Mixing
Does equal probability of rank imply two random variables symmetric around median?
$begingroup$
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
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
$endgroup$
$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12
add a comment |
$begingroup$
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
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
$endgroup$
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
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
asked Jan 3 at 16:26
jesterIIjesterII
1,20121326
1,20121326
$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12
add a comment |
$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12
$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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}
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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}
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
$endgroup$
add a comment |
$begingroup$
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}
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
$endgroup$
add a comment |
$begingroup$
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}
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
$endgroup$
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}
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
edited Jan 3 at 18:56
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
answered Jan 3 at 18:29
Lee David Chung LinLee David Chung Lin
3,86531140
3,86531140
add a comment |
add a comment |
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$begingroup$
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?
$endgroup$
– Lee David Chung Lin
Jan 3 at 17:36
$begingroup$
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.
$endgroup$
– jesterII
Jan 3 at 18:12