Dgdrxrt

Why are vacuum tubes still used in amateur radios?

My mentor says to set image to Fine instead of RAW — how is this different from JPG?

Sally's older brother

A proverb that is used to imply that you have unexpectedly faced a big problem

I got rid of Mac OSX and replaced it with linux but now I can't change it back to OSX or windows

Simple HTTP Server

Weaponising the Grasp-at-a-Distance spell

What initially awakened the Balrog?

Does the Mueller report show a conspiracy between Russia and the Trump Campaign?

In musical terms, what properties are varied by the human voice to produce different words / syllables?

Project Euler #1 in C++

After Sam didn't return home in the end, were he and Al still friends?

Relating to the President and obstruction, were Mueller's conclusions preordained?

As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?

Should a wizard buy fine inks every time he want to copy spells into his spellbook?

Why is a lens darker than other ones when applying the same settings?

Simple Line in LaTeX Help!

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

Why do early math courses focus on the cross sections of a cone and not on other 3D objects?

What are the main differences between the original Stargate SG-1 and the Final Cut edition?

A term for a woman complaining about things/begging in a cute/childish way

Constant factor of an array

Why is it faster to reheat something than it is to cook it?

i2c bus hangs in master RPi access to MSP430G uC ~1 in 1000 accesses

Conditional probability with independent events

0

$begingroup$

Given two independent events A,B.
Does the following equation hold?

P[(A & B) | C] = P[A|C]*P[B|C]

I assume it does because P[A & B] = P[A]P[B] or did I miss something?

probability

share|cite|improve this question

asked Apr 2 at 10:36

durnovvdurnovv

11

$endgroup$

add a comment | 

0

$begingroup$

Given two independent events A,B.
Does the following equation hold?

P[(A & B) | C] = P[A|C]*P[B|C]

I assume it does because P[A & B] = P[A]P[B] or did I miss something?

probability

share|cite|improve this question

asked Apr 2 at 10:36

durnovvdurnovv

11

$endgroup$

add a comment | 

$begingroup$

Given two independent events A,B.
Does the following equation hold?

P[(A & B) | C] = P[A|C]*P[B|C]

I assume it does because P[A & B] = P[A]P[B] or did I miss something?

probability

share|cite|improve this question

asked Apr 2 at 10:36

durnovvdurnovv

11

$endgroup$

share|cite|improve this question

asked Apr 2 at 10:36

durnovvdurnovv

11

share|cite|improve this question

asked Apr 2 at 10:36

durnovvdurnovv

11

asked Apr 2 at 10:36

durnovvdurnovv

11

asked Apr 2 at 10:36

durnovvdurnovv

11

4 Answers
4

active

oldest

votes

1

$begingroup$

In general it does not. Independence of $A$ and $B$ are not enough to ensure this result.

Under the additional assumptions that $C$ is independent of $A$, of $B$ and of $Aland B $, then yes.

In that case we get:

$P(A land B /C)= P(A land B) = P(A) P(B) = P(A/C) P(B/C)$

It is easy to construct a counter-example where any one of the additional assumptions is not valid (while the other two are), in which case the equation will not hold, which I think will give you the definite answer you need.

share|cite|improve this answer

edited Apr 2 at 16:17

answered Apr 2 at 15:51

CrispostCrispost

265

$endgroup$

add a comment | 

0

$begingroup$

Not necessarily. Independence does not imply conditional independence. The first example that comes to my mind is that of independent numbers of calls arriving at two call centres. But if there is a breakdown in the telephone lines they are not independent anymore. See also
https://en.wikipedia.org/wiki/Conditional_independence.

share|cite|improve this answer

edited Apr 2 at 11:09

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make sense, thanks
  $endgroup$
  – durnovv
  Apr 2 at 11:07
  

  
  
  
  

add a comment | 

0

$begingroup$

No. For example, let A be the event that you get Heads on a coin in your left hand, B the event that you get Heads on a coin in your right hand, and C the event that you get heads on at least one of the two coins.

share|cite|improve this answer

answered Apr 2 at 11:18

SimonSimon

793513

$endgroup$

add a comment | 

0

$begingroup$

There is a well known example of three events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. In this case the equation does not hold.

share|cite|improve this answer

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171702%2fconditional-probability-with-independent-events%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

In general it does not. Independence of $A$ and $B$ are not enough to ensure this result.

Under the additional assumptions that $C$ is independent of $A$, of $B$ and of $Aland B $, then yes.

In that case we get:

$P(A land B /C)= P(A land B) = P(A) P(B) = P(A/C) P(B/C)$

It is easy to construct a counter-example where any one of the additional assumptions is not valid (while the other two are), in which case the equation will not hold, which I think will give you the definite answer you need.

share|cite|improve this answer

edited Apr 2 at 16:17

answered Apr 2 at 15:51

CrispostCrispost

265

$endgroup$

add a comment | 

1

$begingroup$

In general it does not. Independence of $A$ and $B$ are not enough to ensure this result.

