Prove that $A, B, C$ are independent (under certain conditions) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)To prove the mutual independence of eventsShow that events $B$ and $C$ are independent.How to rigorously determine whether two events are independent?Let $A,B,C$ be mutually independent events. Show that $B$ is independent from ($A cap C$)'.Does $mathbbP(Xin A, Yin A)=mathbbP(Xin A)mathbbP(Yin A)$ for all $A$ borel sets imply $X$ and $Y$ are independent?Suppose that events $A$, $B$ and $C$ satisfy $P(A cap B cap C) = 0$ and each of them has probability not smaller than $frac23$. Find $P(A)$.Independent Events Conceptual Meaning - probability theoryGiven $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Prove that intersection of complement of event A with union of event B and C complement are independentQuestions about the independence of three events
Is it true that "carbohydrates are of no use for the basal metabolic need"?
Single word antonym of "flightless"
What to do with chalk when deepwater soloing?
Storing hydrofluoric acid before the invention of plastics
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
Should I discuss the type of campaign with my players?
Using audio cues to encourage good posture
How do I keep my slimes from escaping their pens?
Echoing a tail command produces unexpected output?
Fundamental Solution of the Pell Equation
Do I really need recursive chmod to restrict access to a folder?
Identify plant with long narrow paired leaves and reddish stems
What causes the vertical darker bands in my photo?
What's the purpose of writing one's academic biography in the third person?
Bete Noir -- no dairy
English words in a non-english sci-fi novel
What does this icon in iOS Stardew Valley mean?
3 doors, three guards, one stone
Sci-Fi book where patients in a coma ward all live in a subconscious world linked together
How does the particle を relate to the verb 行く in the structure「A を + B に行く」?
2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?
If a contract sometimes uses the wrong name, is it still valid?
What is Arya's weapon design?
Why did the Falcon Heavy center core fall off the ASDS OCISLY barge?
Prove that $A, B, C$ are independent (under certain conditions)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)To prove the mutual independence of eventsShow that events $B$ and $C$ are independent.How to rigorously determine whether two events are independent?Let $A,B,C$ be mutually independent events. Show that $B$ is independent from ($A cap C$)'.Does $mathbbP(Xin A, Yin A)=mathbbP(Xin A)mathbbP(Yin A)$ for all $A$ borel sets imply $X$ and $Y$ are independent?Suppose that events $A$, $B$ and $C$ satisfy $P(A cap B cap C) = 0$ and each of them has probability not smaller than $frac23$. Find $P(A)$.Independent Events Conceptual Meaning - probability theoryGiven $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Prove that intersection of complement of event A with union of event B and C complement are independentQuestions about the independence of three events
$begingroup$
Suppose:
$A$ is independent from both $Bcap C$ and $B cup C$.
$B$ is independent from $A cap C$.
$C$ is independent from $A cap B$.
$P(A), P(B), P(C)$ are all greater than zero.
Prove that $A, B, C$ are independent.
I know I have to show that the 3 events are all pairwise independent and that $P(Acap B cap C) = P(A) P(B) P(C)$.
Working with 1., I got to the equality $P(A cap B) + P(A cap C) = P(A) [P(Bcap C) + P(B cup C)]$, but I don't find it very useful. I tried several other approaches and they didn't work.
Could someone give me any hints? I'd also appreciate any intuition into how to solve these type of exercises. I always solve them by trial and error. Thanks!
probability independence
asked Mar 26 at 16:52
John DoeJohn Doe
1134
$endgroup$
add a comment |
$begingroup$
Suppose:
$A$ is independent from both $Bcap C$ and $B cup C$.
$B$ is independent from $A cap C$.
$C$ is independent from $A cap B$.
$P(A), P(B), P(C)$ are all greater than zero.
Prove that $A, B, C$ are independent.
I know I have to show that the 3 events are all pairwise independent and that $P(Acap B cap C) = P(A) P(B) P(C)$.
