Showing $int_{mathbb R} mid F(x)-G(x)mid dx = int_0^1 mid F^{-1}(u)-G^{-1}(u)mid du$ with $F$, $G$ CDF...











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Let's $X$ and $Y$ have CDF functions admitting moment of order $1$.
Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.



I want to show that $$int_{mathbb R} mid F(x)-G(x)mid dx = int_{0}^{1} mid F^{-1}(u)-G^{-1}(u)mid du,.$$










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  • What is $F^{-1}=$?
    – Daniel Camarena Perez
    Nov 17 at 13:09










  • $F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
    – Farouk Deutsch
    Nov 17 at 13:19






  • 1




    I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
    – irchans
    Nov 17 at 15:19










  • thank you i now visualizing the thing but i'm still stuck to prove it with words
    – Farouk Deutsch
    Nov 17 at 16:07















up vote
4
down vote

favorite
2












Let's $X$ and $Y$ have CDF functions admitting moment of order $1$.
Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.



I want to show that $$int_{mathbb R} mid F(x)-G(x)mid dx = int_{0}^{1} mid F^{-1}(u)-G^{-1}(u)mid du,.$$










share|cite|improve this question
























  • What is $F^{-1}=$?
    – Daniel Camarena Perez
    Nov 17 at 13:09










  • $F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
    – Farouk Deutsch
    Nov 17 at 13:19






  • 1




    I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
    – irchans
    Nov 17 at 15:19










  • thank you i now visualizing the thing but i'm still stuck to prove it with words
    – Farouk Deutsch
    Nov 17 at 16:07













up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





Let's $X$ and $Y$ have CDF functions admitting moment of order $1$.
Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.



I want to show that $$int_{mathbb R} mid F(x)-G(x)mid dx = int_{0}^{1} mid F^{-1}(u)-G^{-1}(u)mid du,.$$










share|cite|improve this question















Let's $X$ and $Y$ have CDF functions admitting moment of order $1$.
Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.



I want to show that $$int_{mathbb R} mid F(x)-G(x)mid dx = int_{0}^{1} mid F^{-1}(u)-G^{-1}(u)mid du,.$$







probability probability-theory probability-distributions definite-integrals inverse-function






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edited Nov 17 at 17:18









Batominovski

31.8k23190




31.8k23190










asked Nov 17 at 12:59









Farouk Deutsch

1189




1189












  • What is $F^{-1}=$?
    – Daniel Camarena Perez
    Nov 17 at 13:09










  • $F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
    – Farouk Deutsch
    Nov 17 at 13:19






  • 1




    I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
    – irchans
    Nov 17 at 15:19










  • thank you i now visualizing the thing but i'm still stuck to prove it with words
    – Farouk Deutsch
    Nov 17 at 16:07


















  • What is $F^{-1}=$?
    – Daniel Camarena Perez
    Nov 17 at 13:09










  • $F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
    – Farouk Deutsch
    Nov 17 at 13:19






  • 1




    I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
    – irchans
    Nov 17 at 15:19










  • thank you i now visualizing the thing but i'm still stuck to prove it with words
    – Farouk Deutsch
    Nov 17 at 16:07
















What is $F^{-1}=$?
– Daniel Camarena Perez
Nov 17 at 13:09




What is $F^{-1}=$?
– Daniel Camarena Perez
Nov 17 at 13:09












$F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
– Farouk Deutsch
Nov 17 at 13:19




$F^{-1}$ is the quantile function it is the inverse of the CDF. In this case, since i don't have an explicit CDF i can't know ecplicitely$F^{-1}$
– Farouk Deutsch
Nov 17 at 13:19




1




1




I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
– irchans
Nov 17 at 15:19




I think that en.wikipedia.org/wiki/Integral_of_inverse_functions would be helpful.
– irchans
Nov 17 at 15:19












thank you i now visualizing the thing but i'm still stuck to prove it with words
– Farouk Deutsch
Nov 17 at 16:07




thank you i now visualizing the thing but i'm still stuck to prove it with words
– Farouk Deutsch
Nov 17 at 16:07










3 Answers
3






active

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up vote
2
down vote



accepted










If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
$$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$



Then note that



$$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
thus



$$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$



returning to the 1D integral notation, this is saying that
$$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$



Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )



enter image description here



$^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.






share|cite|improve this answer



















  • 1




    If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
    – Batominovski
    Nov 17 at 17:16












  • @Batominovski I think you didn't make a mistake :)
    – Calvin Khor
    Nov 17 at 17:20


















up vote
3
down vote













This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
$$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
so we have
$$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$



Now, because $F$ and $G$ admit the first moments, the integral
$$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
is finite due to (*). From Calvin Khor's answer, we have
$$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
$$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
and
$$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
Observe that
$$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
leq G^{-1}(y)big}$$

and
$$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
leq F^{-1}(y)big},.$$

Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
$$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$






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  • The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
    – Batominovski
    Nov 17 at 17:11




















up vote
2
down vote













Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.



This means that one has to prove that
$$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
and
$$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
are in fact the same sets. This is "pure logic"; one just has to go through the motions.






