Mixing
Bayesian probability: proving $P(x|y,x) = 1$
$begingroup$
I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$
I've come up with the following but I'm not sure if it feels right to me:
$P(x|y,x) =$
$P(x,y,x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $
We now have
$P(x|y,x) = P(x|y,x)$
So:
$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$
Is this correct?
bayesian bayes-theorem
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
$endgroup$
add a comment |
$begingroup$
I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$
I've come up with the following but I'm not sure if it feels right to me:
$P(x|y,x) =$
$P(x,y,x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $
We now have
$P(x|y,x) = P(x|y,x)$
So:
$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$
Is this correct?
bayesian bayes-theorem
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
$endgroup$
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40
add a comment |
$begingroup$
I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$
I've come up with the following but I'm not sure if it feels right to me:
$P(x|y,x) =$
$P(x,y,x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $
We now have
$P(x|y,x) = P(x|y,x)$
So:
$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$
Is this correct?
bayesian bayes-theorem
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
$endgroup$
I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$
I've come up with the following but I'm not sure if it feels right to me:
$P(x|y,x) =$
$P(x,y,x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y,x) = $
$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $
We now have
$P(x|y,x) = P(x|y,x)$
So:
$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$
Is this correct?
bayesian bayes-theorem
bayesian bayes-theorem
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
edited Jan 16 at 13:31
ronaldfisher
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
asked Jan 16 at 13:21
ronaldfisherronaldfisher
11
11
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40
add a comment |
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
$endgroup$
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
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
});
}
});
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%2f3075715%2fbayesian-probability-proving-pxy-x-1%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
$begingroup$
Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
$endgroup$
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
add a comment |
$begingroup$
Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
$endgroup$
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
add a comment |
$begingroup$
Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
$endgroup$
Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
answered Jan 16 at 13:32
trancelocationtrancelocation
12.1k1826
12.1k1826
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
add a comment |
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36
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%2f3075715%2fbayesian-probability-proving-pxy-x-1%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
$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27
$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40