Working with 1., I got to the equality $P(A cap B) + P(A cap C) = P(A) [P(Bcap C) + P(B cup C)]$, but I don't find it very useful. I tried several other approaches and they didn't work.
Could someone give me any hints? I'd also appreciate any intuition into how to solve these type of exercises. I always solve them by trial and error. Thanks!
probability independence
asked Mar 26 at 16:52
John DoeJohn Doe
1134
$endgroup$
add a comment |
$begingroup$
Suppose:
$A$ is independent from both $Bcap C$ and $B cup C$.
$B$ is independent from $A cap C$.
$C$ is independent from $A cap B$.
$P(A), P(B), P(C)$ are all greater than zero.
Prove that $A, B, C$ are independent.
I know I have to show that the 3 events are all pairwise independent and that $P(Acap B cap C) = P(A) P(B) P(C)$.
Working with 1., I got to the equality $P(A cap B) + P(A cap C) = P(A) [P(Bcap C) + P(B cup C)]$, but I don't find it very useful. I tried several other approaches and they didn't work.
Could someone give me any hints? I'd also appreciate any intuition into how to solve these type of exercises. I always solve them by trial and error. Thanks!
probability independence
asked Mar 26 at 16:52
John DoeJohn Doe
1134
$endgroup$
Suppose:
$A$ is independent from both $Bcap C$ and $B cup C$.
$B$ is independent from $A cap C$.
$C$ is independent from $A cap B$.
$P(A), P(B), P(C)$ are all greater than zero.
Prove that $A, B, C$ are independent.
I know I have to show that the 3 events are all pairwise independent and that $P(Acap B cap C) = P(A) P(B) P(C)$.
Working with 1., I got to the equality $P(A cap B) + P(A cap C) = P(A) [P(Bcap C) + P(B cup C)]$, but I don't find it very useful. I tried several other approaches and they didn't work.
Could someone give me any hints? I'd also appreciate any intuition into how to solve these type of exercises. I always solve them by trial and error. Thanks!
probability independence
probability independence
asked Mar 26 at 16:52
John DoeJohn Doe
1134
asked Mar 26 at 16:52
John DoeJohn Doe
1134
edited Mar 26 at 20:17
John Doe
asked Mar 26 at 16:52
John DoeJohn Doe
1134
asked Mar 26 at 16:52
John DoeJohn Doe
1134
asked Mar 26 at 16:52
John DoeJohn Doe
1134
1134
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you are on the right track.
Let us start from the equation you got from hypothesis 1.
begineqnarray
P(Acap B) &=& P(A) cdot P(Bcup C) + P(A)cdot P(Bcap C)-P(Acap C)=\
&=&P(A) [P(B) + P(C) - P(Bcap C)]+ P(A)cdot P(Bcap C)-P(Acap C)=\
&=& P(A)cdot P(B) + P(A)cdot P(C) -P(Acap C).
endeqnarray
So we have
beginequation
P(Acap B) -P(A)cdot P(B) = P(A)cdot P(C) - P(Acap C).tag1label1
endequation
Now condition 4. makes me think that I could multiply both sides of eqref1 by $P(B)cdot P(C)$
and get
beginequation
P(B)cdot P(C)cdot P(Acap B) -P(A)cdot P^2(B)cdot P(C) =\ =P(A)cdot P(B)cdot P^2(C) - P(B)cdot P(C)cdot P(Acap C).tag2label2
endequation
By hypothesis 2. and 3.
beginequation
P(Acap Bcap C) = P(B) cdot P(Acap C) = P(C) cdot P(Acap B),tag3label3
endequation
so that eqref2 yields
beginequation
P(B) cdot P(Acap Bcap C) -P(A)cdot P^2(B)cdot P(C) =\
=P(A)cdot P(B)cdot P^2(C)-P(C) cdot P(Acap Bcap C),
endequation
that is
beginequation
[P(B) +P(C)]cdot [P(Acap Bcap C) -P(A)cdot P(B)cdot P(C)] = 0.
endequation
From the last equation you get
beginequation
P(Acap Bcap C) =P(A)cdot P(B)cdot P(C)tag4label4.
endequation
Use now eqref4 and eqref3 to get pairwise independence.
