Durrett Example 1.9 - Pairwise independence does not imply mutual independence? The Next CEO of Stack Overflowmutual independence implies pairwise independence show that the converse is not true.If three events are pairwise independent, are they independent “collectively”?Does independence from independent events imply joint independence?Having birthday at the same dayProbability that grows from the first of January until the end of December. How to compute?Birthday paradoxSolving birthday problem without complementInspired by the Birthday ProblemPairwise independence + additional condition imply independenceProviding Example that independence of events $A, B$ is dependent on Probability measureTennis Paradox ProbabilityWhy does my approach to the birthday problem(with 3 people) produce a wrong result?Probability with Alice and Betty
Find a path from s to t using as few red nodes as possible
Gödel's incompleteness theorems - what are the religious implications?
How can I replace x-axis labels with pre-determined symbols?
What steps are necessary to read a Modern SSD in Medieval Europe?
How can the PCs determine if an item is a phylactery?
How can I separate the number from the unit in argument?
Can I cast Thunderwave and be at the center of its bottom face, but not be affected by it?
Is it OK to decorate a log book cover?
What does this strange code stamp on my passport mean?
Planeswalker Ability and Death Timing
Ising model simulation
Arity of Primitive Recursive Functions
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
Is the offspring between a demon and a celestial possible? If so what is it called and is it in a book somewhere?
Man transported from Alternate World into ours by a Neutrino Detector
Avoiding the "not like other girls" trope?
Is it okay to majorly distort historical facts while writing a fiction story?
Why doesn't Shulchan Aruch include the laws of destroying fruit trees?
Gauss' Posthumous Publications?
Creating a script with console commands
How to coordinate airplane tickets?
How to show a landlord what we have in savings?
Could a dragon use its wings to swim?
How to pronounce fünf in 45
Durrett Example 1.9 - Pairwise independence does not imply mutual independence?
The Next CEO of Stack Overflowmutual independence implies pairwise independence show that the converse is not true.If three events are pairwise independent, are they independent “collectively”?Does independence from independent events imply joint independence?Having birthday at the same dayProbability that grows from the first of January until the end of December. How to compute?Birthday paradoxSolving birthday problem without complementInspired by the Birthday ProblemPairwise independence + additional condition imply independenceProviding Example that independence of events $A, B$ is dependent on Probability measureTennis Paradox ProbabilityWhy does my approach to the birthday problem(with 3 people) produce a wrong result?Probability with Alice and Betty
$begingroup$
The example in question is from Rick Durrett's "Elementrary Probability for Applications", and the setup is something like this:
Let $A$ be the event "Alice and Betty have the same birthday", $B$ be the event "Betty and Carol have the same birthday", and $C$ be the event "Carol and Alice have the same birthday".
Durrett goes on to demonstrate that each pair is independent, since for example,
$P(A cap B) = P(A)P(B)$.
However, he concludes that $A, B$, and $C$ are not independent, since
$P(A cap B cap C) = frac1365^2 neq frac1365^3 = P(A)P(B)P(C)$.
I understand the reasoning here, and that one can generally show that arbitrary events $X$ and $Y$ are not independent by showing that $P(Xcap Y) neq P(X)P(Y)$.
I am a little new to probability, though, and don't understand why exactly $P(A cap B cap C) = frac1365^2$.
My progress so far:
I do see why $P(A) = P(B) =P(C) = frac1365$, and thus why $P(A)P(B)P(C) = frac1365^3$.
It seems like the sample space $Omega = (a, b, c) mid a,b,c in [365] $ -- i.e., all of the possible triples of numbers from 1 to 365, where 1 denotes January 1st, 2 denotes January 2nd, etc. From that, I can conclude $|Omega| = 365^3$, but I'm not sure where to go from here.
It seems like once a single birthday is chosen, the rest are completely determined if they're all equal to each other - is this a good direction to go in?
probability independence
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
$endgroup$
add a comment |
$begingroup$
The example in question is from Rick Durrett's "Elementrary Probability for Applications", and the setup is something like this:
Let $A$ be the event "Alice and Betty have the same birthday", $B$ be the event "Betty and Carol have the same birthday", and $C$ be the event "Carol and Alice have the same birthday".
Durrett goes on to demonstrate that each pair is independent, since for example,
$P(A cap B) = P(A)P(B)$.
However, he concludes that $A, B$, and $C$ are not independent, since
$P(A cap B cap C) = frac1365^2 neq frac1365^3 = P(A)P(B)P(C)$.
I understand the reasoning here, and that one can generally show that arbitrary events $X$ and $Y$ are not independent by showing that $P(Xcap Y) neq P(X)P(Y)$.
I am a little new to probability, though, and don't understand why exactly $P(A cap B cap C) = frac1365^2$.
My progress so far:
I do see why $P(A) = P(B) =P(C) = frac1365$, and thus why $P(A)P(B)P(C) = frac1365^3$.
