Is it possible for an event A to be independent from event B, but not the other way around? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Zero probability and impossibilityExchangeable Random Variable but not independent?Probability of being away from mean for independent random variablesAn example of two random variables that are mean independent but not independentCalculating probability when order matters only sometimesIf X is independent to Y and Z, does it imply that X is independent to YZ ?Representing pairwise-independent but not independent occurrences with venn diagramPairwise independent events but not mutually independentExamples of situation in which two events are independent but one event can be predicted perfectly once we know if the other happened or not.Suppose $A $ and $B$ are independent events. For an event $C $ such that $P(C) > 0$ , prove that the event of $A$ given $C $
Why is a lens darker than other ones when applying the same settings?
Does the Mueller report show a conspiracy between Russia and the Trump Campaign?
How does light 'choose' between wave and particle behaviour?
How can I prevent/balance waiting and turtling as a response to cooldown mechanics
What is the chair depicted in Cesare Maccari's 1889 painting "Cicerone denuncia Catilina"?
What initially awakened the Balrog?
Why is std::move not [[nodiscard]] in C++20?
New Order #6: Easter Egg
The test team as an enemy of development? And how can this be avoided?
Why weren't discrete x86 CPUs ever used in game hardware?
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
How can I save and copy a screenhot at the same time?
Trying to understand entropy as a novice in thermodynamics
Co-worker has annoying ringtone
Mounting TV on a weird wall that has some material between the drywall and stud
Can an iPhone 7 be made to function as a NFC Tag?
Where is the Next Backup Size entry on iOS 12?
Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?
What is the "studentd" process?
What does 丫 mean? 丫是什么意思?
Tips to organize LaTeX presentations for a semester
What does Turing mean by this statement?
Rationale for describing kurtosis as "peakedness"?
How to write capital alpha?
Is it possible for an event A to be independent from event B, but not the other way around?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Zero probability and impossibilityExchangeable Random Variable but not independent?Probability of being away from mean for independent random variablesAn example of two random variables that are mean independent but not independentCalculating probability when order matters only sometimesIf X is independent to Y and Z, does it imply that X is independent to YZ ?Representing pairwise-independent but not independent occurrences with venn diagramPairwise independent events but not mutually independentExamples of situation in which two events are independent but one event can be predicted perfectly once we know if the other happened or not.Suppose $A $ and $B$ are independent events. For an event $C $ such that $P(C) > 0$ , prove that the event of $A$ given $C $
$begingroup$
I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ doesn't make sense, so then $B$ wouldn't be independent from $A$?
Thank you
probability-theory independence
New contributor
$endgroup$
add a comment |
$begingroup$
I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ doesn't make sense, so then $B$ wouldn't be independent from $A$?
Thank you
probability-theory independence
New contributor
$endgroup$
add a comment |
$begingroup$
I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ doesn't make sense, so then $B$ wouldn't be independent from $A$?
Thank you
probability-theory independence
New contributor
$endgroup$
I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ doesn't make sense, so then $B$ wouldn't be independent from $A$?
Thank you
probability-theory independence
probability-theory independence
New contributor
New contributor
New contributor
asked 2 hours ago
MashpaMashpa
273
273
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.
$endgroup$
add a comment |
$begingroup$
$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$
$endgroup$
add a comment |
$begingroup$
$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.
$endgroup$
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
);
);
Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3195328%2fis-it-possible-for-an-event-a-to-be-independent-from-event-b-but-not-the-other%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.
$endgroup$
add a comment |
$begingroup$
$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.
$endgroup$
add a comment |
$begingroup$
$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.
$endgroup$
$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
1,78828
add a comment |
add a comment |
$begingroup$
$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$
$endgroup$
add a comment |
$begingroup$
$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$
$endgroup$
add a comment |
$begingroup$
$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$
$endgroup$
$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
add a comment |
add a comment |
$begingroup$
$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.
$endgroup$
add a comment |
$begingroup$
$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.
$endgroup$
add a comment |
$begingroup$
$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.
$endgroup$
$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.
answered 2 hours ago
amdamd
32k21053
32k21053
add a comment |
add a comment |
Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3195328%2fis-it-possible-for-an-event-a-to-be-independent-from-event-b-but-not-the-other%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