Autocorrelation of Geometric Process
$begingroup$
I have this scenario where $S[n] = 0$ if an email does not arrive and $S[n] = 1$ if an email arrives in the $n^{th}$ minute. Therefore, $S[n]$ follows a Bernoulli distribution. Now $U[m]$ is the number of minutes between the $m^{th}$ and $(m+1)^{th}$ email where $m = 0, 1, 2, 3...$. Therefore, $U[m]$ follows a Geometric distribution. Also $P(S[n] = 0) = p$, therefore, I have concluded that,
$P_{U[m]}(m) = p^{m}(1-p)$
I am required to determine the autocorrelation of $U[m]$, and so far, I have:
$R_{UU}[m, m+p] = E[U[m]] times E[U[m+p]] = frac{p^2}{(1-p)^2}$ for $p ne 0$
But for $p = 0$, I need to find:
$E[U^2[m]] = sum_{m=0}^{infty}u_m^2 cdot P_{U[m]}(m)$
Which is also:
$E[U^2[m]] = sum_{m=0}^{infty}m^2 cdot (1-p) cdot p^m$
$E[U^2[m]] = (1-p) cdot p sum_{m=0}^{infty}m^2 cdot p^{m-1}$
Now I am not sure how to proceed?
statistics correlation expected-value
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
$endgroup$
add a comment |
$begingroup$
I have this scenario where $S[n] = 0$ if an email does not arrive and $S[n] = 1$ if an email arrives in the $n^{th}$ minute. Therefore, $S[n]$ follows a Bernoulli distribution. Now $U[m]$ is the number of minutes between the $m^{th}$ and $(m+1)^{th}$ email where $m = 0, 1, 2, 3...$. Therefore, $U[m]$ follows a Geometric distribution. Also $P(S[n] = 0) = p$, therefore, I have concluded that,
$P_{U[m]}(m) = p^{m}(1-p)$
I am required to determine the autocorrelation of $U[m]$, and so far, I have:
$R_{UU}[m, m+p] = E[U[m]] times E[U[m+p]] = frac{p^2}{(1-p)^2}$ for $p ne 0$
But for $p = 0$, I need to find:
$E[U^2[m]] = sum_{m=0}^{infty}u_m^2 cdot P_{U[m]}(m)$
Which is also:
$E[U^2[m]] = sum_{m=0}^{infty}m^2 cdot (1-p) cdot p^m$
$E[U^2[m]] = (1-p) cdot p sum_{m=0}^{infty}m^2 cdot p^{m-1}$
Now I am not sure how to proceed?
statistics correlation expected-value
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
$endgroup$
$begingroup$
You should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58
add a comment |
$begingroup$
I have this scenario where $S[n] = 0$ if an email does not arrive and $S[n] = 1$ if an email arrives in the $n^{th}$ minute. Therefore, $S[n]$ follows a Bernoulli distribution. Now $U[m]$ is the number of minutes between the $m^{th}$ and $(m+1)^{th}$ email where $m = 0, 1, 2, 3...$. Therefore, $U[m]$ follows a Geometric distribution. Also $P(S[n] = 0) = p$, therefore, I have concluded that,
$P_{U[m]}(m) = p^{m}(1-p)$
I am required to determine the autocorrelation of $U[m]$, and so far, I have:
$R_{UU}[m, m+p] = E[U[m]] times E[U[m+p]] = frac{p^2}{(1-p)^2}$ for $p ne 0$
But for $p = 0$, I need to find:
$E[U^2[m]] = sum_{m=0}^{infty}u_m^2 cdot P_{U[m]}(m)$
Which is also:
$E[U^2[m]] = sum_{m=0}^{infty}m^2 cdot (1-p) cdot p^m$
$E[U^2[m]] = (1-p) cdot p sum_{m=0}^{infty}m^2 cdot p^{m-1}$
Now I am not sure how to proceed?
statistics correlation expected-value
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
$endgroup$
I have this scenario where $S[n] = 0$ if an email does not arrive and $S[n] = 1$ if an email arrives in the $n^{th}$ minute. Therefore, $S[n]$ follows a Bernoulli distribution. Now $U[m]$ is the number of minutes between the $m^{th}$ and $(m+1)^{th}$ email where $m = 0, 1, 2, 3...$. Therefore, $U[m]$ follows a Geometric distribution. Also $P(S[n] = 0) = p$, therefore, I have concluded that,
$P_{U[m]}(m) = p^{m}(1-p)$
I am required to determine the autocorrelation of $U[m]$, and so far, I have:
$R_{UU}[m, m+p] = E[U[m]] times E[U[m+p]] = frac{p^2}{(1-p)^2}$ for $p ne 0$
But for $p = 0$, I need to find:
$E[U^2[m]] = sum_{m=0}^{infty}u_m^2 cdot P_{U[m]}(m)$
Which is also:
$E[U^2[m]] = sum_{m=0}^{infty}m^2 cdot (1-p) cdot p^m$
$E[U^2[m]] = (1-p) cdot p sum_{m=0}^{infty}m^2 cdot p^{m-1}$
Now I am not sure how to proceed?
statistics correlation expected-value
statistics correlation expected-value
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 3:14
Micah MungalMicah Mungal
83
83
$begingroup$
You should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58
add a comment |
$begingroup$
You should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58
$begingroup$
You should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
You should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58
add a comment |
0
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',
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%2f3031957%2fautocorrelation-of-geometric-process%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
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.
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%2f3031957%2fautocorrelation-of-geometric-process%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 should have that $mathbb P(U[m]=k) = p^{k-1}(1-p)$ for $kgeqslant1$. It is still a geometric distribution, but there must be at least one minute between each mail, so the index starts at one.
$endgroup$
– Math1000
Dec 9 '18 at 3:31
$begingroup$
More importantly, there shouldn't be any autocorrelation in this process. The random variables in the sequence $U[m]$ are the increments of a Bernoulli process, which are independent.
$endgroup$
– Math1000
Dec 9 '18 at 3:47
$begingroup$
The question states that $m=0, 1, 2, 3, …$, that is why I used P $P_{U[m]}(m) = p^m (1-p)$. The question also has an answer, (given below), I am just trying to figure out how to reach that answer. The answer is $R_{UU}(m, m + p) = frac{p^2}{(1-p)^2} + frac{p}{(1-p)^2}delta[-p].$
$endgroup$
– Micah Mungal
Dec 9 '18 at 10:58