Explain why $E(X) = int_0^infty (1-F_X (t)) , dt$ for every nonnegative random variable $X$
$begingroup$
Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,
$$E(X) = int_0^infty (1-F_X (t)) , dt$$
when $X$ has : a) a discrete distribution, b) a continuous distribution.
I assumed that for the case of a continuous distribution, since $F_X (t) = mathbb{P}(Xleq t)$, then $1-F_X (t) = 1- mathbb{P}(Xleq t) = mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea.
probability probability-theory expected-value faq
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
$endgroup$
add a comment |
$begingroup$
Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,
$$E(X) = int_0^infty (1-F_X (t)) , dt$$
when $X$ has : a) a discrete distribution, b) a continuous distribution.
I assumed that for the case of a continuous distribution, since $F_X (t) = mathbb{P}(Xleq t)$, then $1-F_X (t) = 1- mathbb{P}(Xleq t) = mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea.
probability probability-theory expected-value faq
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
$endgroup$
1
$begingroup$
In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
$endgroup$
– Davide Giraudo
Jul 19 '12 at 13:42
3
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
$endgroup$
– Michael Chernick
Jul 19 '12 at 14:21
2
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
$endgroup$
– Dilip Sarwate
Jul 19 '12 at 15:38
2
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
$endgroup$
– cantorhead
Oct 26 '15 at 21:12
add a comment |
$begingroup$
Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,
$$E(X) = int_0^infty (1-F_X (t)) , dt$$
when $X$ has : a) a discrete distribution, b) a continuous distribution.
I assumed that for the case of a continuous distribution, since $F_X (t) = mathbb{P}(Xleq t)$, then $1-F_X (t) = 1- mathbb{P}(Xleq t) = mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea.
probability probability-theory expected-value faq
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
$endgroup$
Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,
$$E(X) = int_0^infty (1-F_X (t)) , dt$$
when $X$ has : a) a discrete distribution, b) a continuous distribution.
I assumed that for the case of a continuous distribution, since $F_X (t) = mathbb{P}(Xleq t)$, then $1-F_X (t) = 1- mathbb{P}(Xleq t) = mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea.
probability probability-theory expected-value faq
probability probability-theory expected-value faq
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
edited Nov 13 '18 at 11:55
Lee David Chung Lin
4,24531141
4,24531141
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
asked Jul 19 '12 at 13:37
mercurialmercurial
5762714
5762714
1
$begingroup$
In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
$endgroup$
– Davide Giraudo
Jul 19 '12 at 13:42
3
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
$endgroup$
– Michael Chernick
Jul 19 '12 at 14:21
2
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
$endgroup$
– Dilip Sarwate
Jul 19 '12 at 15:38
2
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
$endgroup$
– cantorhead
Oct 26 '15 at 21:12
add a comment |
1
$begingroup$
In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
$endgroup$
– Davide Giraudo
Jul 19 '12 at 13:42
3
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
$endgroup$
– Michael Chernick
Jul 19 '12 at 14:21
2
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
$endgroup$
– Dilip Sarwate
Jul 19 '12 at 15:38
2
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
$endgroup$
– cantorhead
Oct 26 '15 at 21:12
1
1
$begingroup$
In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
$endgroup$
– Davide Giraudo
Jul 19 '12 at 13:42
$begingroup$
In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
$endgroup$
– Davide Giraudo
Jul 19 '12 at 13:42
3
3
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
$endgroup$
– Michael Chernick
Jul 19 '12 at 14:21
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
$endgroup$
– Michael Chernick
Jul 19 '12 at 14:21
2
2
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
$endgroup$
– Dilip Sarwate
Jul 19 '12 at 15:38
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
$endgroup$
– Dilip Sarwate
Jul 19 '12 at 15:38
2
2
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
$endgroup$
– cantorhead
Oct 26 '15 at 21:12
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
$endgroup$
– cantorhead
Oct 26 '15 at 21:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For every nonnegative random variable $X$, whether discrete or continuous or a mix of these,
$$
X=int_0^Xmathrm dt=int_0^{+infty}mathbf 1_{Xgt t},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},mathrm dt,
$$
hence
$$
mathrm E(X)=int_0^{+infty}mathrm P(Xgt t),mathrm dt=int_0^{+infty}mathrm P(Xgeqslant t),mathrm dt.
