On the properties of a finite mixture












1












$begingroup$


Let $Y,X,W$ be real-valued random variables, respectively with supports denoted by $mathcal{Y},mathcal{X},mathcal{W}$.



(A1) Assume that $mathcal{X},mathcal{W}$ are finite. Without loss of generality, assume that $mathcal{X}equiv {x_1,x_2}$ and $mathcal{W}equiv {w_1,w_2}$.



(A2) For each realisation $xin mathcal{X}$ of $X$, let $epsilon_x$ be another random variable. Assume that, $forall x in mathcal{X}$, $epsilon_x$ is stochastically independent of $X,W$.



(A3) Assume that the following relation holds
$$
Y=h(X,W)+epsilon_{X}
$$





Consider now the cdf $F(cdot)$ of $Y$ evaluated at $yin mathcal{Y}$. Following here, we can write



$$F(y)=p(x_1,w_1)times F(y| X=x_1, W=w_1)
$$

$$
+p(x_2,w_1)times F(y| X=x_2, W=w_1)
$$

$$
+p(x_1,w_2)times F(y| X=x_1, W=w_2)$$

$$
+p(x_2,w_2)times F(y| X=x_2, W=w_2) $$



where $p(x,w)$ is the probability mass function of $(X,W)$ evaluated at $(x,w)$ and $F(cdot| X=x, W=w)$ is the cdf of $Y$ conditional on $X=x,W=w$.



The lines above highlight that $F(cdot)$ can be expressed as a finite mixture.





Let's focus on the relation between $F(cdot| X=x, W=w)$, (A3), and the cdf of $epsilon_x$.



For any $(x,w)$, $F(cdot| X=x, W=w)$ is determined by (A3) and the cdf of $epsilon_x$.



For example, if $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.



In my exercise, I want to remain non-parametric about the distribution of $epsilon_x$ and I'm looking for non-parametric features of $F(cdot |X=x,W=w)$ that are compatible with (A3). Specifically, this is my question:





Question: is (A3) compatible with writing $F(cdot)$ at any $yin mathcal{Y}$ as
$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
$$

where $G: mathbb{R}rightarrow [0,1]$ is a cdf symmetric around zero [i.e., $G(y)=1-G(-y)$] and ${mu_{x,w}}_{x,w}$ are real numbers all different between each other?



In other words, I'm wondering whether the differences across
$$
F(y| X=x_1, W=w_1), F(y| X=x_1, W=w_2),F(y| X=x_2, W=w_1) ,F(y| X=x_2, W=w_2)
$$

could be captured by a location shift $mu_{x,w}$ differing across $(x,w)$.





Further thoughts: notice that, as explained here, the cdf's ${H_{x,w}}_{x,w}$ with
$$
H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
$$

are characterised by equivalent central moments.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $Y,X,W$ be real-valued random variables, respectively with supports denoted by $mathcal{Y},mathcal{X},mathcal{W}$.



    (A1) Assume that $mathcal{X},mathcal{W}$ are finite. Without loss of generality, assume that $mathcal{X}equiv {x_1,x_2}$ and $mathcal{W}equiv {w_1,w_2}$.



    (A2) For each realisation $xin mathcal{X}$ of $X$, let $epsilon_x$ be another random variable. Assume that, $forall x in mathcal{X}$, $epsilon_x$ is stochastically independent of $X,W$.



    (A3) Assume that the following relation holds
    $$
    Y=h(X,W)+epsilon_{X}
    $$





    Consider now the cdf $F(cdot)$ of $Y$ evaluated at $yin mathcal{Y}$. Following here, we can write



    $$F(y)=p(x_1,w_1)times F(y| X=x_1, W=w_1)
    $$

    $$
    +p(x_2,w_1)times F(y| X=x_2, W=w_1)
    $$

    $$
    +p(x_1,w_2)times F(y| X=x_1, W=w_2)$$

    $$
    +p(x_2,w_2)times F(y| X=x_2, W=w_2) $$



    where $p(x,w)$ is the probability mass function of $(X,W)$ evaluated at $(x,w)$ and $F(cdot| X=x, W=w)$ is the cdf of $Y$ conditional on $X=x,W=w$.



    The lines above highlight that $F(cdot)$ can be expressed as a finite mixture.





    Let's focus on the relation between $F(cdot| X=x, W=w)$, (A3), and the cdf of $epsilon_x$.



    For any $(x,w)$, $F(cdot| X=x, W=w)$ is determined by (A3) and the cdf of $epsilon_x$.



    For example, if $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.



    In my exercise, I want to remain non-parametric about the distribution of $epsilon_x$ and I'm looking for non-parametric features of $F(cdot |X=x,W=w)$ that are compatible with (A3). Specifically, this is my question:





    Question: is (A3) compatible with writing $F(cdot)$ at any $yin mathcal{Y}$ as
    $$
    F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
    $$

    where $G: mathbb{R}rightarrow [0,1]$ is a cdf symmetric around zero [i.e., $G(y)=1-G(-y)$] and ${mu_{x,w}}_{x,w}$ are real numbers all different between each other?



