Necessary and sufficient conditions on a trivariate probability distribution for being the probability...












2














Consider a trivariate probability distribution $P: mathbb{R}^3rightarrow [0,1]$. I have the following questions:



(1) Are there necessary conditions on the cumulative distribution function (CDF) associated with $P$ ensuring that
$$
exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
$$



(2) Are there necessary and sufficient conditions on the CDF associated with $P$ ensuring that
$$
exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
$$



(3) The conditions that you propose can be "approximated" as a linear constraint on the CDF?





I'm providing more details on my question also thanks to/inspired by the answers below. The answers below help, but I'm still not satisfied. Please help if you can.



If there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $P$ should satisfy: for every $begin{pmatrix}
a_1\
b_1\
c_1
end{pmatrix}leq begin{pmatrix}
a_2\
b_2\
c_2
end{pmatrix}$




  • If $a_2geq b_2+c_2$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\
    P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & forall a_3geq a_2\
    end{cases}
    $$


  • If $b_1leq a_1-c_2$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\
    P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & forall b_3leq b_1\
    end{cases}
    $$


  • If $a_1 leq b_1+c_1$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([b_1+c_1,a_2],[b_1,b_2],[c_1,c_2])\
    P([a_3,a_1], [b_1, b_2], [c_1, c_2])=0 & forall a_3 leq a_1
    end{cases}
    $$


  • If $b_2geq a_2-c_1$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\
    P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & forall b_3geq b_2
    end{cases}
    $$


  • If $c_2 geq a_2-b_1$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\
    P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & forall c_3geq c_2
    end{cases}
    $$


  • If $c_1leq a_1-b_2$
    $$
    begin{cases}
    P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\
    P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & forall c_3leq c_1
    end{cases}
    $$



All the implications above can be re-written as linear function of the CDF associated with $P$.



However: are these implications also sufficient? If yes, I don't know how to prove it; If not, I don't know how to find a counterexample.










share|cite|improve this question





























    2














    Consider a trivariate probability distribution $P: mathbb{R}^3rightarrow [0,1]$. I have the following questions:



    (1) Are there necessary conditions on the cumulative distribution function (CDF) associated with $P$ ensuring that
    $$
    exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
    $$



    (2) Are there necessary and sufficient conditions on the CDF associated with $P$ ensuring that
    $$
    exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
    $$



    (3) The conditions that you propose can be "approximated" as a linear constraint on the CDF?





    I'm providing more details on my question also thanks to/inspired by the answers below. The answers below help, but I'm still not satisfied. Please help if you can.



    If there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $P$ should satisfy: for every $begin{pmatrix}
    a_1\
    b_1\
    c_1
    end{pmatrix}leq begin{pmatrix}
    a_2\
    b_2\
    c_2
    end{pmatrix}$




    • If $a_2geq b_2+c_2$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\
      P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & forall a_3geq a_2\
      end{cases}
      $$


    • If $b_1leq a_1-c_2$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\
      P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & forall b_3leq b_1\
      end{cases}
      $$


    • If $a_1 leq b_1+c_1$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([b_1+c_1,a_2],[b_1,b_2],[c_1,c_2])\
      P([a_3,a_1], [b_1, b_2], [c_1, c_2])=0 & forall a_3 leq a_1
      end{cases}
      $$


    • If $b_2geq a_2-c_1$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\
      P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & forall b_3geq b_2
      end{cases}
      $$


    • If $c_2 geq a_2-b_1$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\
      P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & forall c_3geq c_2
      end{cases}
      $$


    • If $c_1leq a_1-b_2$
      $$
      begin{cases}
      P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\
      P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & forall c_3leq c_1
      end{cases}
      $$



    All the implications above can be re-written as linear function of the CDF associated with $P$.



    However: are these implications also sufficient? If yes, I don't know how to prove it; If not, I don't know how to find a counterexample.










    share|cite|improve this question



























      2












      2








      2


      2





      Consider a trivariate probability distribution $P: mathbb{R}^3rightarrow [0,1]$. I have the following questions:



      (1) Are there necessary conditions on the cumulative distribution function (CDF) associated with $P$ ensuring that
      $$
      exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
      $$



      (2) Are there necessary and sufficient conditions on the CDF associated with $P$ ensuring that
      $$
      exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
      $$



      (3) The conditions that you propose can be "approximated" as a linear constraint on the CDF?





      I'm providing more details on my question also thanks to/inspired by the answers below. The answers below help, but I'm still not satisfied. Please help if you can.



