample spaceShow that the collection of all subsets of a finite sample space satisfies the 3 conditions to be called a collection of events.How to find union and intersection of events?Number of possible events is $2^N$ ( $N$ is the number of outcomes )?Populations and Sample SpacesEvents and sample spaceHow can we define the Ordered Union of events?
$begingroup$
In DeGroot and Schervish's 'Probability and Statistics'(4th editon), they define the sample space as the union of all outcomes, and an event as a set of possible outcomes.
Definition 1.3.1 on page 5.
"Experiment and Event.
An experiment is any process, real or hypothetical, in which the possible outcomes can be identified ahead of time. An event is a well-defined set of possible outcomes of the experiment.
"
They then list 3 conditions that an event must satisfy, but I'm confused with the 3rd condition.
Condition 3 on page 10:
"If $A_1$, $A_2$, . . . is a countable collection of events, then $bigcup_i=1^infty A_i$ is also an event.
In other words, if we choose to call each set of outcomes in some countable collection an event, we are required to call their union an event also. We do not require that the union of an arbitrary collection of events be an event. To be clear, let I be an arbitrary set that we use to index a general collection of events $A_i : i in I $. The union of the events in this collection is the set of outcomes that are in at least one of the events in the collection. The notation for this union is $bigcup_i in I A_i$. We do not require that $bigcup_i in I A_i$ be an event unless I is countable.
Condition 3 refers to a countable collection of events. We can prove that the condition also applies to every finite collection of events.
"
What confuses me is that, is there an uncountable union of events, that is not an event?
probability statistics
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
$endgroup$
add a comment |
$begingroup$
In DeGroot and Schervish's 'Probability and Statistics'(4th editon), they define the sample space as the union of all outcomes, and an event as a set of possible outcomes.
Definition 1.3.1 on page 5.
"Experiment and Event.
An experiment is any process, real or hypothetical, in which the possible outcomes can be identified ahead of time. An event is a well-defined set of possible outcomes of the experiment.
"
They then list 3 conditions that an event must satisfy, but I'm confused with the 3rd condition.
Condition 3 on page 10:
"If $A_1$, $A_2$, . . . is a countable collection of events, then $bigcup_i=1^infty A_i$ is also an event.
In other words, if we choose to call each set of outcomes in some countable collection an event, we are required to call their union an event also. We do not require that the union of an arbitrary collection of events be an event. To be clear, let I be an arbitrary set that we use to index a general collection of events $A_i : i in I $. The union of the events in this collection is the set of outcomes that are in at least one of the events in the collection. The notation for this union is $bigcup_i in I A_i$. We do not require that $bigcup_i in I A_i$ be an event unless I is countable.
Condition 3 refers to a countable collection of events. We can prove that the condition also applies to every finite collection of events.
"
What confuses me is that, is there an uncountable union of events, that is not an event?
probability statistics
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
$endgroup$
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57
add a comment |
$begingroup$
In DeGroot and Schervish's 'Probability and Statistics'(4th editon), they define the sample space as the union of all outcomes, and an event as a set of possible outcomes.
Definition 1.3.1 on page 5.
"Experiment and Event.
An experiment is any process, real or hypothetical, in which the possible outcomes can be identified ahead of time. An event is a well-defined set of possible outcomes of the experiment.
"
They then list 3 conditions that an event must satisfy, but I'm confused with the 3rd condition.
Condition 3 on page 10:
"If $A_1$, $A_2$, . . . is a countable collection of events, then $bigcup_i=1^infty A_i$ is also an event.
In other words, if we choose to call each set of outcomes in some countable collection an event, we are required to call their union an event also. We do not require that the union of an arbitrary collection of events be an event. To be clear, let I be an arbitrary set that we use to index a general collection of events $A_i : i in I $. The union of the events in this collection is the set of outcomes that are in at least one of the events in the collection. The notation for this union is $bigcup_i in I A_i$. We do not require that $bigcup_i in I A_i$ be an event unless I is countable.
Condition 3 refers to a countable collection of events. We can prove that the condition also applies to every finite collection of events.
"
What confuses me is that, is there an uncountable union of events, that is not an event?
probability statistics
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
$endgroup$
In DeGroot and Schervish's 'Probability and Statistics'(4th editon), they define the sample space as the union of all outcomes, and an event as a set of possible outcomes.
Definition 1.3.1 on page 5.
"Experiment and Event.
An experiment is any process, real or hypothetical, in which the possible outcomes can be identified ahead of time. An event is a well-defined set of possible outcomes of the experiment.
