Expectation using survival function formula? The 2019 Stack Overflow Developer Survey Results...
Origin of "cooter" meaning "vagina"
Multiply Two Integer Polynomials
During Temple times, who can butcher a kosher animal?
Button changing it's text & action. Good or terrible?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
What is the most effective way of iterating a std::vector and why?
What tool would a Roman-age civilization have for the breaking of silver and other metals into dust?
Is this app Icon Browser Safe/Legit?
How to type this arrow in math mode?
Delete all lines which don't have n characters before delimiter
Are there incongruent pythagorean triangles with the same perimeter and same area?
Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?
What do hard-Brexiteers want with respect to the Irish border?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Why isn't airport relocation done gradually?
How to save as into a customized destination on macOS?
How to notate time signature switching consistently every measure
Is a "Democratic" Oligarchy-Style System Possible?
Write faster on AT24C32
Are there any other methods to apply to solving simultaneous equations?
Why not take a picture of a closer black hole?
A poker game description that does not feel gimmicky
Is "plugging out" electronic devices an American expression?
FPGA - DIY Programming
Expectation using survival function formula?
The 2019 Stack Overflow Developer Survey Results Are InSum of truncated normalsExpectation and Variance of Conditional Sum (using formal definition of conditional expectation)Expectation of a non-negative random variableRandom variable with infinite expectation but finite conditional expectationSimplified Strong Law of Large Number by Using Truncating FunctionSequence of random variables with infinite expectation, but partial sum convergesProof that a Gamma distribution is finite almost surelyFinite expectation induces finite almost surelySupremum of the sample mean without first moment assumption.Upper bound of expected maximum of weighted sub-gaussian r.v.s
$begingroup$
I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.
Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$
We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$
Now, the textbook gives the answer as
$$E[X] = int_e^infty e/xlog x space dx$$
However, surely using the above formula it should actually be
$$E[X] = e+ int_e^infty e/xlog x space dx$$
Since $P(X > y) = 1$ for $y in [0,e]$?
Or am I misunderstanding something basic here?
Edit:
The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_{(|X|leq n)}]$
probability
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
$endgroup$
add a comment |
$begingroup$
I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.
Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$
We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$
Now, the textbook gives the answer as
$$E[X] = int_e^infty e/xlog x space dx$$
However, surely using the above formula it should actually be
$$E[X] = e+ int_e^infty e/xlog x space dx$$
Since $P(X > y) = 1$ for $y in [0,e]$?
Or am I misunderstanding something basic here?
Edit:
The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_{(|X|leq n)}]$
probability
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
$endgroup$
2
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00
add a comment |
$begingroup$
I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.
Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$
We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$
Now, the textbook gives the answer as
$$E[X] = int_e^infty e/xlog x space dx$$
However, surely using the above formula it should actually be
$$E[X] = e+ int_e^infty e/xlog x space dx$$
Since $P(X > y) = 1$ for $y in [0,e]$?
Or am I misunderstanding something basic here?
Edit:
The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_{(|X|leq n)}]$
probability
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
$endgroup$
I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.
Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$
We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$
Now, the textbook gives the answer as
$$E[X] = int_e^infty e/xlog x space dx$$
However, surely using the above formula it should actually be
$$E[X] = e+ int_e^infty e/xlog x space dx$$
Since $P(X > y) = 1$ for $y in [0,e]$?
Or am I misunderstanding something basic here?
Edit:
The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_{(|X|leq n)}]$
probability
probability
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
edited Mar 22 at 1:01
Xiaomi
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
asked Mar 21 at 14:48
XiaomiXiaomi
1,081115
1,081115
2
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00
add a comment |
2
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00
2
2
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00
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%2f3156900%2fexpectation-using-survival-function-formula%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%2f3156900%2fexpectation-using-survival-function-formula%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
$begingroup$
Where does the summand $e$ in your calculation come from?
$endgroup$
– J. Wang
Mar 21 at 15:01
$begingroup$
What is the name of the survival function?
$endgroup$
– callculus
Mar 21 at 16:09
$begingroup$
$E(X)$ does not converge. See here
$endgroup$
– callculus
Mar 21 at 16:35
$begingroup$
I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
$endgroup$
– Xiaomi
Mar 22 at 1:00
$begingroup$
@J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
$endgroup$
– Xiaomi
Mar 22 at 1:00