Is $mathbbE(Xcdot I_X geq a = intlimits_a^infty P(X geq x) dx, a geq 0$ for non negative RV $X$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Does $f$ monotone and $fin L_1([a,infty))$ imply $lim_ttoinfty t f(t)=0$?Relation between integral by parts and Fubini's theoremManipulation of probability integralsAlternate definition of Lebesgue integralProve that if E is positive lebesgue measurabel set, then E − E and E + E contain non-empty open sets.Showing countable additivity for a pre product measureShow that $limlimits_T rightarrow infty frac12T int_-T^T |varphi(t)|^2 dt = sum_x in mathbbR(mu (x))^2$Integration by parts for functions not n times continuously differentibal?A more advanced proof for showing $mathbbE[X]=int_0^inftymathbbP(Xgeq t)dt$Upper bound for some measurable sets given the inequality $sum_n=1^infty mu (A_n) leq mu (bigcup_n=1^infty A_n) + epsilon$
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Is $mathbbE(Xcdot I_X geq a = intlimits_a^infty P(X geq x) dx, a geq 0$ for non negative RV $X$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Does $f$ monotone and $fin L_1([a,infty))$ imply $lim_ttoinfty t f(t)=0$?Relation between integral by parts and Fubini's theoremManipulation of probability integralsAlternate definition of Lebesgue integralProve that if E is positive lebesgue measurabel set, then E − E and E + E contain non-empty open sets.Showing countable additivity for a pre product measureShow that $limlimits_T rightarrow infty frac12T int_-T^T |varphi(t)|^2 dt = sum_x in mathbbR(mu (x))^2$Integration by parts for functions not n times continuously differentibal?A more advanced proof for showing $mathbbE[X]=int_0^inftymathbbP(Xgeq t)dt$Upper bound for some measurable sets given the inequality $sum_n=1^infty mu (A_n) leq mu (bigcup_n=1^infty A_n) + epsilon$
$begingroup$
For $a=0$, the solution is well known and I was able to show it both with integration by parts and Fubini's theorem. For $a > 0$ I could not establish the equality. Any hints?
probability-theory measure-theory
edited Apr 1 at 17:54
Bernard
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
$endgroup$
add a comment |
$begingroup$
For $a=0$, the solution is well known and I was able to show it both with integration by parts and Fubini's theorem. For $a > 0$ I could not establish the equality. Any hints?
probability-theory measure-theory
edited Apr 1 at 17:54
Bernard
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
$endgroup$
add a comment |
$begingroup$
For $a=0$, the solution is well known and I was able to show it both with integration by parts and Fubini's theorem. For $a > 0$ I could not establish the equality. Any hints?
probability-theory measure-theory
edited Apr 1 at 17:54
Bernard
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
$endgroup$
For $a=0$, the solution is well known and I was able to show it both with integration by parts and Fubini's theorem. For $a > 0$ I could not establish the equality. Any hints?
probability-theory measure-theory
probability-theory measure-theory
edited Apr 1 at 17:54
Bernard
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
edited Apr 1 at 17:54
Bernard
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
edited Apr 1 at 17:54
Bernard
124k742117
edited Apr 1 at 17:54
Bernard
124k742117
edited Apr 1 at 17:54
Bernard
124k742117
124k742117
asked Apr 1 at 17:54
Mike D.Mike D.
1
asked Apr 1 at 17:54
Mike D.Mike D.
1
asked Apr 1 at 17:54
Mike D.Mike D.
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes. Put $Y=Xcdot I_Xgeq a$ and integrate from zero.
$$
beginaligned
mathbbE(Xcdot I_Xgeq a)&=mathbbE(Y)\
&=int_0^inftyP(Ygeq x),dx\
&=int_0^inftyP(Xgeq x,hboxand,Xgeq a),dx\
&=int_a^inftyP(Xgeq x),dx
endaligned$$
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. Put $Y=Xcdot I_Xgeq a$ and integrate from zero.
$$
beginaligned
mathbbE(Xcdot I_Xgeq a)&=mathbbE(Y)\
&=int_0^inftyP(Ygeq x),dx\
&=int_0^inftyP(Xgeq x,hboxand,Xgeq a),dx\
&=int_a^inftyP(Xgeq x),dx
endaligned$$
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
$endgroup$
add a comment |
$begingroup$
Yes. Put $Y=Xcdot I_Xgeq a$ and integrate from zero.
$$
beginaligned
mathbbE(Xcdot I_Xgeq a)&=mathbbE(Y)\
&=int_0^inftyP(Ygeq x),dx\
&=int_0^inftyP(Xgeq x,hboxand,Xgeq a),dx\
&=int_a^inftyP(Xgeq x),dx
endaligned$$
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
$endgroup$
add a comment |
$begingroup$
Yes. Put $Y=Xcdot I_Xgeq a$ and integrate from zero.
$$
beginaligned
mathbbE(Xcdot I_Xgeq a)&=mathbbE(Y)\
&=int_0^inftyP(Ygeq x),dx\
&=int_0^inftyP(Xgeq x,hboxand,Xgeq a),dx\
&=int_a^inftyP(Xgeq x),dx
endaligned$$
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
$endgroup$
Yes. Put $Y=Xcdot I_Xgeq a$ and integrate from zero.
$$
beginaligned
mathbbE(Xcdot I_Xgeq a)&=mathbbE(Y)\
&=int_0^inftyP(Ygeq x),dx\
&=int_0^inftyP(Xgeq x,hboxand,Xgeq a),dx\
&=int_a^inftyP(Xgeq x),dx
endaligned$$
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
edited Apr 1 at 18:09
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
answered Apr 1 at 18:03
uniquesolutionuniquesolution
9,7421823
9,7421823
add a comment |
add a comment |
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