If two random variables are independent, why isn't their min and max?
$begingroup$
Suppose $X_1, X_2$ are independent $U(0, 1)$ random variables, and
$$Y = min(X_1, X_2) $$
$$Z = max(X_1, X_2) $$
By this question, they $Y$ and $Z$ should be independent:
Are functions of independent variables also independent?
But by this answer the covariance is not zero:
What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?
How do I reconcile these two things? The $min$ and $max$ are a function of independent random variables, yet they have covariance.
probability probability-theory statistics
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
$endgroup$
add a comment |
$begingroup$
Suppose $X_1, X_2$ are independent $U(0, 1)$ random variables, and
$$Y = min(X_1, X_2) $$
$$Z = max(X_1, X_2) $$
By this question, they $Y$ and $Z$ should be independent:
Are functions of independent variables also independent?
But by this answer the covariance is not zero:
What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?
How do I reconcile these two things? The $min$ and $max$ are a function of independent random variables, yet they have covariance.
probability probability-theory statistics
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
$endgroup$
7
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
3
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
1
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36
add a comment |
$begingroup$
Suppose $X_1, X_2$ are independent $U(0, 1)$ random variables, and
$$Y = min(X_1, X_2) $$
$$Z = max(X_1, X_2) $$
By this question, they $Y$ and $Z$ should be independent:
Are functions of independent variables also independent?
But by this answer the covariance is not zero:
What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?
How do I reconcile these two things? The $min$ and $max$ are a function of independent random variables, yet they have covariance.
probability probability-theory statistics
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
$endgroup$
Suppose $X_1, X_2$ are independent $U(0, 1)$ random variables, and
$$Y = min(X_1, X_2) $$
$$Z = max(X_1, X_2) $$
By this question, they $Y$ and $Z$ should be independent:
Are functions of independent variables also independent?
But by this answer the covariance is not zero:
What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?
How do I reconcile these two things? The $min$ and $max$ are a function of independent random variables, yet they have covariance.
probability probability-theory statistics
probability probability-theory statistics
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
asked Dec 17 '18 at 20:00
badmaxbadmax
551516
551516
7
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
3
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
1
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36
add a comment |
7
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
3
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
1
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36
7
7
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
3
3
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
1
1
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36
add a comment |
2 Answers
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$begingroup$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
$begingroup$
Think of a random vector $bar{X} = [X_1 ; X_2]$. Now, both $Y$ and $Z$ are functions of $bar{X}$. Whether the elements of $bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $bar{X}$.
Alternatively, if $X_1 < X_2$, the $min$ ($=Y$) is $X_1$, which automatically implies that the $max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
$endgroup$
add a comment |
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2 Answers
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$begingroup$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
$begingroup$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
$begingroup$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
$endgroup$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
answered Dec 17 '18 at 20:37
grand_chatgrand_chat
20.3k11326
20.3k11326
add a comment |
add a comment |
$begingroup$
Think of a random vector $bar{X} = [X_1 ; X_2]$. Now, both $Y$ and $Z$ are functions of $bar{X}$. Whether the elements of $bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $bar{X}$.
Alternatively, if $X_1 < X_2$, the $min$ ($=Y$) is $X_1$, which automatically implies that the $max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
$endgroup$
add a comment |
$begingroup$
Think of a random vector $bar{X} = [X_1 ; X_2]$. Now, both $Y$ and $Z$ are functions of $bar{X}$. Whether the elements of $bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $bar{X}$.
Alternatively, if $X_1 < X_2$, the $min$ ($=Y$) is $X_1$, which automatically implies that the $max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
$endgroup$
add a comment |
$begingroup$
Think of a random vector $bar{X} = [X_1 ; X_2]$. Now, both $Y$ and $Z$ are functions of $bar{X}$. Whether the elements of $bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $bar{X}$.
Alternatively, if $X_1 < X_2$, the $min$ ($=Y$) is $X_1$, which automatically implies that the $max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
$endgroup$
Think of a random vector $bar{X} = [X_1 ; X_2]$. Now, both $Y$ and $Z$ are functions of $bar{X}$. Whether the elements of $bar{X}$ are independent or not, there is no reason to believe that $Y$ and $Z$ are independent, since they are both functions of $bar{X}$.
Alternatively, if $X_1 < X_2$, the $min$ ($=Y$) is $X_1$, which automatically implies that the $max$ is $X_2$ ($=Z$), i.e. knowing $Y$ immediately tells you $Z$, and vice-versa.
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
answered Dec 17 '18 at 21:52
Aditya DuaAditya Dua
1,18418
1,18418
add a comment |
add a comment |
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7
$begingroup$
The min and the max cannot be independent. If you know one, you know something about the other.
$endgroup$
– Hans Engler
Dec 17 '18 at 20:05
3
$begingroup$
It is true that functions of independent RV's are again independent. But you are looking at two functions of the same RV, $(X_1,X_2)$, which is of course not independent of itself.
$endgroup$
– Tki Deneb
Dec 17 '18 at 20:10
1
$begingroup$
In your first link when they talk about functions of independent variables, they mean a function of only one variable i.e $f(X_1)$, not a function of two variable like $f(X_1,X_2)$
$endgroup$
– Sauhard Sharma
Dec 17 '18 at 20:36