It seems like the sample space $Omega = (a, b, c) mid a,b,c in [365] $ -- i.e., all of the possible triples of numbers from 1 to 365, where 1 denotes January 1st, 2 denotes January 2nd, etc. From that, I can conclude $|Omega| = 365^3$, but I'm not sure where to go from here.
It seems like once a single birthday is chosen, the rest are completely determined if they're all equal to each other - is this a good direction to go in?
probability independence
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
$endgroup$
add a comment |
$begingroup$
The example in question is from Rick Durrett's "Elementrary Probability for Applications", and the setup is something like this:
Let $A$ be the event "Alice and Betty have the same birthday", $B$ be the event "Betty and Carol have the same birthday", and $C$ be the event "Carol and Alice have the same birthday".
Durrett goes on to demonstrate that each pair is independent, since for example,
$P(A cap B) = P(A)P(B)$.
However, he concludes that $A, B$, and $C$ are not independent, since
$P(A cap B cap C) = frac1365^2 neq frac1365^3 = P(A)P(B)P(C)$.
I understand the reasoning here, and that one can generally show that arbitrary events $X$ and $Y$ are not independent by showing that $P(Xcap Y) neq P(X)P(Y)$.
I am a little new to probability, though, and don't understand why exactly $P(A cap B cap C) = frac1365^2$.
My progress so far:
I do see why $P(A) = P(B) =P(C) = frac1365$, and thus why $P(A)P(B)P(C) = frac1365^3$.
It seems like the sample space $Omega = (a, b, c) mid a,b,c in [365] $ -- i.e., all of the possible triples of numbers from 1 to 365, where 1 denotes January 1st, 2 denotes January 2nd, etc. From that, I can conclude $|Omega| = 365^3$, but I'm not sure where to go from here.
It seems like once a single birthday is chosen, the rest are completely determined if they're all equal to each other - is this a good direction to go in?
probability independence
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
$endgroup$
The example in question is from Rick Durrett's "Elementrary Probability for Applications", and the setup is something like this:
Let $A$ be the event "Alice and Betty have the same birthday", $B$ be the event "Betty and Carol have the same birthday", and $C$ be the event "Carol and Alice have the same birthday".
Durrett goes on to demonstrate that each pair is independent, since for example,
$P(A cap B) = P(A)P(B)$.
However, he concludes that $A, B$, and $C$ are not independent, since
$P(A cap B cap C) = frac1365^2 neq frac1365^3 = P(A)P(B)P(C)$.
I understand the reasoning here, and that one can generally show that arbitrary events $X$ and $Y$ are not independent by showing that $P(Xcap Y) neq P(X)P(Y)$.
I am a little new to probability, though, and don't understand why exactly $P(A cap B cap C) = frac1365^2$.
My progress so far:
I do see why $P(A) = P(B) =P(C) = frac1365$, and thus why $P(A)P(B)P(C) = frac1365^3$.
It seems like the sample space $Omega = (a, b, c) mid a,b,c in [365] $ -- i.e., all of the possible triples of numbers from 1 to 365, where 1 denotes January 1st, 2 denotes January 2nd, etc. From that, I can conclude $|Omega| = 365^3$, but I'm not sure where to go from here.
It seems like once a single birthday is chosen, the rest are completely determined if they're all equal to each other - is this a good direction to go in?
probability independence
probability independence
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
asked Jan 10 '16 at 0:30
D. Zack GarzaD. Zack Garza
8712
8712
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
add a comment |
$begingroup$
The event $Acap Bcap C$ happens if and only if the event $Acap B$ happens. And you know how to find $Pr(Acap B)$.
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
$endgroup$
add a comment |
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
);
);
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%2f1606205%2fdurrett-example-1-9-pairwise-independence-does-not-imply-mutual-independence%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
$begingroup$
There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
add a comment |
$begingroup$
There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
add a comment |
$begingroup$
There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
answered Jan 10 '16 at 0:38
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
add a comment |
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
$begingroup$
Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks!
$endgroup$
– D. Zack Garza
Jan 10 '16 at 0:55
add a comment |
$begingroup$
The event $Acap Bcap C$ happens if and only if the event $Acap B$ happens. And you know how to find $Pr(Acap B)$.
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
$endgroup$
add a comment |
$begingroup$
The event $Acap Bcap C$ happens if and only if the event $Acap B$ happens. And you know how to find $Pr(Acap B)$.
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
$endgroup$
add a comment |
$begingroup$
The event $Acap Bcap C$ happens if and only if the event $Acap B$ happens. And you know how to find $Pr(Acap B)$.
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
$endgroup$
The event $Acap Bcap C$ happens if and only if the event $Acap B$ happens. And you know how to find $Pr(Acap B)$.
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
answered Jan 10 '16 at 0:41
André NicolasAndré Nicolas
455k36432819
455k36432819
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%2f1606205%2fdurrett-example-1-9-pairwise-independence-does-not-imply-mutual-independence%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