"
They then list 3 conditions that an event must satisfy, but I'm confused with the 3rd condition.
Condition 3 on page 10:
"If $A_1$, $A_2$, . . . is a countable collection of events, then $bigcup_i=1^infty A_i$ is also an event.
In other words, if we choose to call each set of outcomes in some countable collection an event, we are required to call their union an event also. We do not require that the union of an arbitrary collection of events be an event. To be clear, let I be an arbitrary set that we use to index a general collection of events $A_i : i in I $. The union of the events in this collection is the set of outcomes that are in at least one of the events in the collection. The notation for this union is $bigcup_i in I A_i$. We do not require that $bigcup_i in I A_i$ be an event unless I is countable.
Condition 3 refers to a countable collection of events. We can prove that the condition also applies to every finite collection of events.
"
What confuses me is that, is there an uncountable union of events, that is not an event?
probability statistics
probability statistics
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
edited Mar 26 at 9:57
Brent Wagner
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
asked Mar 26 at 9:45
Brent WagnerBrent Wagner
304
304
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57
add a comment |
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, there exist sets that cannot be written as countable unions of events. Unfortunatelly, they are not constructible, and you need the axiom of choice to prove them.
An example of such sets are the Vitali sets, which cannot be written as countable unions of intervals.
Note: this is the part of probability theory that most coincides with a field called measure theory. In measure theory, we are interested in whether sets are "measurable" or not. If the "measure" of a space is precisely $1$, then we call that space a probability space, and a "measurable set" in a probability space is called an "event". So, under some assumptions, "non-events" are the same thing as "non-measurable sets".
In short, if you are interested in probability theory, I highly suggest you take a class on measure theory sometime in the future. It makes probability much easier and more interesting at the same time.
answered Mar 26 at 9:53
5xum5xum
92.7k394162
$endgroup$
add a comment |
$begingroup$
Yes. Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be $0$ and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points $mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $alphain (0^circ, 360^circ)$ results in a new set of points $mathscr A_alpha$. None of these $mathscr A_alpha$ have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability $p$ to picking a point in $mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $mathscr A_alpha$. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those $p$'s should be $1$. Thus we have
$$
sum_alphain [0, 360)cap Bbb Qp = 1
$$
But if $p$ is $0$, the sum is $0$, and if $p$ is positive, then the sum is infinite. So it is impossible to assign a probability to this union $mathscr A_0$, and therefore we are better off not calling it an event.
answered Mar 26 at 9:53
ArthurArthur
123k7122211
$endgroup$
add a comment |
Your Answer
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162954%2fwhat-is-an-uncountable-union-of-events%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, there exist sets that cannot be written as countable unions of events. Unfortunatelly, they are not constructible, and you need the axiom of choice to prove them.
An example of such sets are the Vitali sets, which cannot be written as countable unions of intervals.
Note: this is the part of probability theory that most coincides with a field called measure theory. In measure theory, we are interested in whether sets are "measurable" or not. If the "measure" of a space is precisely $1$, then we call that space a probability space, and a "measurable set" in a probability space is called an "event". So, under some assumptions, "non-events" are the same thing as "non-measurable sets".
In short, if you are interested in probability theory, I highly suggest you take a class on measure theory sometime in the future. It makes probability much easier and more interesting at the same time.
answered Mar 26 at 9:53
5xum5xum
92.7k394162
$endgroup$
add a comment |
$begingroup$
Yes, there exist sets that cannot be written as countable unions of events. Unfortunatelly, they are not constructible, and you need the axiom of choice to prove them.
An example of such sets are the Vitali sets, which cannot be written as countable unions of intervals.
Note: this is the part of probability theory that most coincides with a field called measure theory. In measure theory, we are interested in whether sets are "measurable" or not. If the "measure" of a space is precisely $1$, then we call that space a probability space, and a "measurable set" in a probability space is called an "event". So, under some assumptions, "non-events" are the same thing as "non-measurable sets".
In short, if you are interested in probability theory, I highly suggest you take a class on measure theory sometime in the future. It makes probability much easier and more interesting at the same time.
answered Mar 26 at 9:53
5xum5xum
92.7k394162
$endgroup$
add a comment |
$begingroup$
Yes, there exist sets that cannot be written as countable unions of events. Unfortunatelly, they are not constructible, and you need the axiom of choice to prove them.
An example of such sets are the Vitali sets, which cannot be written as countable unions of intervals.
