Given the time of the $n$:th arrival, find $E(S_1|S_n)$.
$begingroup$
Let $S_1,S_2...$ be the arrival times of a Poisson process with
parameter $lambda$. Given the time of the $n$:th arrival, find
$E(S_1|S_n)$.
I know that $S_n=X_1+X_2+...+X_n$ where $X_i$ are the arrival times. Conditioning on $S_n$ I get
$$E(S_1|S_n=t)=E(S_1|S_{n-1}+X_n=t),$$
I'm not sure how to rewrite this. First I thought that
$$E(X_1 | X_1+...+X_n=t)$$
would be a good idea since all the $X_i$ are independent of eachother. But ofcourse $X_1$ can't be independent of itself.
probability conditional-expectation
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
$endgroup$
add a comment |
$begingroup$
Let $S_1,S_2...$ be the arrival times of a Poisson process with
parameter $lambda$. Given the time of the $n$:th arrival, find
$E(S_1|S_n)$.
I know that $S_n=X_1+X_2+...+X_n$ where $X_i$ are the arrival times. Conditioning on $S_n$ I get
$$E(S_1|S_n=t)=E(S_1|S_{n-1}+X_n=t),$$
I'm not sure how to rewrite this. First I thought that
$$E(X_1 | X_1+...+X_n=t)$$
would be a good idea since all the $X_i$ are independent of eachother. But ofcourse $X_1$ can't be independent of itself.
probability conditional-expectation
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
$endgroup$
2
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
add a comment |
$begingroup$
Let $S_1,S_2...$ be the arrival times of a Poisson process with
parameter $lambda$. Given the time of the $n$:th arrival, find
$E(S_1|S_n)$.
I know that $S_n=X_1+X_2+...+X_n$ where $X_i$ are the arrival times. Conditioning on $S_n$ I get
$$E(S_1|S_n=t)=E(S_1|S_{n-1}+X_n=t),$$
I'm not sure how to rewrite this. First I thought that
$$E(X_1 | X_1+...+X_n=t)$$
would be a good idea since all the $X_i$ are independent of eachother. But ofcourse $X_1$ can't be independent of itself.
probability conditional-expectation
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
$endgroup$
Let $S_1,S_2...$ be the arrival times of a Poisson process with
parameter $lambda$. Given the time of the $n$:th arrival, find
$E(S_1|S_n)$.
I know that $S_n=X_1+X_2+...+X_n$ where $X_i$ are the arrival times. Conditioning on $S_n$ I get
$$E(S_1|S_n=t)=E(S_1|S_{n-1}+X_n=t),$$
I'm not sure how to rewrite this. First I thought that
$$E(X_1 | X_1+...+X_n=t)$$
would be a good idea since all the $X_i$ are independent of eachother. But ofcourse $X_1$ can't be independent of itself.
probability conditional-expectation
probability conditional-expectation
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
asked Jan 15 at 22:33
ParsevalParseval
2,8821718
2,8821718
2
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
add a comment |
2
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
2
2
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$S_n = sum_{i=1}^n X_i = X_1 + sum_{i=2}^n X_i = S_1 + sum_{i=2}^n X_i$$.
Thus, $$mathbb{E}[S_n|S_n] = mathbb{E}[S_1|S_n] + sum_{i=2}^n mathbb{E}[X_i|S_n]$$, implying
$$mathbb{E}[S_1|S_n] = S_n - sum_{i=2}^n mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 leq i leq n$, implying $nmathbb{E}[X_i|S_n] = mathbb{E}[S_n|S_n] = S_n$, i.e. $mathbb{E}[X_i|S_n] = {S_n over n}$.
Substituting back, we get:
$$mathbb{E}[S_1|S_n] = S_n - (n-1){S_n over n} = {S_n over n}$$
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
$endgroup$
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
});
}
});
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%2f3075058%2fgiven-the-time-of-the-nth-arrival-find-es-1s-n%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
$begingroup$
$$S_n = sum_{i=1}^n X_i = X_1 + sum_{i=2}^n X_i = S_1 + sum_{i=2}^n X_i$$.