As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
$endgroup$
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%2f3163461%2fprove-that-a-b-c-are-independent-under-certain-conditions%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you are on the right track.
Let us start from the equation you got from hypothesis 1.
begineqnarray
P(Acap B) &=& P(A) cdot P(Bcup C) + P(A)cdot P(Bcap C)-P(Acap C)=\
&=&P(A) [P(B) + P(C) - P(Bcap C)]+ P(A)cdot P(Bcap C)-P(Acap C)=\
&=& P(A)cdot P(B) + P(A)cdot P(C) -P(Acap C).
endeqnarray
So we have
beginequation
P(Acap B) -P(A)cdot P(B) = P(A)cdot P(C) - P(Acap C).tag1label1
endequation
Now condition 4. makes me think that I could multiply both sides of eqref1 by $P(B)cdot P(C)$
and get
beginequation
P(B)cdot P(C)cdot P(Acap B) -P(A)cdot P^2(B)cdot P(C) =\ =P(A)cdot P(B)cdot P^2(C) - P(B)cdot P(C)cdot P(Acap C).tag2label2
endequation
By hypothesis 2. and 3.
beginequation
P(Acap Bcap C) = P(B) cdot P(Acap C) = P(C) cdot P(Acap B),tag3label3
endequation
so that eqref2 yields
beginequation
P(B) cdot P(Acap Bcap C) -P(A)cdot P^2(B)cdot P(C) =\
=P(A)cdot P(B)cdot P^2(C)-P(C) cdot P(Acap Bcap C),
endequation
that is
beginequation
[P(B) +P(C)]cdot [P(Acap Bcap C) -P(A)cdot P(B)cdot P(C)] = 0.
endequation
From the last equation you get
beginequation
P(Acap Bcap C) =P(A)cdot P(B)cdot P(C)tag4label4.
endequation
Use now eqref4 and eqref3 to get pairwise independence.
As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
$endgroup$
add a comment |
$begingroup$
I think you are on the right track.
Let us start from the equation you got from hypothesis 1.
begineqnarray
P(Acap B) &=& P(A) cdot P(Bcup C) + P(A)cdot P(Bcap C)-P(Acap C)=\
&=&P(A) [P(B) + P(C) - P(Bcap C)]+ P(A)cdot P(Bcap C)-P(Acap C)=\
&=& P(A)cdot P(B) + P(A)cdot P(C) -P(Acap C).
endeqnarray
So we have
beginequation
P(Acap B) -P(A)cdot P(B) = P(A)cdot P(C) - P(Acap C).tag1label1
endequation
Now condition 4. makes me think that I could multiply both sides of eqref1 by $P(B)cdot P(C)$
and get
beginequation
P(B)cdot P(C)cdot P(Acap B) -P(A)cdot P^2(B)cdot P(C) =\ =P(A)cdot P(B)cdot P^2(C) - P(B)cdot P(C)cdot P(Acap C).tag2label2
endequation
By hypothesis 2. and 3.
beginequation
P(Acap Bcap C) = P(B) cdot P(Acap C) = P(C) cdot P(Acap B),tag3label3
endequation
so that eqref2 yields
beginequation
P(B) cdot P(Acap Bcap C) -P(A)cdot P^2(B)cdot P(C) =\
=P(A)cdot P(B)cdot P^2(C)-P(C) cdot P(Acap Bcap C),
endequation
that is
beginequation
[P(B) +P(C)]cdot [P(Acap Bcap C) -P(A)cdot P(B)cdot P(C)] = 0.
endequation
From the last equation you get
beginequation
P(Acap Bcap C) =P(A)cdot P(B)cdot P(C)tag4label4.
endequation
Use now eqref4 and eqref3 to get pairwise independence.
