Understanding $ int_c^infty (x-c) dF(x) $ through integration by parts
$begingroup$
Following this answer, it is claimed that we can solve the problem in the following way:
$$displaystyle int_{c}^{infty} (x-c) dF(x) = lim_{y rightarrow infty} (y-c) F(y) - displaystyle int_{c}^{infty} F(x) dx.$$
where $F$ is the cumulative distribution function of a given random variable.
Does this limit exist though? In my understanding, since distribution function saturates at 1, the first limit should converge to infinity? What am I missing?
integration probability-theory probability-distributions expected-value
edited Dec 18 '18 at 15:07
Did
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
$endgroup$
|
show 5 more comments
$begingroup$
Following this answer, it is claimed that we can solve the problem in the following way:
$$displaystyle int_{c}^{infty} (x-c) dF(x) = lim_{y rightarrow infty} (y-c) F(y) - displaystyle int_{c}^{infty} F(x) dx.$$
where $F$ is the cumulative distribution function of a given random variable.
Does this limit exist though? In my understanding, since distribution function saturates at 1, the first limit should converge to infinity? What am I missing?
integration probability-theory probability-distributions expected-value
edited Dec 18 '18 at 15:07
Did
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
$endgroup$
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03
|
show 5 more comments
$begingroup$
Following this answer, it is claimed that we can solve the problem in the following way:
$$displaystyle int_{c}^{infty} (x-c) dF(x) = lim_{y rightarrow infty} (y-c) F(y) - displaystyle int_{c}^{infty} F(x) dx.$$
where $F$ is the cumulative distribution function of a given random variable.
Does this limit exist though? In my understanding, since distribution function saturates at 1, the first limit should converge to infinity? What am I missing?
integration probability-theory probability-distributions expected-value
edited Dec 18 '18 at 15:07
Did
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
$endgroup$
Following this answer, it is claimed that we can solve the problem in the following way:
$$displaystyle int_{c}^{infty} (x-c) dF(x) = lim_{y rightarrow infty} (y-c) F(y) - displaystyle int_{c}^{infty} F(x) dx.$$
where $F$ is the cumulative distribution function of a given random variable.
Does this limit exist though? In my understanding, since distribution function saturates at 1, the first limit should converge to infinity? What am I missing?
integration probability-theory probability-distributions expected-value
integration probability-theory probability-distributions expected-value
edited Dec 18 '18 at 15:07
Did
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
edited Dec 18 '18 at 15:07
Did
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
edited Dec 18 '18 at 15:07
Did
248k23224463
edited Dec 18 '18 at 15:07
Did
248k23224463
edited Dec 18 '18 at 15:07
Did
248k23224463
248k23224463
asked Dec 18 '18 at 11:54
NutleNutle
320110
asked Dec 18 '18 at 11:54
NutleNutle
320110
asked Dec 18 '18 at 11:54
NutleNutle
320110
320110
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03
|
show 5 more comments
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Shouldn't the formula be,
$displaystyle int_c^infty dP(x) = lim_{ytoinfty} left{ (y-c) P(y) - int_c^y P(x) dx right} $
Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $infty$.
Not very helpful in finding the answer but perhaps explains the confusion.
answered Dec 18 '18 at 12:45
WA DonWA Don
261
$endgroup$
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
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%2f3045074%2funderstanding-int-c-infty-x-c-dfx-through-integration-by-parts%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
$begingroup$
Shouldn't the formula be,
$displaystyle int_c^infty dP(x) = lim_{ytoinfty} left{ (y-c) P(y) - int_c^y P(x) dx right} $
Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $infty$.
Not very helpful in finding the answer but perhaps explains the confusion.
answered Dec 18 '18 at 12:45
WA DonWA Don
261
$endgroup$
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
add a comment |
$begingroup$
Shouldn't the formula be,
$displaystyle int_c^infty dP(x) = lim_{ytoinfty} left{ (y-c) P(y) - int_c^y P(x) dx right} $
Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $infty$.
Not very helpful in finding the answer but perhaps explains the confusion.
answered Dec 18 '18 at 12:45
WA DonWA Don
261
$endgroup$
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
add a comment |
$begingroup$
Shouldn't the formula be,
$displaystyle int_c^infty dP(x) = lim_{ytoinfty} left{ (y-c) P(y) - int_c^y P(x) dx right} $
Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $infty$.
Not very helpful in finding the answer but perhaps explains the confusion.
answered Dec 18 '18 at 12:45
WA DonWA Don
261
$endgroup$
Shouldn't the formula be,
$displaystyle int_c^infty dP(x) = lim_{ytoinfty} left{ (y-c) P(y) - int_c^y P(x) dx right} $
Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $infty$.
Not very helpful in finding the answer but perhaps explains the confusion.
answered Dec 18 '18 at 12:45
WA DonWA Don
261
answered Dec 18 '18 at 12:45
WA DonWA Don
261
answered Dec 18 '18 at 12:45
WA DonWA Don
261
answered Dec 18 '18 at 12:45
WA DonWA Don
261
261
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
add a comment |
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
$begingroup$
Thanks. You're right, missed that the second term would approach $-infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out?
$endgroup$
– Nutle
Dec 18 '18 at 13:00
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%2f3045074%2funderstanding-int-c-infty-x-c-dfx-through-integration-by-parts%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
$begingroup$
@CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it
$endgroup$
– Nutle
Dec 18 '18 at 14:13
$begingroup$
Oh I may have misunderstood, is P the cumulative distribution function?
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:16
$begingroup$
@CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks!
$endgroup$
– Nutle
Dec 18 '18 at 14:18
$begingroup$
If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below
$endgroup$
– Calvin Khor
Dec 18 '18 at 14:21
$begingroup$
A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-)
$endgroup$
– Did
Dec 18 '18 at 15:03