$operatorname{supp}(f_{X,Y})={(x,y)inmathbb{R}^2||x|+|y|leq1}$ then $X,Y$ are not independent












0












$begingroup$


Let $Z=(X,Y)$ be a absolutely coninuous random variable such that
$$
\operatorname{supp}(f_Z)={(x,y)inmathbb{R}^2||x|+|y|leq1}
$$
Show that $X,Y$ are not independent.




I don't have a good idea how to continue the proof. I need to find $(t,s)inmathbb{R}^2$ that setasfies
$$
\ int_{-1}^{s}int_{-1}^{t}f_Z(x,y)dx dy=int_{-1}^{1}f_Z(t,y)dycdotint_{-1}^{1}f_Z(x,s)dx
$$






probability integration random-variables absolute-continuity







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $Z=(X,Y)$ be a absolutely coninuous random variable such that
    $$
    \operatorname{supp}(f_Z)={(x,y)inmathbb{R}^2||x|+|y|leq1}
    $$
    Show that $X,Y$ are not independent.




    I don't have a good idea how to continue the proof. I need to find $(t,s)inmathbb{R}^2$ that setasfies
    $$
    \ int_{-1}^{s}int_{-1}^{t}f_Z(x,y)dx dy=int_{-1}^{1}f_Z(t,y)dycdotint_{-1}^{1}f_Z(x,s)dx
    $$






    probability integration random-variables absolute-continuity







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $Z=(X,Y)$ be a absolutely coninuous random variable such that
      $$
      \operatorname{supp}(f_Z)={(x,y)inmathbb{R}^2||x|+|y|leq1}
      $$
      Show that $X,Y$ are not independent.




      I don't have a good idea how to continue the proof. I need to find $(t,s)inmathbb{R}^2$ that setasfies
      $$
      \ int_{-1}^{s}int_{-1}^{t}f_Z(x,y)dx dy=int_{-1}^{1}f_Z(t,y)dycdotint_{-1}^{1}f_Z(x,s)dx
      $$






      probability integration random-variables absolute-continuity







      share|cite|improve this question









      $endgroup$




      Let $Z=(X,Y)$ be a absolutely coninuous random variable such that
      $$
      \operatorname{supp}(f_Z)={(x,y)inmathbb{R}^2||x|+|y|leq1}
      $$
      Show that $X,Y$ are not independent.




      I don't have a good idea how to continue the proof. I need to find $(t,s)inmathbb{R}^2$ that setasfies
      $$
      \ int_{-1}^{s}int_{-1}^{t}f_Z(x,y)dx dy=int_{-1}^{1}f_Z(t,y)dycdotint_{-1}^{1}f_Z(x,s)dx
      $$






      probability integration random-variables absolute-continuity




      probability integration random-variables absolute-continuity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 12 '18 at 19:00









      J. DoeJ. Doe

      16111




      16111






















          1 Answer
          1






          active

          oldest

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          4












          $begingroup$

          Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?




          Hint 2: I mean the following. Note that since the set ${x > frac{1}{2}, y > frac{1}{2}}$ intersects the support of $f_Z$ trivially, we must have that $P[X > frac{1}{2}, Y > frac{1}{2}] = 0,.$ To show that they're not independent, you'd just now need to show that $P[X > frac{1}{2}] > 0$ and $P[Y > frac{1}{2}] > 0$, since we'd then have $$Pleft[X > frac{1}{2}, Y > frac{1}{2}right] neq Pleft[X > frac{1}{2}right] cdot Pleft[Y > frac{1}{2}right],. $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
            $endgroup$
            – J. Doe
            Dec 12 '18 at 19:40








          • 1




            $begingroup$
            @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
            $endgroup$
            – Marcus M
            Dec 12 '18 at 19:53











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          4












          $begingroup$

          Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?




