Mixing
Restricting the median of marginals has implications for the joint distribution?
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Consider a bivariate cumulative distribution function (cdf) $F: barmathbbR^2rightarrow [0,1]$ where $barmathbbR^2equiv mathbbRcup-infty, inftytimes mathbbRcup-infty, infty$
Let $F_1, F_2$ denote the marginal cdf from $F$ and suppose they both have median zero, i.e.,
$$
F_1(0)=frac12 text, F_2(0)=frac12
$$
Question:
1) Can we conclude that $F(0,0)=frac12$? Or any other value?
2) Can we say something about $F(0,a)$ or $F(b,0)$ for any $(a,b)in barmathbbR^2$?
probability probability-theory probability-distributions random-variables median
asked Jul 25 at 18:59
TEX
2419
add a comment |Â
up vote
1
down vote
favorite
Consider a bivariate cumulative distribution function (cdf) $F: barmathbbR^2rightarrow [0,1]$ where $barmathbbR^2equiv mathbbRcup-infty, inftytimes mathbbRcup-infty, infty$
Let $F_1, F_2$ denote the marginal cdf from $F$ and suppose they both have median zero, i.e.,
$$
F_1(0)=frac12 text, F_2(0)=frac12
$$
Question:
1) Can we conclude that $F(0,0)=frac12$? Or any other value?
2) Can we say something about $F(0,a)$ or $F(b,0)$ for any $(a,b)in barmathbbR^2$?
probability probability-theory probability-distributions random-variables median
asked Jul 25 at 18:59
TEX
2419
1
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
Thank you, I see.
– TEX
Jul 25 at 20:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider a bivariate cumulative distribution function (cdf) $F: barmathbbR^2rightarrow [0,1]$ where $barmathbbR^2equiv mathbbRcup-infty, inftytimes mathbbRcup-infty, infty$
Let $F_1, F_2$ denote the marginal cdf from $F$ and suppose they both have median zero, i.e.,
$$
F_1(0)=frac12 text, F_2(0)=frac12
$$
Question:
1) Can we conclude that $F(0,0)=frac12$? Or any other value?
2) Can we say something about $F(0,a)$ or $F(b,0)$ for any $(a,b)in barmathbbR^2$?
probability probability-theory probability-distributions random-variables median
asked Jul 25 at 18:59
TEX
2419
Consider a bivariate cumulative distribution function (cdf) $F: barmathbbR^2rightarrow [0,1]$ where $barmathbbR^2equiv mathbbRcup-infty, inftytimes mathbbRcup-infty, infty$
Let $F_1, F_2$ denote the marginal cdf from $F$ and suppose they both have median zero, i.e.,
$$
F_1(0)=frac12 text, F_2(0)=frac12
$$
Question:
1) Can we conclude that $F(0,0)=frac12$? Or any other value?
2) Can we say something about $F(0,a)$ or $F(b,0)$ for any $(a,b)in barmathbbR^2$?
probability probability-theory probability-distributions random-variables median
asked Jul 25 at 18:59
TEX
2419
edited Jul 25 at 20:30
asked Jul 25 at 18:59
TEX
2419
asked Jul 25 at 18:59
TEX
2419
asked Jul 25 at 18:59
TEX
2419
2419
1
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
Thank you, I see.
– TEX
Jul 25 at 20:31
add a comment |Â
1
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
Thank you, I see.
– TEX
Jul 25 at 20:31
1
1
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
Thank you, I see.
– TEX
Jul 25 at 20:31
Thank you, I see.
– TEX
Jul 25 at 20:31
add a comment |Â
1 Answer
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2
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accepted
Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.
With regard to 1)
The answer is No! If $X$ and $Y$ are independent, $P(Xleq 0, Yleq 0) = P(Xleq 0)P(Yleq 0) = frac14$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(Xleq 0, Yleq 0) = P(Xleq 0, Xgeq 0) = 0.$$
With regard to 2):
You certainly can say "something". In the example above you can explicitly calculate $F$. Hence
$ F(0,a) = P(Xleq 0, -Xleq a)$ which is zero if $aleq 0$ and $frac12 - P(Xleq -a)$ if $ageq 0$.