      If there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $P$ should satisfy: for every $begin{pmatrix}
      a_1\
      b_1\
      c_1
      end{pmatrix}leq begin{pmatrix}
      a_2\
      b_2\
      c_2
      end{pmatrix}$




      • If $a_2geq b_2+c_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\
        P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & forall a_3geq a_2\
        end{cases}
        $$


      • If $b_1leq a_1-c_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\
        P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & forall b_3leq b_1\
        end{cases}
        $$


      • If $a_1 leq b_1+c_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([b_1+c_1,a_2],[b_1,b_2],[c_1,c_2])\
        P([a_3,a_1], [b_1, b_2], [c_1, c_2])=0 & forall a_3 leq a_1
        end{cases}
        $$


      • If $b_2geq a_2-c_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\
        P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & forall b_3geq b_2
        end{cases}
        $$


      • If $c_2 geq a_2-b_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\
        P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & forall c_3geq c_2
        end{cases}
        $$


      • If $c_1leq a_1-b_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\
        P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & forall c_3leq c_1
        end{cases}
        $$



      All the implications above can be re-written as linear function of the CDF associated with $P$.



      However: are these implications also sufficient? If yes, I don't know how to prove it; If not, I don't know how to find a counterexample.










      share|cite|improve this question















      Consider a trivariate probability distribution $P: mathbb{R}^3rightarrow [0,1]$. I have the following questions:



      (1) Are there necessary conditions on the cumulative distribution function (CDF) associated with $P$ ensuring that
      $$
      exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
      $$



      (2) Are there necessary and sufficient conditions on the CDF associated with $P$ ensuring that
      $$
      exists text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$}
      $$



      (3) The conditions that you propose can be "approximated" as a linear constraint on the CDF?





      I'm providing more details on my question also thanks to/inspired by the answers below. The answers below help, but I'm still not satisfied. Please help if you can.



      If there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $P$ should satisfy: for every $begin{pmatrix}
      a_1\
      b_1\
      c_1
      end{pmatrix}leq begin{pmatrix}
      a_2\
      b_2\
      c_2
      end{pmatrix}$




      • If $a_2geq b_2+c_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\
        P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & forall a_3geq a_2\
        end{cases}
        $$


      • If $b_1leq a_1-c_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\
        P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & forall b_3leq b_1\
        end{cases}
        $$


      • If $a_1 leq b_1+c_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([b_1+c_1,a_2],[b_1,b_2],[c_1,c_2])\
        P([a_3,a_1], [b_1, b_2], [c_1, c_2])=0 & forall a_3 leq a_1
        end{cases}
        $$


      • If $b_2geq a_2-c_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\
        P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & forall b_3geq b_2
        end{cases}
        $$


      • If $c_2 geq a_2-b_1$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\
        P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & forall c_3geq c_2
        end{cases}
        $$


      • If $c_1leq a_1-b_2$
        $$
        begin{cases}
        P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\
        P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & forall c_3leq c_1
        end{cases}
        $$



      All the implications above can be re-written as linear function of the CDF associated with $P$.



      However: are these implications also sufficient? If yes, I don't know how to prove it; If not, I don't know how to find a counterexample.







      probability probability-theory probability-distributions random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 at 16:59

























      asked Nov 30 at 17:14









      STF

      381420




      381420






















          2 Answers
          2






          active

          oldest

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          3





          +50









          Edited to fix a bug, and expand. Still only a rough outline...



          If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.



          For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.



          Anyway, let $F(a,b,c) = Prob(X le a, Y le b, Z le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S={(a,b,c)inmathbb R^3 colon a=b+c}$ must have probability $0$. So you can easily come up one Necessary Condition:



          [NC1] $forall b, c: F(infty,b,c) = F(b+c,b,c)$, because given $Yle b$ and $Z le c$ this restricts $X le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $infty$ to denote positive infinity.]



          Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Yle b, Z le c)$ and is saying $P(B(b,c)) = F(infty,b,c) - F(b+c,b,c)= 0$.



          The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).



          The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.



          First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $bigcup_{b,c} B(b,c)$ also have $P(bigcup_{b,c} B(b,c)) = 0$. The union $bigcup_{b,c} B(b,c)$ is (?) in fact ${(a,b,c)inmathbb R^3 colon a > b+c}$, i.e. the open region above the plane $S$.



          So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(Xle b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write





          • $P(Q(b,c)) = F(b+c, infty, infty) - F(b+c, b, infty) - F(b+c, infty, c) + F(b+c, b, c)$.