$$
Likewise, for every $p>0$, $$
X^p=int_0^Xp,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgt t},p,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},p,t^{p-1},mathrm dt,
$$
hence
$$
mathrm E(X^p)=int_0^{+infty}p,t^{p-1},mathrm P(Xgt t),mathrm dt=int_0^{+infty}p,t^{p-1},mathrm P(Xgeqslant t),mathrm dt.
$$
answered Jul 19 '12 at 14:28
DidDid
248k23223460
$endgroup$
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
|
show 4 more comments
$begingroup$
Copied from Cross Validated / stats.stackexchange:
where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.
edited Apr 13 '17 at 12:44
Community♦
1
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
$endgroup$
add a comment |
$begingroup$
Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$int_{0}^{1} F^{-1}(U) 1 du = int_{-infty}^{infty} (1-F(U)) du = int_{-infty}^{infty} P(X geq x) dx$$
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For every nonnegative random variable $X$, whether discrete or continuous or a mix of these,
$$
X=int_0^Xmathrm dt=int_0^{+infty}mathbf 1_{Xgt t},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},mathrm dt,
$$
hence
$$
mathrm E(X)=int_0^{+infty}mathrm P(Xgt t),mathrm dt=int_0^{+infty}mathrm P(Xgeqslant t),mathrm dt.
$$
Likewise, for every $p>0$, $$
X^p=int_0^Xp,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgt t},p,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},p,t^{p-1},mathrm dt,
$$
hence
$$
mathrm E(X^p)=int_0^{+infty}p,t^{p-1},mathrm P(Xgt t),mathrm dt=int_0^{+infty}p,t^{p-1},mathrm P(Xgeqslant t),mathrm dt.
$$
answered Jul 19 '12 at 14:28
DidDid
248k23223460
$endgroup$
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
|
show 4 more comments
$begingroup$
For every nonnegative random variable $X$, whether discrete or continuous or a mix of these,
$$
X=int_0^Xmathrm dt=int_0^{+infty}mathbf 1_{Xgt t},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},mathrm dt,
$$
hence
$$
mathrm E(X)=int_0^{+infty}mathrm P(Xgt t),mathrm dt=int_0^{+infty}mathrm P(Xgeqslant t),mathrm dt.
$$
Likewise, for every $p>0$, $$
X^p=int_0^Xp,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgt t},p,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},p,t^{p-1},mathrm dt,
$$
hence
$$
mathrm E(X^p)=int_0^{+infty}p,t^{p-1},mathrm P(Xgt t),mathrm dt=int_0^{+infty}p,t^{p-1},mathrm P(Xgeqslant t),mathrm dt.
$$
answered Jul 19 '12 at 14:28
DidDid
248k23223460
$endgroup$
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
|
show 4 more comments
$begingroup$
For every nonnegative random variable $X$, whether discrete or continuous or a mix of these,
$$
X=int_0^Xmathrm dt=int_0^{+infty}mathbf 1_{Xgt t},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},mathrm dt,
$$
hence
$$
mathrm E(X)=int_0^{+infty}mathrm P(Xgt t),mathrm dt=int_0^{+infty}mathrm P(Xgeqslant t),mathrm dt.
$$
Likewise, for every $p>0$, $$
X^p=int_0^Xp,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgt t},p,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},p,t^{p-1},mathrm dt,
$$
hence
$$
mathrm E(X^p)=int_0^{+infty}p,t^{p-1},mathrm P(Xgt t),mathrm dt=int_0^{+infty}p,t^{p-1},mathrm P(Xgeqslant t),mathrm dt.
$$
answered Jul 19 '12 at 14:28
DidDid
248k23223460
$endgroup$
For every nonnegative random variable $X$, whether discrete or continuous or a mix of these,
$$
X=int_0^Xmathrm dt=int_0^{+infty}mathbf 1_{Xgt t},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},mathrm dt,
$$
hence
$$
mathrm E(X)=int_0^{+infty}mathrm P(Xgt t),mathrm dt=int_0^{+infty}mathrm P(Xgeqslant t),mathrm dt.
