An independent collection, where the pairwise products are also independent












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Can you give me a non-trivial example of a sequence ${X_k}_{k=1}^n$ of independent, and identically distributed random variables, such that, the pairwise products, ${X_kX_{ell}}_{1leq k<ell leq n}$ are all independent?



Of course, if ${k,ell}cap {k',ell'}=varnothing$, we automatically have this, and since $(k,ell)neq {k',ell'}$, it boils down to giving an example where $X_kX_ell$ and $X_kX_{ell'}$ are always independent, whenever $ellneq ell'$ and $ell,ell'neq k$.






probability probability-theory probability-distributions independence conditional-probability







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  • 1




    $begingroup$
    When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
    $endgroup$
    – kimchi lover
    Dec 5 '18 at 20:44










  • $begingroup$
    kimchi lover, I means pairwise. However, even the other direction seems interesting.
    $endgroup$
    – Aaron
    Dec 5 '18 at 22:35
















0












$begingroup$


Can you give me a non-trivial example of a sequence ${X_k}_{k=1}^n$ of independent, and identically distributed random variables, such that, the pairwise products, ${X_kX_{ell}}_{1leq k<ell leq n}$ are all independent?



Of course, if ${k,ell}cap {k',ell'}=varnothing$, we automatically have this, and since $(k,ell)neq {k',ell'}$, it boils down to giving an example where $X_kX_ell$ and $X_kX_{ell'}$ are always independent, whenever $ellneq ell'$ and $ell,ell'neq k$.






probability probability-theory probability-distributions independence conditional-probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
    $endgroup$
    – kimchi lover
    Dec 5 '18 at 20:44










  • $begingroup$
    kimchi lover, I means pairwise. However, even the other direction seems interesting.
    $endgroup$
    – Aaron
    Dec 5 '18 at 22:35














0












0








0





$begingroup$


Can you give me a non-trivial example of a sequence ${X_k}_{k=1}^n$ of independent, and identically distributed random variables, such that, the pairwise products, ${X_kX_{ell}}_{1leq k<ell leq n}$ are all independent?



Of course, if ${k,ell}cap {k',ell'}=varnothing$, we automatically have this, and since $(k,ell)neq {k',ell'}$, it boils down to giving an example where $X_kX_ell$ and $X_kX_{ell'}$ are always independent, whenever $ellneq ell'$ and $ell,ell'neq k$.






probability probability-theory probability-distributions independence conditional-probability







share|cite|improve this question









$endgroup$




Can you give me a non-trivial example of a sequence ${X_k}_{k=1}^n$ of independent, and identically distributed random variables, such that, the pairwise products, ${X_kX_{ell}}_{1leq k<ell leq n}$ are all independent?



Of course, if ${k,ell}cap {k',ell'}=varnothing$, we automatically have this, and since $(k,ell)neq {k',ell'}$, it boils down to giving an example where $X_kX_ell$ and $X_kX_{ell'}$ are always independent, whenever $ellneq ell'$ and $ell,ell'neq k$.






probability probability-theory probability-distributions independence conditional-probability




probability probability-theory probability-distributions independence conditional-probability






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asked Dec 5 '18 at 20:31









AaronAaron

1,828415




1,828415








  • 1




    $begingroup$
    When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
    $endgroup$
    – kimchi lover
    Dec 5 '18 at 20:44










  • $begingroup$
    kimchi lover, I means pairwise. However, even the other direction seems interesting.
    $endgroup$
    – Aaron
    Dec 5 '18 at 22:35














  • 1




    $begingroup$
    When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
    $endgroup$
    – kimchi lover
    Dec 5 '18 at 20:44










  • $begingroup$
    kimchi lover, I means pairwise. However, even the other direction seems interesting.
    $endgroup$
    – Aaron
    Dec 5 '18 at 22:35








1




1




$begingroup$
When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
$endgroup$
– kimchi lover
Dec 5 '18 at 20:44




$begingroup$
When you ask for the pairwise products to be independent, do you mean to ask for the pairwise products be jointly independent or only pair-wise independent? Your "boils down" statement makes it sound like the latter.
$endgroup$
– kimchi lover
Dec 5 '18 at 20:44












$begingroup$
kimchi lover, I means pairwise. However, even the other direction seems interesting.
$endgroup$
– Aaron
Dec 5 '18 at 22:35




$begingroup$
kimchi lover, I means pairwise. However, even the other direction seems interesting.
$endgroup$
– Aaron
Dec 5 '18 at 22:35










2 Answers
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I don't know what you mean by non-trivial but consider this example:
$P(X = 1) = P(X = -1) = frac 1 2$



Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = frac 1 2$ because 1*1 and -1*-1 happen with the same probability.



Then we can write
$$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = frac 1 2 $$



$$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = frac 1 2 $$



$$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = frac 1 2 $$



Since the conditional probability is the same as unconditional, you can say they are independent.



Hope it helps






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    @Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.



    EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1leq i <jleq n}$ forms an independent collection of random variables?



    Take $X_isim {rm Unif}{1,2,dots,p-1}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_jpmod{p}$. We have,
    $$
    mathbb{P}(X_iX_j = k,X_iX_kell)=sum_{t=1}^{p-1}mathbb{P}(X_iX_j=k,X_iX_k=ell|X_i=t)mathbb{P}(X_i=t)=sum_{t=1}^{p-1}mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}ell)mathbb{P}(X_i=t),
    $$
    which evaluates as, $frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $kin{1,2,dots,p-1}$, $mathbb{P}(X_iX_j=k)=frac{1}{p-1}$, proving the independence.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
      $endgroup$
      – kimchi lover
      Dec 5 '18 at 22:46












    • $begingroup$
      How about pairwise independence of pairwise products?
      $endgroup$
      – Aaron
      Dec 5 '18 at 23:53










    • $begingroup$
      In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
      $endgroup$
      – kimchi lover
      Dec 6 '18 at 0:17











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    2 Answers
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    2 Answers
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    $begingroup$

    I don't know what you mean by non-trivial but consider this example:
    $P(X = 1) = P(X = -1) = frac 1 2$



    Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = frac 1 2$ because 1*1 and -1*-1 happen with the same probability.



    Then we can write
    $$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = frac 1 2 $$



    $$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = frac 1 2 $$



    $$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = frac 1 2 $$



    Since the conditional probability is the same as unconditional, you can say they are independent.



    Hope it helps






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I don't know what you mean by non-trivial but consider this example:
      $P(X = 1) = P(X = -1) = frac 1 2$



      Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = frac 1 2$ because 1*1 and -1*-1 happen with the same probability.



      Then we can write
      $$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = frac 1 2 $$



      $$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = frac 1 2 $$



      $$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = frac 1 2 $$



      Since the conditional probability is the same as unconditional, you can say they are independent.



      Hope it helps






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I don't know what you mean by non-trivial but consider this example:
        $P(X = 1) = P(X = -1) = frac 1 2$



        Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = frac 1 2$ because 1*1 and -1*-1 happen with the same probability.



        Then we can write
        $$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = frac 1 2 $$



        $$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = frac 1 2 $$



        $$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = frac 1 2 $$



        Since the conditional probability is the same as unconditional, you can say they are independent.



        Hope it helps






        share|cite|improve this answer









        $endgroup$



        I don't know what you mean by non-trivial but consider this example:
        $P(X = 1) = P(X = -1) = frac 1 2$



        Knowing that the product is -1 or 1 doesn't change the probability for other products. $P(X_i = 1|X_i.X_j = 1) = frac 1 2$ and $P(X_i = -1|X_i.X_j = 1) = frac 1 2$ because 1*1 and -1*-1 happen with the same probability.



        Then we can write
        $$P(X_i.X_k = 1|X_i.X_j = 1) = P(X_k = 1|X_i.X_j = 1, X_i=1 ).P(X_i = 1|X_i.X_j = 1) + P(X_k = -1|X_i.X_j = 1, X_i=-1 ).P(X_i = -1|X_i.X_j = 1) = frac 1 2 $$



        $$P(X_i.X_k = 1|X_i.X_j = -1) = P(X_k = 1|X_i.X_j = -1, X_i=1 ).P(X_i = 1|X_i.X_j = -1) + P(X_k = -1|X_i.X_j = -1, X_i=-1 ).P(X_i = -1|X_i.X_j = -1) = frac 1 2 $$



        $$P(X_i.X_k = 1) = P(X_k = 1| X_i=1 ).P(X_i = 1) + P(X_k = -1|X_i=-1 ).P(X_i = -1) = frac 1 2 $$



        Since the conditional probability is the same as unconditional, you can say they are independent.



        Hope it helps







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 20:46









        OfyaOfya

        5048




        5048























            0












            $begingroup$

            @Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.



            EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1leq i <jleq n}$ forms an independent collection of random variables?



            Take $X_isim {rm Unif}{1,2,dots,p-1}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_jpmod{p}$. We have,
            $$
            mathbb{P}(X_iX_j = k,X_iX_kell)=sum_{t=1}^{p-1}mathbb{P}(X_iX_j=k,X_iX_k=ell|X_i=t)mathbb{P}(X_i=t)=sum_{t=1}^{p-1}mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}ell)mathbb{P}(X_i=t),
            $$
            which evaluates as, $frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $kin{1,2,dots,p-1}$, $mathbb{P}(X_iX_j=k)=frac{1}{p-1}$, proving the independence.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
              $endgroup$
              – kimchi lover
              Dec 5 '18 at 22:46












            • $begingroup$
              How about pairwise independence of pairwise products?
              $endgroup$
              – Aaron
              Dec 5 '18 at 23:53










            • $begingroup$
              In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
              $endgroup$
              – kimchi lover
              Dec 6 '18 at 0:17
















            0












            $begingroup$

            @Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.



            EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1leq i <jleq n}$ forms an independent collection of random variables?



            Take $X_isim {rm Unif}{1,2,dots,p-1}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_jpmod{p}$. We have,
            $$
            mathbb{P}(X_iX_j = k,X_iX_kell)=sum_{t=1}^{p-1}mathbb{P}(X_iX_j=k,X_iX_k=ell|X_i=t)mathbb{P}(X_i=t)=sum_{t=1}^{p-1}mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}ell)mathbb{P}(X_i=t),
            $$
            which evaluates as, $frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $kin{1,2,dots,p-1}$, $mathbb{P}(X_iX_j=k)=frac{1}{p-1}$, proving the independence.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
              $endgroup$
              – kimchi lover
              Dec 5 '18 at 22:46












            • $begingroup$
              How about pairwise independence of pairwise products?
              $endgroup$
              – Aaron
              Dec 5 '18 at 23:53










            • $begingroup$
              In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
              $endgroup$
              – kimchi lover
              Dec 6 '18 at 0:17














            0












            0








            0





            $begingroup$

            @Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.



            EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1leq i <jleq n}$ forms an independent collection of random variables?



            Take $X_isim {rm Unif}{1,2,dots,p-1}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_jpmod{p}$. We have,
            $$
            mathbb{P}(X_iX_j = k,X_iX_kell)=sum_{t=1}^{p-1}mathbb{P}(X_iX_j=k,X_iX_k=ell|X_i=t)mathbb{P}(X_i=t)=sum_{t=1}^{p-1}mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}ell)mathbb{P}(X_i=t),
            $$
            which evaluates as, $frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $kin{1,2,dots,p-1}$, $mathbb{P}(X_iX_j=k)=frac{1}{p-1}$, proving the independence.






            share|cite|improve this answer











            $endgroup$



            @Ofya, thanks for your answer, I was seeking for something where I can control the support, and I guess, I've found my answer.



            EDIT (Extra question) Suppose that, $X_i$ are independent, standard normal random variables. Is it true that, $(X_iX_j)_{1leq i <jleq n}$ forms an independent collection of random variables?



            Take $X_isim {rm Unif}{1,2,dots,p-1}$ with $p$ being a prime, and consider $X_iX_j$ as $X_iX_jpmod{p}$. We have,
            $$
            mathbb{P}(X_iX_j = k,X_iX_kell)=sum_{t=1}^{p-1}mathbb{P}(X_iX_j=k,X_iX_k=ell|X_i=t)mathbb{P}(X_i=t)=sum_{t=1}^{p-1}mathbb{P}(X_j=t^{-1}k,X_k=t^{-1}ell)mathbb{P}(X_i=t),
            $$
            which evaluates as, $frac{1}{(p-1)^2}$. Finally, it is easy to see that, for any pair $(i ,j)$ and any $kin{1,2,dots,p-1}$, $mathbb{P}(X_iX_j=k)=frac{1}{p-1}$, proving the independence.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 22:40

























            answered Dec 5 '18 at 21:55









            AaronAaron

            1,828415




            1,828415












            • $begingroup$
              As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
              $endgroup$
              – kimchi lover
              Dec 5 '18 at 22:46












            • $begingroup$
              How about pairwise independence of pairwise products?
              $endgroup$
              – Aaron
              Dec 5 '18 at 23:53










            • $begingroup$
              In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
              $endgroup$
              – kimchi lover
              Dec 6 '18 at 0:17


















            • $begingroup$
              As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
              $endgroup$
              – kimchi lover
              Dec 5 '18 at 22:46












            • $begingroup$
              How about pairwise independence of pairwise products?
              $endgroup$
              – Aaron
              Dec 5 '18 at 23:53










            • $begingroup$
              In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
              $endgroup$
              – kimchi lover
              Dec 6 '18 at 0:17
















            $begingroup$
            As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
            $endgroup$
            – kimchi lover
            Dec 5 '18 at 22:46






            $begingroup$
            As for your extra credit problem: take the product of $X_1X_2, X_2X_3,$ and $X_3X_1.$ The product is positive with probability 1, so the 3 pairwise products cannot be jointly independent.
            $endgroup$
            – kimchi lover
            Dec 5 '18 at 22:46














            $begingroup$
            How about pairwise independence of pairwise products?
            $endgroup$
            – Aaron
            Dec 5 '18 at 23:53




            $begingroup$
            How about pairwise independence of pairwise products?
            $endgroup$
            – Aaron
            Dec 5 '18 at 23:53












            $begingroup$
            In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
            $endgroup$
            – kimchi lover
            Dec 6 '18 at 0:17




            $begingroup$
            In general, Ofya's answer puts paid to that one. In the gaussian case, I think one can show that $XY$ and $YZ$ are not independent, if $X,Y,Z$ are iid $N(0,1)$ rv.s.
            $endgroup$
            – kimchi lover
            Dec 6 '18 at 0:17


















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