Note: this is the part of probability theory that most coincides with a field called measure theory. In measure theory, we are interested in whether sets are "measurable" or not. If the "measure" of a space is precisely $1$, then we call that space a probability space, and a "measurable set" in a probability space is called an "event". So, under some assumptions, "non-events" are the same thing as "non-measurable sets".
In short, if you are interested in probability theory, I highly suggest you take a class on measure theory sometime in the future. It makes probability much easier and more interesting at the same time.
answered Mar 26 at 9:53
5xum5xum
92.7k394162
$endgroup$
Yes, there exist sets that cannot be written as countable unions of events. Unfortunatelly, they are not constructible, and you need the axiom of choice to prove them.
An example of such sets are the Vitali sets, which cannot be written as countable unions of intervals.
Note: this is the part of probability theory that most coincides with a field called measure theory. In measure theory, we are interested in whether sets are "measurable" or not. If the "measure" of a space is precisely $1$, then we call that space a probability space, and a "measurable set" in a probability space is called an "event". So, under some assumptions, "non-events" are the same thing as "non-measurable sets".
In short, if you are interested in probability theory, I highly suggest you take a class on measure theory sometime in the future. It makes probability much easier and more interesting at the same time.
answered Mar 26 at 9:53
5xum5xum
92.7k394162
answered Mar 26 at 9:53
5xum5xum
92.7k394162
answered Mar 26 at 9:53
5xum5xum
92.7k394162
answered Mar 26 at 9:53
5xum5xum
92.7k394162
92.7k394162
add a comment |
add a comment |
$begingroup$
Yes. Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be $0$ and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points $mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $alphain (0^circ, 360^circ)$ results in a new set of points $mathscr A_alpha$. None of these $mathscr A_alpha$ have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability $p$ to picking a point in $mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $mathscr A_alpha$. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those $p$'s should be $1$. Thus we have
$$
sum_alphain [0, 360)cap Bbb Qp = 1
$$
But if $p$ is $0$, the sum is $0$, and if $p$ is positive, then the sum is infinite. So it is impossible to assign a probability to this union $mathscr A_0$, and therefore we are better off not calling it an event.
answered Mar 26 at 9:53
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
Yes. Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be $0$ and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points $mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $alphain (0^circ, 360^circ)$ results in a new set of points $mathscr A_alpha$. None of these $mathscr A_alpha$ have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability $p$ to picking a point in $mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $mathscr A_alpha$. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those $p$'s should be $1$. Thus we have
$$
sum_alphain [0, 360)cap Bbb Qp = 1
$$
But if $p$ is $0$, the sum is $0$, and if $p$ is positive, then the sum is infinite. So it is impossible to assign a probability to this union $mathscr A_0$, and therefore we are better off not calling it an event.
answered Mar 26 at 9:53
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
Yes. Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be $0$ and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points $mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $alphain (0^circ, 360^circ)$ results in a new set of points $mathscr A_alpha$. None of these $mathscr A_alpha$ have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability $p$ to picking a point in $mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $mathscr A_alpha$. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those $p$'s should be $1$. Thus we have
$$
sum_alphain [0, 360)cap Bbb Qp = 1
$$
But if $p$ is $0$, the sum is $0$, and if $p$ is positive, then the sum is infinite. So it is impossible to assign a probability to this union $mathscr A_0$, and therefore we are better off not calling it an event.
answered Mar 26 at 9:53
ArthurArthur
123k7122211
$endgroup$
Yes. Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be $0$ and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points $mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $alphain (0^circ, 360^circ)$ results in a new set of points $mathscr A_alpha$. None of these $mathscr A_alpha$ have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability $p$ to picking a point in $mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $mathscr A_alpha$. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those $p$'s should be $1$. Thus we have
$$
sum_alphain [0, 360)cap Bbb Qp = 1
$$
But if $p$ is $0$, the sum is $0$, and if $p$ is positive, then the sum is infinite. So it is impossible to assign a probability to this union $mathscr A_0$, and therefore we are better off not calling it an event.
answered Mar 26 at 9:53
ArthurArthur
123k7122211
edited Mar 26 at 10:09
answered Mar 26 at 9:53
ArthurArthur
123k7122211
answered Mar 26 at 9:53
ArthurArthur
123k7122211
answered Mar 26 at 9:53
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162954%2fwhat-is-an-uncountable-union-of-events%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
You are going to want to talk about the probability of events. This is based on measure theory. If you allow the arbitrary construction of events as the uncountable union of other events then you may leave open the possibility of non-measurable sets and so events which do not have a probability.
$endgroup$
– Henry
Mar 26 at 9:57