As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
$endgroup$
add a comment |
$begingroup$
I think you are on the right track.
Let us start from the equation you got from hypothesis 1.
begineqnarray
P(Acap B) &=& P(A) cdot P(Bcup C) + P(A)cdot P(Bcap C)-P(Acap C)=\
&=&P(A) [P(B) + P(C) - P(Bcap C)]+ P(A)cdot P(Bcap C)-P(Acap C)=\
&=& P(A)cdot P(B) + P(A)cdot P(C) -P(Acap C).
endeqnarray
So we have
beginequation
P(Acap B) -P(A)cdot P(B) = P(A)cdot P(C) - P(Acap C).tag1label1
endequation
Now condition 4. makes me think that I could multiply both sides of eqref1 by $P(B)cdot P(C)$
and get
beginequation
P(B)cdot P(C)cdot P(Acap B) -P(A)cdot P^2(B)cdot P(C) =\ =P(A)cdot P(B)cdot P^2(C) - P(B)cdot P(C)cdot P(Acap C).tag2label2
endequation
By hypothesis 2. and 3.
beginequation
P(Acap Bcap C) = P(B) cdot P(Acap C) = P(C) cdot P(Acap B),tag3label3
endequation
so that eqref2 yields
beginequation
P(B) cdot P(Acap Bcap C) -P(A)cdot P^2(B)cdot P(C) =\
=P(A)cdot P(B)cdot P^2(C)-P(C) cdot P(Acap Bcap C),
endequation
that is
beginequation
[P(B) +P(C)]cdot [P(Acap Bcap C) -P(A)cdot P(B)cdot P(C)] = 0.
endequation
From the last equation you get
beginequation
P(Acap Bcap C) =P(A)cdot P(B)cdot P(C)tag4label4.
endequation
Use now eqref4 and eqref3 to get pairwise independence.
As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
$endgroup$
I think you are on the right track.
Let us start from the equation you got from hypothesis 1.
begineqnarray
P(Acap B) &=& P(A) cdot P(Bcup C) + P(A)cdot P(Bcap C)-P(Acap C)=\
&=&P(A) [P(B) + P(C) - P(Bcap C)]+ P(A)cdot P(Bcap C)-P(Acap C)=\
&=& P(A)cdot P(B) + P(A)cdot P(C) -P(Acap C).
endeqnarray
So we have
beginequation
P(Acap B) -P(A)cdot P(B) = P(A)cdot P(C) - P(Acap C).tag1label1
endequation
Now condition 4. makes me think that I could multiply both sides of eqref1 by $P(B)cdot P(C)$
and get
beginequation
P(B)cdot P(C)cdot P(Acap B) -P(A)cdot P^2(B)cdot P(C) =\ =P(A)cdot P(B)cdot P^2(C) - P(B)cdot P(C)cdot P(Acap C).tag2label2
endequation
By hypothesis 2. and 3.
beginequation
P(Acap Bcap C) = P(B) cdot P(Acap C) = P(C) cdot P(Acap B),tag3label3
endequation
so that eqref2 yields
beginequation
P(B) cdot P(Acap Bcap C) -P(A)cdot P^2(B)cdot P(C) =\
=P(A)cdot P(B)cdot P^2(C)-P(C) cdot P(Acap Bcap C),
endequation
that is
beginequation
[P(B) +P(C)]cdot [P(Acap Bcap C) -P(A)cdot P(B)cdot P(C)] = 0.
endequation
From the last equation you get
beginequation
P(Acap Bcap C) =P(A)cdot P(B)cdot P(C)tag4label4.
endequation
Use now eqref4 and eqref3 to get pairwise independence.
As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
edited Mar 26 at 22:13
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
answered Mar 26 at 19:49
MatteoMatteo
1,3121313
1,3121313
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%2f3163461%2fprove-that-a-b-c-are-independent-under-certain-conditions%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