Under the additional assumptions that $C$ is independent of $A$, of $B$ and of $Aland B $, then yes.

In that case we get:

$P(A land B /C)= P(A land B) = P(A) P(B) = P(A/C) P(B/C)$

It is easy to construct a counter-example where any one of the additional assumptions is not valid (while the other two are), in which case the equation will not hold, which I think will give you the definite answer you need.

share|cite|improve this answer

edited Apr 2 at 16:17

answered Apr 2 at 15:51

CrispostCrispost

265

$endgroup$

add a comment | 

$begingroup$

In general it does not. Independence of $A$ and $B$ are not enough to ensure this result.

Under the additional assumptions that $C$ is independent of $A$, of $B$ and of $Aland B $, then yes.

In that case we get:

$P(A land B /C)= P(A land B) = P(A) P(B) = P(A/C) P(B/C)$

It is easy to construct a counter-example where any one of the additional assumptions is not valid (while the other two are), in which case the equation will not hold, which I think will give you the definite answer you need.

share|cite|improve this answer

edited Apr 2 at 16:17

answered Apr 2 at 15:51

CrispostCrispost

265

$endgroup$

share|cite|improve this answer

edited Apr 2 at 16:17

answered Apr 2 at 15:51

CrispostCrispost

265

answered Apr 2 at 15:51

CrispostCrispost

265

answered Apr 2 at 15:51

CrispostCrispost

265

0

$begingroup$

Not necessarily. Independence does not imply conditional independence. The first example that comes to my mind is that of independent numbers of calls arriving at two call centres. But if there is a breakdown in the telephone lines they are not independent anymore. See also
https://en.wikipedia.org/wiki/Conditional_independence.

share|cite|improve this answer

edited Apr 2 at 11:09

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make sense, thanks
  $endgroup$
  – durnovv
  Apr 2 at 11:07
  

  
  
  
  

add a comment | 

0

$begingroup$

Not necessarily. Independence does not imply conditional independence. The first example that comes to my mind is that of independent numbers of calls arriving at two call centres. But if there is a breakdown in the telephone lines they are not independent anymore. See also
https://en.wikipedia.org/wiki/Conditional_independence.

share|cite|improve this answer

edited Apr 2 at 11:09

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make sense, thanks
  $endgroup$
  – durnovv
  Apr 2 at 11:07
  

  
  
  
  

add a comment | 

$begingroup$

Not necessarily. Independence does not imply conditional independence. The first example that comes to my mind is that of independent numbers of calls arriving at two call centres. But if there is a breakdown in the telephone lines they are not independent anymore. See also
https://en.wikipedia.org/wiki/Conditional_independence.

share|cite|improve this answer

edited Apr 2 at 11:09

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

$endgroup$

share|cite|improve this answer

edited Apr 2 at 11:09

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

answered Apr 2 at 11:04

Marco StamazzaMarco Stamazza

1063

0

$begingroup$

No. For example, let A be the event that you get Heads on a coin in your left hand, B the event that you get Heads on a coin in your right hand, and C the event that you get heads on at least one of the two coins.

share|cite|improve this answer

answered Apr 2 at 11:18

SimonSimon

793513

$endgroup$

add a comment | 

0

$begingroup$

No. For example, let A be the event that you get Heads on a coin in your left hand, B the event that you get Heads on a coin in your right hand, and C the event that you get heads on at least one of the two coins.

share|cite|improve this answer

answered Apr 2 at 11:18

SimonSimon

793513

$endgroup$

add a comment | 

$begingroup$

No. For example, let A be the event that you get Heads on a coin in your left hand, B the event that you get Heads on a coin in your right hand, and C the event that you get heads on at least one of the two coins.

share|cite|improve this answer

answered Apr 2 at 11:18

SimonSimon

793513

$endgroup$

share|cite|improve this answer

answered Apr 2 at 11:18

SimonSimon

793513

answered Apr 2 at 11:18

SimonSimon

793513

answered Apr 2 at 11:18

SimonSimon

793513

0

$begingroup$

There is a well known example of three events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. In this case the equation does not hold.

share|cite|improve this answer

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

0

$begingroup$

There is a well known example of three events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. In this case the equation does not hold.

share|cite|improve this answer

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

$begingroup$

There is a well known example of three events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. In this case the equation does not hold.

share|cite|improve this answer

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

share|cite|improve this answer

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered Apr 2 at 11:58

Kavi Rama MurthyKavi Rama Murthy

76.4k53370