    In other words, I'm wondering whether the differences across
    $$
    F(y| X=x_1, W=w_1), F(y| X=x_1, W=w_2),F(y| X=x_2, W=w_1) ,F(y| X=x_2, W=w_2)
    $$

    could be captured by a location shift $mu_{x,w}$ differing across $(x,w)$.





    Further thoughts: notice that, as explained here, the cdf's ${H_{x,w}}_{x,w}$ with
    $$
    H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
    $$

    are characterised by equivalent central moments.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $Y,X,W$ be real-valued random variables, respectively with supports denoted by $mathcal{Y},mathcal{X},mathcal{W}$.



      (A1) Assume that $mathcal{X},mathcal{W}$ are finite. Without loss of generality, assume that $mathcal{X}equiv {x_1,x_2}$ and $mathcal{W}equiv {w_1,w_2}$.



      (A2) For each realisation $xin mathcal{X}$ of $X$, let $epsilon_x$ be another random variable. Assume that, $forall x in mathcal{X}$, $epsilon_x$ is stochastically independent of $X,W$.



      (A3) Assume that the following relation holds
      $$
      Y=h(X,W)+epsilon_{X}
      $$





      Consider now the cdf $F(cdot)$ of $Y$ evaluated at $yin mathcal{Y}$. Following here, we can write



      $$F(y)=p(x_1,w_1)times F(y| X=x_1, W=w_1)
      $$

      $$
      +p(x_2,w_1)times F(y| X=x_2, W=w_1)
      $$

      $$
      +p(x_1,w_2)times F(y| X=x_1, W=w_2)$$

      $$
      +p(x_2,w_2)times F(y| X=x_2, W=w_2) $$



      where $p(x,w)$ is the probability mass function of $(X,W)$ evaluated at $(x,w)$ and $F(cdot| X=x, W=w)$ is the cdf of $Y$ conditional on $X=x,W=w$.



      The lines above highlight that $F(cdot)$ can be expressed as a finite mixture.





      Let's focus on the relation between $F(cdot| X=x, W=w)$, (A3), and the cdf of $epsilon_x$.



      For any $(x,w)$, $F(cdot| X=x, W=w)$ is determined by (A3) and the cdf of $epsilon_x$.



      For example, if $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.



      In my exercise, I want to remain non-parametric about the distribution of $epsilon_x$ and I'm looking for non-parametric features of $F(cdot |X=x,W=w)$ that are compatible with (A3). Specifically, this is my question:





      Question: is (A3) compatible with writing $F(cdot)$ at any $yin mathcal{Y}$ as
      $$
      F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
      $$

      where $G: mathbb{R}rightarrow [0,1]$ is a cdf symmetric around zero [i.e., $G(y)=1-G(-y)$] and ${mu_{x,w}}_{x,w}$ are real numbers all different between each other?



      In other words, I'm wondering whether the differences across
      $$
      F(y| X=x_1, W=w_1), F(y| X=x_1, W=w_2),F(y| X=x_2, W=w_1) ,F(y| X=x_2, W=w_2)
      $$

      could be captured by a location shift $mu_{x,w}$ differing across $(x,w)$.





      Further thoughts: notice that, as explained here, the cdf's ${H_{x,w}}_{x,w}$ with
      $$
      H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
      $$

      are characterised by equivalent central moments.










      share|cite|improve this question











      $endgroup$




      Let $Y,X,W$ be real-valued random variables, respectively with supports denoted by $mathcal{Y},mathcal{X},mathcal{W}$.



      (A1) Assume that $mathcal{X},mathcal{W}$ are finite. Without loss of generality, assume that $mathcal{X}equiv {x_1,x_2}$ and $mathcal{W}equiv {w_1,w_2}$.



      (A2) For each realisation $xin mathcal{X}$ of $X$, let $epsilon_x$ be another random variable. Assume that, $forall x in mathcal{X}$, $epsilon_x$ is stochastically independent of $X,W$.



      (A3) Assume that the following relation holds
      $$
      Y=h(X,W)+epsilon_{X}
      $$





      Consider now the cdf $F(cdot)$ of $Y$ evaluated at $yin mathcal{Y}$. Following here, we can write



      $$F(y)=p(x_1,w_1)times F(y| X=x_1, W=w_1)
      $$

      $$
      +p(x_2,w_1)times F(y| X=x_2, W=w_1)
      $$

      $$
      +p(x_1,w_2)times F(y| X=x_1, W=w_2)$$

      $$
      +p(x_2,w_2)times F(y| X=x_2, W=w_2) $$



      where $p(x,w)$ is the probability mass function of $(X,W)$ evaluated at $(x,w)$ and $F(cdot| X=x, W=w)$ is the cdf of $Y$ conditional on $X=x,W=w$.



      The lines above highlight that $F(cdot)$ can be expressed as a finite mixture.





      Let's focus on the relation between $F(cdot| X=x, W=w)$, (A3), and the cdf of $epsilon_x$.



