Does equal probability of rank imply two random variables symmetric around median?












1














Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X ne F_Y$, and that



$$mathbb{P}[X<Y] = mathbb{P}[X>Y] = 1/2. qquad (*)$$



It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m in mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?






probability probability-theory probability-distributions random-variables examples-counterexamples







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  • Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
    – Lee David Chung Lin
    2 days ago










  • The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
    – jesterII
    2 days ago
















1














Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X ne F_Y$, and that



$$mathbb{P}[X<Y] = mathbb{P}[X>Y] = 1/2. qquad (*)$$



It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m in mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?






probability probability-theory probability-distributions random-variables examples-counterexamples







share|cite|improve this question






















  • Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
    – Lee David Chung Lin
    2 days ago










  • The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
    – jesterII
    2 days ago














1












1








1







Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X ne F_Y$, and that



$$mathbb{P}[X<Y] = mathbb{P}[X>Y] = 1/2. qquad (*)$$



It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m in mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?






probability probability-theory probability-distributions random-variables examples-counterexamples







share|cite|improve this question













Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X ne F_Y$, and that



$$mathbb{P}[X<Y] = mathbb{P}[X>Y] = 1/2. qquad (*)$$



It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m in mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?






probability probability-theory probability-distributions random-variables examples-counterexamples




probability probability-theory probability-distributions random-variables examples-counterexamples






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asked 2 days ago









jesterII

1,20121326




1,20121326












  • Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
    – Lee David Chung Lin
    2 days ago










  • The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
    – jesterII
    2 days ago


















  • Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
    – Lee David Chung Lin
    2 days ago










  • The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
    – jesterII
    2 days ago
















Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
– Lee David Chung Lin
2 days ago




Are you interested only in distributions which density is nonzero over the entire $mathbb{R}$? Or will you take those with zero density outside a bounded interval?
– Lee David Chung Lin
2 days ago












The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
– jesterII
2 days ago




The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric.
– jesterII
2 days ago










1 Answer
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Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.



begin{align}
F_X(t) &= t^2 \
F_Y(t) &= 1 - (1 - t)^a quad text{where} quad a = frac{-3 + sqrt{17}}2 approx 0.561553
end{align}
That is, $X$ has a linear density $f_X(t) = 2t$ and it's easy to check that
$$mathbb{P}[X > Y] = int_{x = 0}^1 f_X(x) cdot F_Y(x) ,mathrm{d}x = frac12 $$
Or, more explicitly
begin{align}
mathbb{P}[X > Y] &= int_{x = 0}^1 int_{y = 0}^x 2x cdot frac{d}{dt}left( 1 - (1-t)^aright)Big|_{t=y} ,mathrm{d}y ,mathrm{d}x \
&= int_{x = 0}^1 2xleft( 1 - (1-x)^aright) ,mathrm{d}x \
&= 1-frac2{a^2 + 3a + 2} \
&= frac12
end{align}
by design of having the power $a$ being the root to $a^2 + 3a + 2 = 4$ that is larger than $-1$.



For the record, the medians are
begin{align}
m_X &= frac1{ sqrt{2} } approx 0.707107 \
m_Y &= 1 - frac1{ 2^{1/a} } approx 0.708973
end{align}






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    3














    Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.



    begin{align}
    F_X(t) &= t^2 \
    F_Y(t) &= 1 - (1 - t)^a quad text{where} quad a = frac{-3 + sqrt{17}}2 approx 0.561553
    end{align}
    That is, $X$ has a linear density $f_X(t) = 2t$ and it's easy to check that
    $$mathbb{P}[X > Y] = int_{x = 0}^1 f_X(x) cdot F_Y(x) ,mathrm{d}x = frac12 $$
    Or, more explicitly
    begin{align}
    mathbb{P}[X > Y] &= int_{x = 0}^1 int_{y = 0}^x 2x cdot frac{d}{dt}left( 1 - (1-t)^aright)Big|_{t=y} ,mathrm{d}y ,mathrm{d}x \
    &= int_{x = 0}^1 2xleft( 1 - (1-x)^aright) ,mathrm{d}x \
    &= 1-frac2{a^2 + 3a + 2} \
    &= frac12
    end{align}
    by design of having the power $a$ being the root to $a^2 + 3a + 2 = 4$ that is larger than $-1$.