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    3 Answers
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    active

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    3 Answers
    3






    active

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    active

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    active

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    up vote
    2
    down vote



    accepted










    If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$



    Then note that



    $$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
    and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
    thus



    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$



    returning to the 1D integral notation, this is saying that
    $$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$



    Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )



    enter image description here



    $^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.






    share|cite|improve this answer



















    • 1




      If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
      – Batominovski
      Nov 17 at 17:16












    • @Batominovski I think you didn't make a mistake :)
      – Calvin Khor
      Nov 17 at 17:20















    up vote
    2
    down vote



    accepted










    If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$



    Then note that



    $$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
    and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
    thus



    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$



    returning to the 1D integral notation, this is saying that
    $$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$



    Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )



    enter image description here



    $^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.






    share|cite|improve this answer



















    • 1




      If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
      – Batominovski
      Nov 17 at 17:16












    • @Batominovski I think you didn't make a mistake :)
      – Calvin Khor
      Nov 17 at 17:20













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$



    Then note that



    $$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
    and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
    thus



    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$



    returning to the 1D integral notation, this is saying that
    $$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$



    Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )



    enter image description here



    $^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.






    share|cite|improve this answer














    If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$



    Then note that



    $$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
    and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
    thus



    $$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$



    returning to the 1D integral notation, this is saying that
    $$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$



    Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )



    enter image description here



    $^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 17 at 16:35

























    answered Nov 17 at 16:18









    Calvin Khor

    10.8k21437




    10.8k21437








    • 1




      If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
      – Batominovski
      Nov 17 at 17:16












    • @Batominovski I think you didn't make a mistake :)
      – Calvin Khor
      Nov 17 at 17:20














    • 1




      If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
      – Batominovski
      Nov 17 at 17:16












    • @Batominovski I think you didn't make a mistake :)
      – Calvin Khor
      Nov 17 at 17:20








    1




    1




    If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
    – Batominovski
    Nov 17 at 17:16






    If I didn't make any mistake, then your guess is correct. The claim works even for $F$ and $G$ without inverses. See my answer.
    – Batominovski
    Nov 17 at 17:16














    @Batominovski I think you didn't make a mistake :)
    – Calvin Khor
    Nov 17 at 17:20




    @Batominovski I think you didn't make a mistake :)
    – Calvin Khor
    Nov 17 at 17:20










    up vote
    3
    down vote













    This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
    $$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
    for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
    so we have
    $$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$



    Now, because $F$ and $G$ admit the first moments, the integral
    $$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
    is finite due to (*). From Calvin Khor's answer, we have
    $$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
    $$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
    and
    $$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
    Observe that
    $$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
    leq G^{-1}(y)big}$$

    and
    $$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
    leq F^{-1}(y)big},.$$

    Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
    $$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$






    share|cite|improve this answer























    • The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
      – Batominovski
      Nov 17 at 17:11

















    up vote
    3
    down vote













    This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
    $$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
    for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
    so we have
    $$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$



    Now, because $F$ and $G$ admit the first moments, the integral
    $$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
    is finite due to (*). From Calvin Khor's answer, we have
    $$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
    $$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
    and
    $$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
    Observe that
    $$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
    leq G^{-1}(y)big}$$

    and
    $$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
    leq F^{-1}(y)big},.$$

    Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
    $$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$






    share|cite|improve this answer























    • The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
      – Batominovski
      Nov 17 at 17:11















    up vote
    3
    down vote










    up vote
    3
    down vote









    This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
    $$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
    for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
    so we have
    $$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$



    Now, because $F$ and $G$ admit the first moments, the integral
    $$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
    is finite due to (*). From Calvin Khor's answer, we have
    $$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
    $$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
    and
    $$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
    Observe that
    $$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
    leq G^{-1}(y)big}$$

    and
    $$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
    leq F^{-1}(y)big},.$$

    Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
    $$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$






    share|cite|improve this answer














    This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
    $$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
    for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
    so we have
    $$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$



    Now, because $F$ and $G$ admit the first moments, the integral
    $$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
    is finite due to (*). From Calvin Khor's answer, we have
    $$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
    $$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
    and
    $$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
    Observe that
    $$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
    leq G^{-1}(y)big}$$

    and
    $$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
    leq F^{-1}(y)big},.$$

    Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
    $$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 18 at 15:38

























    answered Nov 17 at 17:07









    Batominovski

    31.8k23190




    31.8k23190












    • The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
      – Batominovski
      Nov 17 at 17:11




















    • The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
      – Batominovski
      Nov 17 at 17:11


















    The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
    – Batominovski
    Nov 17 at 17:11






    The proof still works, by the way, if we instead define $T^{-1}:(0,1)tomathbb{R}$ to be $$T^{-1}(u):=infbig{vinmathbb{R},|,T(v)geq ubig},,$$ for each distribution function $T$.
    – Batominovski
    Nov 17 at 17:11












    up vote
    2
    down vote













    Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.



    This means that one has to prove that
    $$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
    and
    $$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
    are in fact the same sets. This is "pure logic"; one just has to go through the motions.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.



      This means that one has to prove that
      $$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
      and
      $$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
      are in fact the same sets. This is "pure logic"; one just has to go through the motions.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.



        This means that one has to prove that
        $$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
        and
        $$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
        are in fact the same sets. This is "pure logic"; one just has to go through the motions.






        share|cite|improve this answer














        Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.



        This means that one has to prove that
        $$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
        and
        $$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
        are in fact the same sets. This is "pure logic"; one just has to go through the motions.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 17 at 16:33

























        answered Nov 17 at 15:20









        Christian Blatter

        171k7111325




        171k7111325






























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