Dtjytyk

How do hiring committees for research positions view getting "scooped"?

How to get the n-th line after a grepped one?

Pronounciation of the combination "st" in spanish accents

Would it be believable to defy demographics in a story?

Differential and Linear trail propagation in Noekeon

Why didn't Héctor fade away after this character died in the movie Coco?

Does .bashrc contain syntax errors?

Writing in a Christian voice

Brake pads destroying wheels

Probably overheated black color SMD pads

How is the partial sum of a geometric sequence calculated?

What should I install to correct "ld: cannot find -lgbm and -linput" so that I can compile a Rust program?

Do US professors/group leaders only get a salary, but no group budget?

My friend is being a hypocrite

How to define limit operations in general topological spaces? Are nets able to do this?

What (if any) is the reason to buy in small local stores?

Is it insecure to send a password in a `curl` command?

Wrapping homogeneous Python objects

Variable completely messes up echoed string

What is the term when voters “dishonestly” choose something that they do not want to choose?

What are substitutions for coconut in curry?

Print last inputted byte

Calculate the frequency of characters in a string

Can a medieval gyroplane be built?

Is there a more general form of the conditional probability?

Intuition behind throwing two dice and conditional probability in generalFor a 2-dimensional random vector, express the probability of being in a rectangle in terms of CDFHow to do a Proof by induction.If $X_1,X_2,…,X_n $ are independent random variables. Then $a_1X_1+b_1,a_2X2+b_2,…,a_nX_n+b_n$ are independent.PDF of the difference between two independent beta random variablesCaculate the probability of sum of normal distributionweighted sequence of arbitrary sequence of random variables can converge to zero almost surelyMaking a sequence of random variables converge almost surely to $0$Independence among random vectorsFind the conditional joint PMF of the random vector $(X_1,…,X_n)$ given $S_n=X_1+…+X_n=k$ where $X_i$~Bernoulli(p)Necessary and sufficient conditions on a trivariate probability distribution for being the probability distribution of $(X,Y,X-Y)$

1

1

$begingroup$

Say you have $P(A_1, A_2, dots, A_n | B_1, B_2, dots, B_m)$. Is there a general way to break this down into combinations of $P(X|Y)$'s and $P(Z)$'s?

I understand that $P(X|Y) = frac{P(X,Y)}{P(Y)}$ and $P(X_1, dots, X_n) = P(X_1)P(X_2 | X_1)P(X_3 | X_2, X_1)dots P(X_n|X_{n-1}, dots, X_1)$, but I'm not sure if there's a way to reduce this using these formulas.

Does anyone have any ideas?

probability conditional-probability

share|cite|improve this question

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

$endgroup$

This question has an open bounty worth +50
reputation from Oliver G ending in 3 days.

Looking for an answer drawing from credible and/or official sources.

add a comment | 

1

1

$begingroup$

Say you have $P(A_1, A_2, dots, A_n | B_1, B_2, dots, B_m)$. Is there a general way to break this down into combinations of $P(X|Y)$'s and $P(Z)$'s?

I understand that $P(X|Y) = frac{P(X,Y)}{P(Y)}$ and $P(X_1, dots, X_n) = P(X_1)P(X_2 | X_1)P(X_3 | X_2, X_1)dots P(X_n|X_{n-1}, dots, X_1)$, but I'm not sure if there's a way to reduce this using these formulas.

Does anyone have any ideas?

probability conditional-probability

share|cite|improve this question

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

$endgroup$

This question has an open bounty worth +50
reputation from Oliver G ending in 3 days.

Looking for an answer drawing from credible and/or official sources.

add a comment | 

$begingroup$

Say you have $P(A_1, A_2, dots, A_n | B_1, B_2, dots, B_m)$. Is there a general way to break this down into combinations of $P(X|Y)$'s and $P(Z)$'s?

I understand that $P(X|Y) = frac{P(X,Y)}{P(Y)}$ and $P(X_1, dots, X_n) = P(X_1)P(X_2 | X_1)P(X_3 | X_2, X_1)dots P(X_n|X_{n-1}, dots, X_1)$, but I'm not sure if there's a way to reduce this using these formulas.

Does anyone have any ideas?

probability conditional-probability

share|cite|improve this question

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

$endgroup$

share|cite|improve this question

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

share|cite|improve this question

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

asked Mar 11 at 19:05

Oliver GOliver G

1,4381632

1 Answer
1

active

oldest

votes

0

$begingroup$

Any probability of the form P(.|X) forms a probability law. They obey all the axioms.

So, $𝑃(𝐴_1,𝐴_2,…,𝐴_n|𝐵_1,𝐵_2,…,𝐵_𝑚) = P(A_1|𝐵_1,𝐵_2,…,𝐵_𝑚) . P(A_2 | 𝐵_1,𝐵_2,…,𝐵_𝑚,A_1) . P(A_3 | 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2) ... P(A_n| 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2 ... A_n)$
.

share|cite|improve this answer

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

$endgroup$

add a comment | 

Your Answer

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
});


}
});

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%2f3144089%2fis-there-a-more-general-form-of-the-conditional-probability%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Any probability of the form P(.|X) forms a probability law. They obey all the axioms.

So, $𝑃(𝐴_1,𝐴_2,…,𝐴_n|𝐵_1,𝐵_2,…,𝐵_𝑚) = P(A_1|𝐵_1,𝐵_2,…,𝐵_𝑚) . P(A_2 | 𝐵_1,𝐵_2,…,𝐵_𝑚,A_1) . P(A_3 | 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2) ... P(A_n| 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2 ... A_n)$
.

share|cite|improve this answer

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

$endgroup$

add a comment | 

0

$begingroup$

Any probability of the form P(.|X) forms a probability law. They obey all the axioms.

So, $𝑃(𝐴_1,𝐴_2,…,𝐴_n|𝐵_1,𝐵_2,…,𝐵_𝑚) = P(A_1|𝐵_1,𝐵_2,…,𝐵_𝑚) . P(A_2 | 𝐵_1,𝐵_2,…,𝐵_𝑚,A_1) . P(A_3 | 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2) ... P(A_n| 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2 ... A_n)$
.

share|cite|improve this answer

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

$endgroup$

add a comment | 

$begingroup$

Any probability of the form P(.|X) forms a probability law. They obey all the axioms.

So, $𝑃(𝐴_1,𝐴_2,…,𝐴_n|𝐵_1,𝐵_2,…,𝐵_𝑚) = P(A_1|𝐵_1,𝐵_2,…,𝐵_𝑚) . P(A_2 | 𝐵_1,𝐵_2,…,𝐵_𝑚,A_1) . P(A_3 | 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2) ... P(A_n| 𝐵_1,𝐵_2,…,𝐵_𝑚, A_1, A_2 ... A_n)$
.

share|cite|improve this answer

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

$endgroup$

share|cite|improve this answer

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114

answered Mar 14 at 0:31

Neo M HackerNeo M Hacker

1114