Thus, $$mathbb{E}[S_n|S_n] = mathbb{E}[S_1|S_n] + sum_{i=2}^n mathbb{E}[X_i|S_n]$$, implying
$$mathbb{E}[S_1|S_n] = S_n - sum_{i=2}^n mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 leq i leq n$, implying $nmathbb{E}[X_i|S_n] = mathbb{E}[S_n|S_n] = S_n$, i.e. $mathbb{E}[X_i|S_n] = {S_n over n}$.
Substituting back, we get:
$$mathbb{E}[S_1|S_n] = S_n - (n-1){S_n over n} = {S_n over n}$$
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
$endgroup$
add a comment |
$begingroup$
$$S_n = sum_{i=1}^n X_i = X_1 + sum_{i=2}^n X_i = S_1 + sum_{i=2}^n X_i$$.
Thus, $$mathbb{E}[S_n|S_n] = mathbb{E}[S_1|S_n] + sum_{i=2}^n mathbb{E}[X_i|S_n]$$, implying
$$mathbb{E}[S_1|S_n] = S_n - sum_{i=2}^n mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 leq i leq n$, implying $nmathbb{E}[X_i|S_n] = mathbb{E}[S_n|S_n] = S_n$, i.e. $mathbb{E}[X_i|S_n] = {S_n over n}$.
Substituting back, we get:
$$mathbb{E}[S_1|S_n] = S_n - (n-1){S_n over n} = {S_n over n}$$
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
$endgroup$
add a comment |
$begingroup$
$$S_n = sum_{i=1}^n X_i = X_1 + sum_{i=2}^n X_i = S_1 + sum_{i=2}^n X_i$$.
Thus, $$mathbb{E}[S_n|S_n] = mathbb{E}[S_1|S_n] + sum_{i=2}^n mathbb{E}[X_i|S_n]$$, implying
$$mathbb{E}[S_1|S_n] = S_n - sum_{i=2}^n mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 leq i leq n$, implying $nmathbb{E}[X_i|S_n] = mathbb{E}[S_n|S_n] = S_n$, i.e. $mathbb{E}[X_i|S_n] = {S_n over n}$.
Substituting back, we get:
$$mathbb{E}[S_1|S_n] = S_n - (n-1){S_n over n} = {S_n over n}$$
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
$endgroup$
$$S_n = sum_{i=1}^n X_i = X_1 + sum_{i=2}^n X_i = S_1 + sum_{i=2}^n X_i$$.
Thus, $$mathbb{E}[S_n|S_n] = mathbb{E}[S_1|S_n] + sum_{i=2}^n mathbb{E}[X_i|S_n]$$, implying
$$mathbb{E}[S_1|S_n] = S_n - sum_{i=2}^n mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 leq i leq n$, implying $nmathbb{E}[X_i|S_n] = mathbb{E}[S_n|S_n] = S_n$, i.e. $mathbb{E}[X_i|S_n] = {S_n over n}$.
Substituting back, we get:
$$mathbb{E}[S_1|S_n] = S_n - (n-1){S_n over n} = {S_n over n}$$
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
answered Jan 15 at 23:59
Aditya DuaAditya Dua
1,11418
1,11418
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%2f3075058%2fgiven-the-time-of-the-nth-arrival-find-es-1s-n%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
2
$begingroup$
By the exchangeability of the family $(X_k)$, $E(X_kmid S_n)$ does not depend on $1leqslant kleqslant n$. Since $E(S_nmid S_n)=S_n$, this remark yields $E(X_kmid S_n)=frac1nS_n$ for every $1leqslant kleqslant n$, in particular $E(S_1mid S_n)=frac1nS_n$.
$endgroup$
– Did
Jan 15 at 23:00
$begingroup$
$X_i$ are the inter-arrival times.
$endgroup$
– Aditya Dua
Jan 15 at 23:46