Conditioning on random sets?
up vote
1
down vote
favorite
Let $( X_i)_{i=1}^n$ be a random vector of (possibly dependent) real random variables.
Let $A_i$ be any measurable set such that $P(X_i in A_i ,|, X_{i-1} = x_{i-1},ldots,X_1 = x_1) = beta_i$. What I want to work out is the probability
$$P(X_n in A_n,ldots, X_1in A_1).$$
The answer (I think) is just $P(X_n in A_n,ldots, X_1in A_1) = prod_{i=1}^nbeta_i$, since
$$P(X_n in A_n,ldots, X_1in A_1) = P(X_n in A_n|X_{n-1}in A_{n-1},ldots,X_1in A_1)ldots P(X_1in A_1),$$
but I am unsure how to make this formal? Is the following proof valid, where $p(x_n,ldots,x_1)$ is the pdf of $X_n,ldots,X_1$:
begin{align}
P(X_n in A_n,ldots, X_1in A_1) &= int_{A_1}ldotsint_{A_n} p(x_n,ldots,x_1) mathrm{d}x_nldots mathrm{d}x_1 \
&=int_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1)int_{A_n} p(x_n,|,x_{n-1}ldots,x_1), mathrm{d}x_n,mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&=beta_nint_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1),mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&= prod_{i=1}^nbeta_i.
end{align}
Maybe this is still not well formulated, since the $A_i$ are really any measurable set with constant probability for any $x_1,ldots,x_{i-1}$. Is the correct mathematical formulation that the $A_i$ are random sets? I am quite new to probability theory so am unsure...
probability probability-distributions conditional-probability
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
add a comment |
up vote
1
down vote
favorite
Let $( X_i)_{i=1}^n$ be a random vector of (possibly dependent) real random variables.
Let $A_i$ be any measurable set such that $P(X_i in A_i ,|, X_{i-1} = x_{i-1},ldots,X_1 = x_1) = beta_i$. What I want to work out is the probability
$$P(X_n in A_n,ldots, X_1in A_1).$$
The answer (I think) is just $P(X_n in A_n,ldots, X_1in A_1) = prod_{i=1}^nbeta_i$, since
$$P(X_n in A_n,ldots, X_1in A_1) = P(X_n in A_n|X_{n-1}in A_{n-1},ldots,X_1in A_1)ldots P(X_1in A_1),$$
but I am unsure how to make this formal? Is the following proof valid, where $p(x_n,ldots,x_1)$ is the pdf of $X_n,ldots,X_1$:
begin{align}
P(X_n in A_n,ldots, X_1in A_1) &= int_{A_1}ldotsint_{A_n} p(x_n,ldots,x_1) mathrm{d}x_nldots mathrm{d}x_1 \
&=int_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1)int_{A_n} p(x_n,|,x_{n-1}ldots,x_1), mathrm{d}x_n,mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&=beta_nint_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1),mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&= prod_{i=1}^nbeta_i.
end{align}
Maybe this is still not well formulated, since the $A_i$ are really any measurable set with constant probability for any $x_1,ldots,x_{i-1}$. Is the correct mathematical formulation that the $A_i$ are random sets? I am quite new to probability theory so am unsure...
probability probability-distributions conditional-probability
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
3
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $( X_i)_{i=1}^n$ be a random vector of (possibly dependent) real random variables.
Let $A_i$ be any measurable set such that $P(X_i in A_i ,|, X_{i-1} = x_{i-1},ldots,X_1 = x_1) = beta_i$. What I want to work out is the probability
$$P(X_n in A_n,ldots, X_1in A_1).$$
The answer (I think) is just $P(X_n in A_n,ldots, X_1in A_1) = prod_{i=1}^nbeta_i$, since
$$P(X_n in A_n,ldots, X_1in A_1) = P(X_n in A_n|X_{n-1}in A_{n-1},ldots,X_1in A_1)ldots P(X_1in A_1),$$
but I am unsure how to make this formal? Is the following proof valid, where $p(x_n,ldots,x_1)$ is the pdf of $X_n,ldots,X_1$:
begin{align}
P(X_n in A_n,ldots, X_1in A_1) &= int_{A_1}ldotsint_{A_n} p(x_n,ldots,x_1) mathrm{d}x_nldots mathrm{d}x_1 \
&=int_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1)int_{A_n} p(x_n,|,x_{n-1}ldots,x_1), mathrm{d}x_n,mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&=beta_nint_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1),mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&= prod_{i=1}^nbeta_i.
end{align}
Maybe this is still not well formulated, since the $A_i$ are really any measurable set with constant probability for any $x_1,ldots,x_{i-1}$. Is the correct mathematical formulation that the $A_i$ are random sets? I am quite new to probability theory so am unsure...
probability probability-distributions conditional-probability
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
Let $( X_i)_{i=1}^n$ be a random vector of (possibly dependent) real random variables.
Let $A_i$ be any measurable set such that $P(X_i in A_i ,|, X_{i-1} = x_{i-1},ldots,X_1 = x_1) = beta_i$. What I want to work out is the probability
$$P(X_n in A_n,ldots, X_1in A_1).$$
The answer (I think) is just $P(X_n in A_n,ldots, X_1in A_1) = prod_{i=1}^nbeta_i$, since
$$P(X_n in A_n,ldots, X_1in A_1) = P(X_n in A_n|X_{n-1}in A_{n-1},ldots,X_1in A_1)ldots P(X_1in A_1),$$
but I am unsure how to make this formal? Is the following proof valid, where $p(x_n,ldots,x_1)$ is the pdf of $X_n,ldots,X_1$:
begin{align}
P(X_n in A_n,ldots, X_1in A_1) &= int_{A_1}ldotsint_{A_n} p(x_n,ldots,x_1) mathrm{d}x_nldots mathrm{d}x_1 \
&=int_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1)int_{A_n} p(x_n,|,x_{n-1}ldots,x_1), mathrm{d}x_n,mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&=beta_nint_{A_1}p(x_1)ldotsint_{A_{n-1}}p(x_{n-1},|,x_{n-2}ldots,x_1),mathrm{d}x_{n-1}ldots mathrm{d}x_1 \
&= prod_{i=1}^nbeta_i.
end{align}
Maybe this is still not well formulated, since the $A_i$ are really any measurable set with constant probability for any $x_1,ldots,x_{i-1}$. Is the correct mathematical formulation that the $A_i$ are random sets? I am quite new to probability theory so am unsure...
probability probability-distributions conditional-probability
probability probability-distributions conditional-probability
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
edited Nov 28 at 14:44
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
asked Nov 28 at 0:50
Timothy Hedgeworth
1366
1366
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
3
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46
add a comment |
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
3
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
3
3
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46
add a comment |
active
oldest
votes
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',
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%2f3016548%2fconditioning-on-random-sets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016548%2fconditioning-on-random-sets%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
Intuitively, no. In fact, if you know all about $Y_1 = X_1,$ so you know $Y_2 = X_2 mid X_1.$
– Will M.
Nov 28 at 0:53
3
What does $Y_2 = X_2 | X_1$ mean? There is no such thing defined in probability theory. Do you mean $Y_2 = E[X_2|X_1]$?
– Michael
Nov 28 at 1:09
Thanks Michael for the comment. Yes, I now realise that this didn't make sense. I have tried to rewrite the question so it is closer to what I really wanted to know.
– Timothy Hedgeworth
Nov 28 at 14:46