$$
Likewise, for every $p>0$, $$
X^p=int_0^Xp,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgt t},p,t^{p-1},mathrm dt=int_0^{+infty}mathbf 1_{Xgeqslant t},p,t^{p-1},mathrm dt,
$$
hence
$$
mathrm E(X^p)=int_0^{+infty}p,t^{p-1},mathrm P(Xgt t),mathrm dt=int_0^{+infty}p,t^{p-1},mathrm P(Xgeqslant t),mathrm dt.
$$
answered Jul 19 '12 at 14:28
DidDid
248k23223460
edited May 21 '17 at 13:52
answered Jul 19 '12 at 14:28
DidDid
248k23223460
answered Jul 19 '12 at 14:28
DidDid
248k23223460
answered Jul 19 '12 at 14:28
DidDid
248k23223460
248k23223460
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
|
show 4 more comments
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
$begingroup$
may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $mathbb{R}$) and the right side is an integral and therefore a number. Am I right?
$endgroup$
– Cupitor
Jun 2 '14 at 15:26
2
2
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
$begingroup$
@Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $omega$ of the right-hand-side is $$int_0^{+infty}mathbf 1_{X(omega)geqslant t},mathrm dt.$$
$endgroup$
– Did
Jun 2 '14 at 19:35
4
4
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
$begingroup$
$U=mathbf 1_{Xgeqslant t}$ is the function defined on $Omega$ by $U(omega)=1$ if $X(omega)geqslant t$ and $U(omega)=0$ otherwise.
$endgroup$
– Did
Jun 3 '14 at 10:09
2
2
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
$begingroup$
The second step is to consider the expectation of each side (that is, its integral with respect to $P$).
$endgroup$
– Did
Jun 3 '14 at 15:14
1
1
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
$begingroup$
@see Yes, your reading of these formulas and the proof in your first comment are both correct.
$endgroup$
– Did
Feb 12 '17 at 21:35
|
show 4 more comments
$begingroup$
Copied from Cross Validated / stats.stackexchange:
where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.
edited Apr 13 '17 at 12:44
Community♦
1
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
$endgroup$
add a comment |
$begingroup$
Copied from Cross Validated / stats.stackexchange:
where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.
edited Apr 13 '17 at 12:44
Community♦
1
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
$endgroup$
add a comment |
$begingroup$
Copied from Cross Validated / stats.stackexchange:
where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.
edited Apr 13 '17 at 12:44
Community♦
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answered Jul 19 '12 at 17:53
HenryHenry
100k480165
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Copied from Cross Validated / stats.stackexchange:
where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.
edited Apr 13 '17 at 12:44
Community♦
1
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
edited Apr 13 '17 at 12:44
Community♦
1
edited Apr 13 '17 at 12:44
Community♦
1
edited Apr 13 '17 at 12:44
Community♦
1
1
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
answered Jul 19 '12 at 17:53
HenryHenry
100k480165
100k480165
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$begingroup$
Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$int_{0}^{1} F^{-1}(U) 1 du = int_{-infty}^{infty} (1-F(U)) du = int_{-infty}^{infty} P(X geq x) dx$$
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
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add a comment |
$begingroup$
Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$int_{0}^{1} F^{-1}(U) 1 du = int_{-infty}^{infty} (1-F(U)) du = int_{-infty}^{infty} P(X geq x) dx$$
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
$endgroup$
add a comment |
$begingroup$
Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$int_{0}^{1} F^{-1}(U) 1 du = int_{-infty}^{infty} (1-F(U)) du = int_{-infty}^{infty} P(X geq x) dx$$
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
$endgroup$
Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$int_{0}^{1} F^{-1}(U) 1 du = int_{-infty}^{infty} (1-F(U)) du = int_{-infty}^{infty} P(X geq x) dx$$
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
edited Dec 11 '18 at 23:51
Xander Henderson
14.4k103554
14.4k103554
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
answered Dec 11 '18 at 23:33
CodingWolfCodingWolf
4319
4319
add a comment |
add a comment |
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In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative.
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– Davide Giraudo
Jul 19 '12 at 13:42
3
$begingroup$
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV.
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– Michael Chernick
Jul 19 '12 at 14:21
2
$begingroup$
See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem.
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– Dilip Sarwate
Jul 19 '12 at 15:38
2
$begingroup$
As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible.
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– cantorhead
Oct 26 '15 at 21:12