          This directly gives us another Necessary Condition:



          [NC2] $forall b, c: P(Q(b,c)) = 0$.



          I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?



          Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)






          share|cite|improve this answer























          • By PDF you mean probability distribution function or probability density function?
            – STF
            Dec 10 at 10:20










          • Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
            – STF
            Dec 10 at 10:30












          • (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
            – STF
            Dec 10 at 10:32












          • Please see my update on the question. I still need more details, sorry.
            – STF
            Dec 11 at 17:00










          • @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
            – antkam
            Dec 11 at 22:53



















          3














          One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $Ssubset mathbb R^3$ given by
          $$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
          which is a two dimensional subspace of $mathbb R^3$ (a plane through the $vec 0$).



          If the probability is not concentrated on this plane, that is if
          $$int_S dP<1,$$
          then $P$ cannot be the probability distribution of such a random vector.






          share|cite|improve this answer























          • Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
            – STF
            Nov 30 at 17:26












          • Is it a discrete random vector?
            – Alejandro Nasif Salum
            Nov 30 at 17:28










          • No, it is a continuous random vector
            – STF
            Nov 30 at 17:28










          • And the study of the distribution is based on the probability density function?
            – Alejandro Nasif Salum
            Nov 30 at 17:30






          • 1




            Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
            – Alejandro Nasif Salum
            Nov 30 at 17:31











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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3





          +50









          Edited to fix a bug, and expand. Still only a rough outline...



          If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.



          For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.



          Anyway, let $F(a,b,c) = Prob(X le a, Y le b, Z le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S={(a,b,c)inmathbb R^3 colon a=b+c}$ must have probability $0$. So you can easily come up one Necessary Condition:



          [NC1] $forall b, c: F(infty,b,c) = F(b+c,b,c)$, because given $Yle b$ and $Z le c$ this restricts $X le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $infty$ to denote positive infinity.]



          Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Yle b, Z le c)$ and is saying $P(B(b,c)) = F(infty,b,c) - F(b+c,b,c)= 0$.



          The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).



          The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.



          First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $bigcup_{b,c} B(b,c)$ also have $P(bigcup_{b,c} B(b,c)) = 0$. The union $bigcup_{b,c} B(b,c)$ is (?) in fact ${(a,b,c)inmathbb R^3 colon a > b+c}$, i.e. the open region above the plane $S$.



          So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(Xle b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write





          • $P(Q(b,c)) = F(b+c, infty, infty) - F(b+c, b, infty) - F(b+c, infty, c) + F(b+c, b, c)$.


          This directly gives us another Necessary Condition:



          [NC2] $forall b, c: P(Q(b,c)) = 0$.



          I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?



          Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)






          share|cite|improve this answer























          • By PDF you mean probability distribution function or probability density function?
            – STF
            Dec 10 at 10:20










          • Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
            – STF
            Dec 10 at 10:30












          • (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
            – STF
            Dec 10 at 10:32












          • Please see my update on the question. I still need more details, sorry.
            – STF
            Dec 11 at 17:00










          • @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
            – antkam
            Dec 11 at 22:53
















          3





          +50









          Edited to fix a bug, and expand. Still only a rough outline...



          If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.



          For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.



          Anyway, let $F(a,b,c) = Prob(X le a, Y le b, Z le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S={(a,b,c)inmathbb R^3 colon a=b+c}$ must have probability $0$. So you can easily come up one Necessary Condition:



          [NC1] $forall b, c: F(infty,b,c) = F(b+c,b,c)$, because given $Yle b$ and $Z le c$ this restricts $X le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $infty$ to denote positive infinity.]



          Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Yle b, Z le c)$ and is saying $P(B(b,c)) = F(infty,b,c) - F(b+c,b,c)= 0$.



          The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).



          The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.



          First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $bigcup_{b,c} B(b,c)$ also have $P(bigcup_{b,c} B(b,c)) = 0$. The union $bigcup_{b,c} B(b,c)$ is (?) in fact ${(a,b,c)inmathbb R^3 colon a > b+c}$, i.e. the open region above the plane $S$.



          So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(Xle b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write





          • $P(Q(b,c)) = F(b+c, infty, infty) - F(b+c, b, infty) - F(b+c, infty, c) + F(b+c, b, c)$.


          This directly gives us another Necessary Condition:



          [NC2] $forall b, c: P(Q(b,c)) = 0$.



          I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?



          Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)






          share|cite|improve this answer























          • By PDF you mean probability distribution function or probability density function?
            – STF
            Dec 10 at 10:20










          • Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
            – STF
            Dec 10 at 10:30












          • (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
            – STF
            Dec 10 at 10:32












          • Please see my update on the question. I still need more details, sorry.
            – STF
            Dec 11 at 17:00










          • @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
            – antkam
            Dec 11 at 22:53














          3





          +50







          3





          +50



          3




          +50




          Edited to fix a bug, and expand. Still only a rough outline...



          If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.



          For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.



          Anyway, let $F(a,b,c) = Prob(X le a, Y le b, Z le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S={(a,b,c)inmathbb R^3 colon a=b+c}$ must have probability $0$. So you can easily come up one Necessary Condition:



          [NC1] $forall b, c: F(infty,b,c) = F(b+c,b,c)$, because given $Yle b$ and $Z le c$ this restricts $X le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $infty$ to denote positive infinity.]



          Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Yle b, Z le c)$ and is saying $P(B(b,c)) = F(infty,b,c) - F(b+c,b,c)= 0$.



          The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).



          The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.



          First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $bigcup_{b,c} B(b,c)$ also have $P(bigcup_{b,c} B(b,c)) = 0$. The union $bigcup_{b,c} B(b,c)$ is (?) in fact ${(a,b,c)inmathbb R^3 colon a > b+c}$, i.e. the open region above the plane $S$.



          So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(Xle b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write





          • $P(Q(b,c)) = F(b+c, infty, infty) - F(b+c, b, infty) - F(b+c, infty, c) + F(b+c, b, c)$.


          This directly gives us another Necessary Condition:



          [NC2] $forall b, c: P(Q(b,c)) = 0$.



          I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?



          Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)






          share|cite|improve this answer














          Edited to fix a bug, and expand. Still only a rough outline...



          If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.



          For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.



          Anyway, let $F(a,b,c) = Prob(X le a, Y le b, Z le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S={(a,b,c)inmathbb R^3 colon a=b+c}$ must have probability $0$. So you can easily come up one Necessary Condition:



          [NC1] $forall b, c: F(infty,b,c) = F(b+c,b,c)$, because given $Yle b$ and $Z le c$ this restricts $X le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $infty$ to denote positive infinity.]



          Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Yle b, Z le c)$ and is saying $P(B(b,c)) = F(infty,b,c) - F(b+c,b,c)= 0$.



          The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).



          The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.



          First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $bigcup_{b,c} B(b,c)$ also have $P(bigcup_{b,c} B(b,c)) = 0$. The union $bigcup_{b,c} B(b,c)$ is (?) in fact ${(a,b,c)inmathbb R^3 colon a > b+c}$, i.e. the open region above the plane $S$.



          So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(Xle b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write





          • $P(Q(b,c)) = F(b+c, infty, infty) - F(b+c, b, infty) - F(b+c, infty, c) + F(b+c, b, c)$.


          This directly gives us another Necessary Condition:



          [NC2] $forall b, c: P(Q(b,c)) = 0$.



          I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?



          Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 at 22:49

























          answered Dec 7 at 23:19









          antkam

          1,518112




          1,518112












          • By PDF you mean probability distribution function or probability density function?
            – STF
            Dec 10 at 10:20










          • Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
            – STF
            Dec 10 at 10:30












          • (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
            – STF
            Dec 10 at 10:32












          • Please see my update on the question. I still need more details, sorry.
            – STF
            Dec 11 at 17:00










          • @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
            – antkam
            Dec 11 at 22:53


















          • By PDF you mean probability distribution function or probability density function?
            – STF
            Dec 10 at 10:20










          • Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
            – STF
            Dec 10 at 10:30












          • (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
            – STF
            Dec 10 at 10:32












          • Please see my update on the question. I still need more details, sorry.
            – STF
            Dec 11 at 17:00










          • @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
            – antkam
            Dec 11 at 22:53
















          By PDF you mean probability distribution function or probability density function?
          – STF
          Dec 10 at 10:20




          By PDF you mean probability distribution function or probability density function?
          – STF
          Dec 10 at 10:20












          Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
          – STF
          Dec 10 at 10:30






          Thanks. (1) How do you get the second bullet point? If $Xleq a, text{ }Zequiv X-Yleq c$, then we have that $Xleq a, text{ }Y geq X-c$ which does not imply $Yleq a-c$.
          – STF
          Dec 10 at 10:30














          (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
          – STF
          Dec 10 at 10:32






          (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."?
          – STF
          Dec 10 at 10:32














          Please see my update on the question. I still need more details, sorry.
          – STF
          Dec 11 at 17:00




          Please see my update on the question. I still need more details, sorry.
          – STF
          Dec 11 at 17:00












          @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
          – antkam
          Dec 11 at 22:53




          @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck!
          – antkam
          Dec 11 at 22:53











          3














          One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $Ssubset mathbb R^3$ given by
          $$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
          which is a two dimensional subspace of $mathbb R^3$ (a plane through the $vec 0$).



