Understanding $ int_c^infty (x-c) dF(x) $ through integration by parts












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:07









Did

248k23225463




248k23225463










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


















  • $begingroup$
    @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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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