          Hint 2: I mean the following. Note that since the set ${x > frac{1}{2}, y > frac{1}{2}}$ intersects the support of $f_Z$ trivially, we must have that $P[X > frac{1}{2}, Y > frac{1}{2}] = 0,.$ To show that they're not independent, you'd just now need to show that $P[X > frac{1}{2}] > 0$ and $P[Y > frac{1}{2}] > 0$, since we'd then have $$Pleft[X > frac{1}{2}, Y > frac{1}{2}right] neq Pleft[X > frac{1}{2}right] cdot Pleft[Y > frac{1}{2}right],. $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
            $endgroup$
            – J. Doe
            Dec 12 '18 at 19:40








          • 1




            $begingroup$
            @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
            $endgroup$
            – Marcus M
            Dec 12 '18 at 19:53
















          4












          $begingroup$

          Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?




          Hint 2: I mean the following. Note that since the set ${x > frac{1}{2}, y > frac{1}{2}}$ intersects the support of $f_Z$ trivially, we must have that $P[X > frac{1}{2}, Y > frac{1}{2}] = 0,.$ To show that they're not independent, you'd just now need to show that $P[X > frac{1}{2}] > 0$ and $P[Y > frac{1}{2}] > 0$, since we'd then have $$Pleft[X > frac{1}{2}, Y > frac{1}{2}right] neq Pleft[X > frac{1}{2}right] cdot Pleft[Y > frac{1}{2}right],. $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
            $endgroup$
            – J. Doe
            Dec 12 '18 at 19:40








          • 1




            $begingroup$
            @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
            $endgroup$
            – Marcus M
            Dec 12 '18 at 19:53














          4












          4








          4





          $begingroup$

          Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?




          Hint 2: I mean the following. Note that since the set ${x > frac{1}{2}, y > frac{1}{2}}$ intersects the support of $f_Z$ trivially, we must have that $P[X > frac{1}{2}, Y > frac{1}{2}] = 0,.$ To show that they're not independent, you'd just now need to show that $P[X > frac{1}{2}] > 0$ and $P[Y > frac{1}{2}] > 0$, since we'd then have $$Pleft[X > frac{1}{2}, Y > frac{1}{2}right] neq Pleft[X > frac{1}{2}right] cdot Pleft[Y > frac{1}{2}right],. $$






          share|cite|improve this answer











          $endgroup$



          Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?




          Hint 2: I mean the following. Note that since the set ${x > frac{1}{2}, y > frac{1}{2}}$ intersects the support of $f_Z$ trivially, we must have that $P[X > frac{1}{2}, Y > frac{1}{2}] = 0,.$ To show that they're not independent, you'd just now need to show that $P[X > frac{1}{2}] > 0$ and $P[Y > frac{1}{2}] > 0$, since we'd then have $$Pleft[X > frac{1}{2}, Y > frac{1}{2}right] neq Pleft[X > frac{1}{2}right] cdot Pleft[Y > frac{1}{2}right],. $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 19:52

























          answered Dec 12 '18 at 19:09









          Marcus MMarcus M

          8,80311047




          8,80311047












          • $begingroup$
            Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
            $endgroup$
            – J. Doe
            Dec 12 '18 at 19:40








          • 1




            $begingroup$
            @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
            $endgroup$
            – Marcus M
            Dec 12 '18 at 19:53


















          • $begingroup$
            Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
            $endgroup$
            – J. Doe
            Dec 12 '18 at 19:40








          • 1




            $begingroup$
            @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
            $endgroup$
            – Marcus M
            Dec 12 '18 at 19:53
















          $begingroup$
          Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
          $endgroup$
          – J. Doe
          Dec 12 '18 at 19:40






          $begingroup$
          Do you mean $lim_{tto1}int_{-1}^{1}f_Z(t,y)dy=int_{-1}^1f_Z(1,y)dy=mathbb{P}(-1<Yleq1)$ and same for $X$? Not sure how it helps me. @Marcuus
          $endgroup$
          – J. Doe
          Dec 12 '18 at 19:40






          1




          1




          $begingroup$
          @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
          $endgroup$
          – Marcus M
          Dec 12 '18 at 19:53




          $begingroup$
          @J.Doe, no that's not what I mean. I added a second, more involved hint that puts you nearly all the way there.
          $endgroup$
          – Marcus M
          Dec 12 '18 at 19:53


















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