          If the probability is not concentrated on this plane, that is if
          $$int_S dP<1,$$
          then $P$ cannot be the probability distribution of such a random vector.






          share|cite|improve this answer























          • Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
            – STF
            Nov 30 at 17:26












          • Is it a discrete random vector?
            – Alejandro Nasif Salum
            Nov 30 at 17:28










          • No, it is a continuous random vector
            – STF
            Nov 30 at 17:28










          • And the study of the distribution is based on the probability density function?
            – Alejandro Nasif Salum
            Nov 30 at 17:30






          • 1




            Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
            – Alejandro Nasif Salum
            Nov 30 at 17:31
















          3














          One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $Ssubset mathbb R^3$ given by
          $$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
          which is a two dimensional subspace of $mathbb R^3$ (a plane through the $vec 0$).



          If the probability is not concentrated on this plane, that is if
          $$int_S dP<1,$$
          then $P$ cannot be the probability distribution of such a random vector.






          share|cite|improve this answer























          • Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
            – STF
            Nov 30 at 17:26












          • Is it a discrete random vector?
            – Alejandro Nasif Salum
            Nov 30 at 17:28










          • No, it is a continuous random vector
            – STF
            Nov 30 at 17:28










          • And the study of the distribution is based on the probability density function?
            – Alejandro Nasif Salum
            Nov 30 at 17:30






          • 1




            Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
            – Alejandro Nasif Salum
            Nov 30 at 17:31














          3












          3








          3






          One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $Ssubset mathbb R^3$ given by
          $$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
          which is a two dimensional subspace of $mathbb R^3$ (a plane through the $vec 0$).



          If the probability is not concentrated on this plane, that is if
          $$int_S dP<1,$$
          then $P$ cannot be the probability distribution of such a random vector.






          share|cite|improve this answer














          One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $Ssubset mathbb R^3$ given by
          $$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
          which is a two dimensional subspace of $mathbb R^3$ (a plane through the $vec 0$).



          If the probability is not concentrated on this plane, that is if
          $$int_S dP<1,$$
          then $P$ cannot be the probability distribution of such a random vector.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 at 3:29

























          answered Nov 30 at 17:23









          Alejandro Nasif Salum

          3,999117




          3,999117












          • Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
            – STF
            Nov 30 at 17:26












          • Is it a discrete random vector?
            – Alejandro Nasif Salum
            Nov 30 at 17:28










          • No, it is a continuous random vector
            – STF
            Nov 30 at 17:28










          • And the study of the distribution is based on the probability density function?
            – Alejandro Nasif Salum
            Nov 30 at 17:30






          • 1




            Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
            – Alejandro Nasif Salum
            Nov 30 at 17:31


















          • Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
            – STF
            Nov 30 at 17:26












          • Is it a discrete random vector?
            – Alejandro Nasif Salum
            Nov 30 at 17:28










          • No, it is a continuous random vector
            – STF
            Nov 30 at 17:28










          • And the study of the distribution is based on the probability density function?
            – Alejandro Nasif Salum
            Nov 30 at 17:30






          • 1




            Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
            – Alejandro Nasif Salum
            Nov 30 at 17:31
















          Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
          – STF
          Nov 30 at 17:26






          Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$
          – STF
          Nov 30 at 17:26














          Is it a discrete random vector?
          – Alejandro Nasif Salum
          Nov 30 at 17:28




          Is it a discrete random vector?
          – Alejandro Nasif Salum
          Nov 30 at 17:28












          No, it is a continuous random vector
          – STF
          Nov 30 at 17:28




          No, it is a continuous random vector
          – STF
          Nov 30 at 17:28












          And the study of the distribution is based on the probability density function?
          – Alejandro Nasif Salum
          Nov 30 at 17:30




          And the study of the distribution is based on the probability density function?
          – Alejandro Nasif Salum
          Nov 30 at 17:30




          1




          1




          Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
          – Alejandro Nasif Salum
          Nov 30 at 17:31




          Ok, that's fine, since if it had a density on $mathbb R^3$ it wouldn't be degenerate. I'll think a little about it.
          – Alejandro Nasif Salum
          Nov 30 at 17:31


















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