Stochastic orders of summands when sum has fixed distribution

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Suppose $X,X',Y,Y'$ are independent random variables. We know that $X+Y oversetd= X'+Y'oversetd=Uniform[0,1]$ and $Xprec X'$ in the sense that $P(X> x)le P(X'>x)$ for any $xin mathbbR$. Is it true that $Y'prec Y$?




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    Suppose $X,X',Y,Y'$ are independent random variables. We know that $X+Y oversetd= X'+Y'oversetd=Uniform[0,1]$ and $Xprec X'$ in the sense that $P(X> x)le P(X'>x)$ for any $xin mathbbR$. Is it true that $Y'prec Y$?




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      Suppose $X,X',Y,Y'$ are independent random variables. We know that $X+Y oversetd= X'+Y'oversetd=Uniform[0,1]$ and $Xprec X'$ in the sense that $P(X> x)le P(X'>x)$ for any $xin mathbbR$. Is it true that $Y'prec Y$?




      probability-distributions random-variables




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      Suppose $X,X',Y,Y'$ are independent random variables. We know that $X+Y oversetd= X'+Y'oversetd=Uniform[0,1]$ and $Xprec X'$ in the sense that $P(X> x)le P(X'>x)$ for any $xin mathbbR$. Is it true that $Y'prec Y$?





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      edited Aug 9 at 23:31

























      asked Aug 8 at 22:02









      Uchiha

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          This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y overset d =X'+Y'$ and $Xoverset d=X'$ implies $Yoverset d=Y'$. It is well known that there exist characteristic functions $phi, phi_1,phi_2$ such that $phi (t) phi_1(t)=phi (t) phi_2(t)$ for all $t$ but $phi_1 neq phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $phi $, $Y$ has characteristic function $phi_1$ and $Y'$ has characteristic function $phi_2$ to get a contradiction.






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          • Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
            – Uchiha
            Aug 9 at 23:30











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          This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y overset d =X'+Y'$ and $Xoverset d=X'$ implies $Yoverset d=Y'$. It is well known that there exist characteristic functions $phi, phi_1,phi_2$ such that $phi (t) phi_1(t)=phi (t) phi_2(t)$ for all $t$ but $phi_1 neq phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $phi $, $Y$ has characteristic function $phi_1$ and $Y'$ has characteristic function $phi_2$ to get a contradiction.






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          • Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
            – Uchiha
            Aug 9 at 23:30















          up vote
          1
          down vote













          This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y overset d =X'+Y'$ and $Xoverset d=X'$ implies $Yoverset d=Y'$. It is well known that there exist characteristic functions $phi, phi_1,phi_2$ such that $phi (t) phi_1(t)=phi (t) phi_2(t)$ for all $t$ but $phi_1 neq phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $phi $, $Y$ has characteristic function $phi_1$ and $Y'$ has characteristic function $phi_2$ to get a contradiction.






          share|cite|improve this answer






















          • Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
            – Uchiha
            Aug 9 at 23:30













          up vote
          1
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          up vote
          1
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          This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y overset d =X'+Y'$ and $Xoverset d=X'$ implies $Yoverset d=Y'$. It is well known that there exist characteristic functions $phi, phi_1,phi_2$ such that $phi (t) phi_1(t)=phi (t) phi_2(t)$ for all $t$ but $phi_1 neq phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $phi $, $Y$ has characteristic function $phi_1$ and $Y'$ has characteristic function $phi_2$ to get a contradiction.






          share|cite|improve this answer














          This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y overset d =X'+Y'$ and $Xoverset d=X'$ implies $Yoverset d=Y'$. It is well known that there exist characteristic functions $phi, phi_1,phi_2$ such that $phi (t) phi_1(t)=phi (t) phi_2(t)$ for all $t$ but $phi_1 neq phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $phi $, $Y$ has characteristic function $phi_1$ and $Y'$ has characteristic function $phi_2$ to get a contradiction.







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          edited Aug 9 at 23:33

























          answered Aug 9 at 12:01









          Kavi Rama Murthy

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          21.4k2830











          • Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
            – Uchiha
            Aug 9 at 23:30

















          • Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
            – Uchiha
            Aug 9 at 23:30
















          Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
          – Uchiha
          Aug 9 at 23:30





          Thanks. The equality $phi(t)phi_1(t)=phi(t)phi_2(t)$ means that $phi(t)$ must be zero somewhere. If we know in addition that $X+Yoversetd=$ Uniform[0,1], would there be a counterexample?
          – Uchiha
          Aug 9 at 23:30













           

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