    For the record, the medians are
    begin{align}
    m_X &= frac1{ sqrt{2} } approx 0.707107 \
    m_Y &= 1 - frac1{ 2^{1/a} } approx 0.708973
    end{align}






    share|cite|improve this answer




























      3














      Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.



      begin{align}
      F_X(t) &= t^2 \
      F_Y(t) &= 1 - (1 - t)^a quad text{where} quad a = frac{-3 + sqrt{17}}2 approx 0.561553
      end{align}
      That is, $X$ has a linear density $f_X(t) = 2t$ and it's easy to check that
      $$mathbb{P}[X > Y] = int_{x = 0}^1 f_X(x) cdot F_Y(x) ,mathrm{d}x = frac12 $$
      Or, more explicitly
      begin{align}
      mathbb{P}[X > Y] &= int_{x = 0}^1 int_{y = 0}^x 2x cdot frac{d}{dt}left( 1 - (1-t)^aright)Big|_{t=y} ,mathrm{d}y ,mathrm{d}x \
      &= int_{x = 0}^1 2xleft( 1 - (1-x)^aright) ,mathrm{d}x \
      &= 1-frac2{a^2 + 3a + 2} \
      &= frac12
      end{align}
      by design of having the power $a$ being the root to $a^2 + 3a + 2 = 4$ that is larger than $-1$.



      For the record, the medians are
      begin{align}
      m_X &= frac1{ sqrt{2} } approx 0.707107 \
      m_Y &= 1 - frac1{ 2^{1/a} } approx 0.708973
      end{align}






      share|cite|improve this answer


























        3












        3








        3






        Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.



        begin{align}
        F_X(t) &= t^2 \
        F_Y(t) &= 1 - (1 - t)^a quad text{where} quad a = frac{-3 + sqrt{17}}2 approx 0.561553
        end{align}
        That is, $X$ has a linear density $f_X(t) = 2t$ and it's easy to check that
        $$mathbb{P}[X > Y] = int_{x = 0}^1 f_X(x) cdot F_Y(x) ,mathrm{d}x = frac12 $$
        Or, more explicitly
        begin{align}
        mathbb{P}[X > Y] &= int_{x = 0}^1 int_{y = 0}^x 2x cdot frac{d}{dt}left( 1 - (1-t)^aright)Big|_{t=y} ,mathrm{d}y ,mathrm{d}x \
        &= int_{x = 0}^1 2xleft( 1 - (1-x)^aright) ,mathrm{d}x \
        &= 1-frac2{a^2 + 3a + 2} \
        &= frac12
        end{align}
        by design of having the power $a$ being the root to $a^2 + 3a + 2 = 4$ that is larger than $-1$.



        For the record, the medians are
        begin{align}
        m_X &= frac1{ sqrt{2} } approx 0.707107 \
        m_Y &= 1 - frac1{ 2^{1/a} } approx 0.708973
        end{align}






        share|cite|improve this answer














        Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.



        begin{align}
        F_X(t) &= t^2 \
        F_Y(t) &= 1 - (1 - t)^a quad text{where} quad a = frac{-3 + sqrt{17}}2 approx 0.561553
        end{align}
        That is, $X$ has a linear density $f_X(t) = 2t$ and it's easy to check that
        $$mathbb{P}[X > Y] = int_{x = 0}^1 f_X(x) cdot F_Y(x) ,mathrm{d}x = frac12 $$
        Or, more explicitly
        begin{align}
        mathbb{P}[X > Y] &= int_{x = 0}^1 int_{y = 0}^x 2x cdot frac{d}{dt}left( 1 - (1-t)^aright)Big|_{t=y} ,mathrm{d}y ,mathrm{d}x \
        &= int_{x = 0}^1 2xleft( 1 - (1-x)^aright) ,mathrm{d}x \
        &= 1-frac2{a^2 + 3a + 2} \
        &= frac12
        end{align}
        by design of having the power $a$ being the root to $a^2 + 3a + 2 = 4$ that is larger than $-1$.



        For the record, the medians are
        begin{align}
        m_X &= frac1{ sqrt{2} } approx 0.707107 \
        m_Y &= 1 - frac1{ 2^{1/a} } approx 0.708973
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Lee David Chung Lin

        3,80531140




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