      For any $(x,w)$, $F(cdot| X=x, W=w)$ is determined by (A3) and the cdf of $epsilon_x$.



      For example, if $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.



      In my exercise, I want to remain non-parametric about the distribution of $epsilon_x$ and I'm looking for non-parametric features of $F(cdot |X=x,W=w)$ that are compatible with (A3). Specifically, this is my question:





      Question: is (A3) compatible with writing $F(cdot)$ at any $yin mathcal{Y}$ as
      $$
      F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
      $$

      where $G: mathbb{R}rightarrow [0,1]$ is a cdf symmetric around zero [i.e., $G(y)=1-G(-y)$] and ${mu_{x,w}}_{x,w}$ are real numbers all different between each other?



      In other words, I'm wondering whether the differences across
      $$
      F(y| X=x_1, W=w_1), F(y| X=x_1, W=w_2),F(y| X=x_2, W=w_1) ,F(y| X=x_2, W=w_2)
      $$

      could be captured by a location shift $mu_{x,w}$ differing across $(x,w)$.





      Further thoughts: notice that, as explained here, the cdf's ${H_{x,w}}_{x,w}$ with
      $$
      H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
      $$

      are characterised by equivalent central moments.







      probability probability-theory probability-distributions random-variables conditional-probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 '18 at 16:28







      STF

















      asked Dec 28 '18 at 16:12









      STFSTF

      541422




      541422






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          $$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
          where $G_{i}$ denotes the CDF of $epsilon_{x_{i}}$.



          The third equality is based on independence.



          Only if the distribution
          of $epsilon_{x_{i}}$ does not depend on $i$ then you can write
          $$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
          F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
          $$

          where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
            $endgroup$
            – STF
            Dec 28 '18 at 16:50












          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – drhab
            Dec 28 '18 at 17:10










          • $begingroup$
            Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
            $endgroup$
            – STF
            Dec 28 '18 at 17:16













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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          $$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
          where $G_{i}$ denotes the CDF of $epsilon_{x_{i}}$.



          The third equality is based on independence.



          Only if the distribution
          of $epsilon_{x_{i}}$ does not depend on $i$ then you can write
          $$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
          F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
          $$

          where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
            $endgroup$
            – STF
            Dec 28 '18 at 16:50












          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – drhab
            Dec 28 '18 at 17:10










          • $begingroup$
            Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
            $endgroup$
            – STF
            Dec 28 '18 at 17:16


















          1












          $begingroup$

          $$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
          where $G_{i}$ denotes the CDF of $epsilon_{x_{i}}$.



          The third equality is based on independence.



          Only if the distribution
          of $epsilon_{x_{i}}$ does not depend on $i$ then you can write
          $$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
          F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
          $$

          where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
            $endgroup$
            – STF
            Dec 28 '18 at 16:50












          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – drhab
            Dec 28 '18 at 17:10










          • $begingroup$
            Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
            $endgroup$
            – STF
            Dec 28 '18 at 17:16
















          1












          1








          1





          $begingroup$

          $$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
          where $G_{i}$ denotes the CDF of $epsilon_{x_{i}}$.



          The third equality is based on independence.



          Only if the distribution
          of $epsilon_{x_{i}}$ does not depend on $i$ then you can write
          $$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
          F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
          $$

          where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.






          share|cite|improve this answer









          $endgroup$



          $$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
          where $G_{i}$ denotes the CDF of $epsilon_{x_{i}}$.



          The third equality is based on independence.



          Only if the distribution
          of $epsilon_{x_{i}}$ does not depend on $i$ then you can write
          $$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
          F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
          $$

          where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 16:38









          drhabdrhab

          103k545136




          103k545136












          • $begingroup$
            Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
            $endgroup$
            – STF
            Dec 28 '18 at 16:50












          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – drhab
            Dec 28 '18 at 17:10










          • $begingroup$
            Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
            $endgroup$
            – STF
            Dec 28 '18 at 17:16




















          • $begingroup$
            Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
            $endgroup$
            – STF
            Dec 28 '18 at 16:50












          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – drhab
            Dec 28 '18 at 17:10










          • $begingroup$
            Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
            $endgroup$
            – STF
            Dec 28 '18 at 17:16


















          $begingroup$
          Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
          $endgroup$
          – STF
          Dec 28 '18 at 16:50






          $begingroup$
          Thanks. Just to clarify "Only if the distribution of $epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $epsilon_{x_1}, epsilon_{x_2}$ are identically distributed"?
          $endgroup$
          – STF
          Dec 28 '18 at 16:50














          $begingroup$
          Yes. That is correct.
          $endgroup$
          – drhab
          Dec 28 '18 at 17:10




          $begingroup$
          Yes. That is correct.
          $endgroup$
          – drhab
          Dec 28 '18 at 17:10












          $begingroup$
          Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
          $endgroup$
          – STF
          Dec 28 '18 at 17:16






          $begingroup$
          Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$?
          $endgroup$
          – STF
          Dec 28 '18 at 17:16




















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