In general it is true that $0leq F(0,0) leq frac12$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 leq F(0,0)leq F_1(0) = frac12$.
Furthermore, every intermediate value in the interval $[0, frac12]$ is possible for $F(0,0).$
answered Jul 25 at 20:43
g g
894415
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.
With regard to 1)
The answer is No! If $X$ and $Y$ are independent, $P(Xleq 0, Yleq 0) = P(Xleq 0)P(Yleq 0) = frac14$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(Xleq 0, Yleq 0) = P(Xleq 0, Xgeq 0) = 0.$$
With regard to 2):
You certainly can say "something". In the example above you can explicitly calculate $F$. Hence
$ F(0,a) = P(Xleq 0, -Xleq a)$ which is zero if $aleq 0$ and $frac12 - P(Xleq -a)$ if $ageq 0$.
In general it is true that $0leq F(0,0) leq frac12$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 leq F(0,0)leq F_1(0) = frac12$.
Furthermore, every intermediate value in the interval $[0, frac12]$ is possible for $F(0,0).$
answered Jul 25 at 20:43
g g
894415
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
add a comment |Â
up vote
2
down vote
accepted
Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.
With regard to 1)
The answer is No! If $X$ and $Y$ are independent, $P(Xleq 0, Yleq 0) = P(Xleq 0)P(Yleq 0) = frac14$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(Xleq 0, Yleq 0) = P(Xleq 0, Xgeq 0) = 0.$$
With regard to 2):
You certainly can say "something". In the example above you can explicitly calculate $F$. Hence
$ F(0,a) = P(Xleq 0, -Xleq a)$ which is zero if $aleq 0$ and $frac12 - P(Xleq -a)$ if $ageq 0$.
In general it is true that $0leq F(0,0) leq frac12$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 leq F(0,0)leq F_1(0) = frac12$.
Furthermore, every intermediate value in the interval $[0, frac12]$ is possible for $F(0,0).$
answered Jul 25 at 20:43
g g
894415
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.
With regard to 1)
The answer is No! If $X$ and $Y$ are independent, $P(Xleq 0, Yleq 0) = P(Xleq 0)P(Yleq 0) = frac14$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(Xleq 0, Yleq 0) = P(Xleq 0, Xgeq 0) = 0.$$
With regard to 2):
You certainly can say "something". In the example above you can explicitly calculate $F$. Hence
$ F(0,a) = P(Xleq 0, -Xleq a)$ which is zero if $aleq 0$ and $frac12 - P(Xleq -a)$ if $ageq 0$.
In general it is true that $0leq F(0,0) leq frac12$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 leq F(0,0)leq F_1(0) = frac12$.
Furthermore, every intermediate value in the interval $[0, frac12]$ is possible for $F(0,0).$
answered Jul 25 at 20:43
g g
894415
Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.
With regard to 1)
The answer is No! If $X$ and $Y$ are independent, $P(Xleq 0, Yleq 0) = P(Xleq 0)P(Yleq 0) = frac14$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(Xleq 0, Yleq 0) = P(Xleq 0, Xgeq 0) = 0.$$
With regard to 2):
You certainly can say "something". In the example above you can explicitly calculate $F$. Hence
$ F(0,a) = P(Xleq 0, -Xleq a)$ which is zero if $aleq 0$ and $frac12 - P(Xleq -a)$ if $ageq 0$.
In general it is true that $0leq F(0,0) leq frac12$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 leq F(0,0)leq F_1(0) = frac12$.
Furthermore, every intermediate value in the interval $[0, frac12]$ is possible for $F(0,0).$
answered Jul 25 at 20:43
g g
894415
edited Jul 25 at 21:02
answered Jul 25 at 20:43
g g
894415
answered Jul 25 at 20:43
g g
894415
answered Jul 25 at 20:43
g g
894415
894415
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
add a comment |Â
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
2
2
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
Also, $F(0,infty)=F(infty,0)=frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=frac12$.
– joriki
Jul 25 at 20:44
add a comment |Â
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1
If anything, we might be able to conclude $F(0,0)=frac14$, not $frac12$?
– joriki
Jul 25 at 20:20
Thank you, I see.
– TEX
Jul 25 at 20:31