Understanding $ int_c^infty (x-c) dF(x) $ through integration by parts
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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
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
$endgroup$
|
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
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
$endgroup$
$begingroup$
@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
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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
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
$endgroup$
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
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
edited Dec 18 '18 at 15:07
Did
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
edited Dec 18 '18 at 15:07
Did
248k23226466
edited Dec 18 '18 at 15:07
Did
248k23226466
edited Dec 18 '18 at 15:07
Did
248k23226466
248k23226466
asked Dec 18 '18 at 11:54
NutleNutle
332110
asked Dec 18 '18 at 11:54
NutleNutle
332110
asked Dec 18 '18 at 11:54
NutleNutle
332110
332110
$begingroup$
@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
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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 |
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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 |
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$begingroup$
@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