A question about variance-covariance matrix.
$begingroup$
If $X$ is a vector of random variables such that no element of $X$ is a linear combination of the remaining elements[i.e. there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$],then $Var(X)$ is a positive-definite matrix.
This is a Theorem from George A.F. Seber's book Linear Regression Analysis.
The proof is as follow:
For any vector $c$,we have $;c'Var[X]c=Var[c'X]ge0;$,the equality hold iff $c'X$ is a constant.
My question is that the two statement 'no element of $X$ is a linear combination of the remaining elements' and 'there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$' in the theorem are equivalent?
statistics random-variables covariance
asked Dec 28 '18 at 5:39
Tao XTao X
865
$endgroup$
add a comment |
$begingroup$
If $X$ is a vector of random variables such that no element of $X$ is a linear combination of the remaining elements[i.e. there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$],then $Var(X)$ is a positive-definite matrix.
This is a Theorem from George A.F. Seber's book Linear Regression Analysis.
The proof is as follow:
For any vector $c$,we have $;c'Var[X]c=Var[c'X]ge0;$,the equality hold iff $c'X$ is a constant.
My question is that the two statement 'no element of $X$ is a linear combination of the remaining elements' and 'there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$' in the theorem are equivalent?
statistics random-variables covariance
asked Dec 28 '18 at 5:39
Tao XTao X
865
$endgroup$
$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57
add a comment |
$begingroup$
If $X$ is a vector of random variables such that no element of $X$ is a linear combination of the remaining elements[i.e. there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$],then $Var(X)$ is a positive-definite matrix.
This is a Theorem from George A.F. Seber's book Linear Regression Analysis.
The proof is as follow:
For any vector $c$,we have $;c'Var[X]c=Var[c'X]ge0;$,the equality hold iff $c'X$ is a constant.
My question is that the two statement 'no element of $X$ is a linear combination of the remaining elements' and 'there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$' in the theorem are equivalent?
statistics random-variables covariance
asked Dec 28 '18 at 5:39
Tao XTao X
865
$endgroup$
If $X$ is a vector of random variables such that no element of $X$ is a linear combination of the remaining elements[i.e. there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$],then $Var(X)$ is a positive-definite matrix.
This is a Theorem from George A.F. Seber's book Linear Regression Analysis.
The proof is as follow:
For any vector $c$,we have $;c'Var[X]c=Var[c'X]ge0;$,the equality hold iff $c'X$ is a constant.
My question is that the two statement 'no element of $X$ is a linear combination of the remaining elements' and 'there do not exist $a(neq0)$ and b such that $a'X=b$ for all values of $X=x$' in the theorem are equivalent?
statistics random-variables covariance
statistics random-variables covariance
asked Dec 28 '18 at 5:39
Tao XTao X
865
asked Dec 28 '18 at 5:39
Tao XTao X
865
asked Dec 28 '18 at 5:39
Tao XTao X
865
asked Dec 28 '18 at 5:39
Tao XTao X
865
asked Dec 28 '18 at 5:39
Tao XTao X
865
865
$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57
add a comment |
$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57
$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57
$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57
add a comment |
1 Answer
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$begingroup$
Actually not. The second statement says that no linear combination of the variables of $X$ is a constant variable (not zero, not any other constant). This is the same as saying that the variables in $X$ together with the r.v. which is $1$ with probability $1$ are linearly independent.
For instance, if $Y$ has, say, a normal standard distribution, then $Y$ and $Y+1$ are linearly independent ($Y$ is not a linear combination of $Y+1$—that is, a scalar multiple—nor is the opposite the case). But the set ${Y,Y+1,1}$ is in fact linearly dependent, since, for instance, $Y=1cdot (Y+1)+(-1)cdot 1$.
In any case, if $vec X=(Y,Y+1)$, then $Var(vec X)$ would be positive semi-definite but not definite. Indeed,
$$var(Y)=var(Y+1)=sigma^2,$$
and also
$$cov(Y,Y+1)=cov(Y,Y)=var(Y).$$
So
$$Var(vec X)=left(begin{matrix}sigma^2 &sigma^2 \ sigma^2 &sigma^2 \end{matrix}right).$$
After all, what the property means is that definiteness of the covariance matrix is equivalent to the probability density or mass of probability of $vec X$ not to be entirely contained in any line, plane or hyperplane (passing through the origin or not).
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
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add a comment |
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$begingroup$
Actually not. The second statement says that no linear combination of the variables of $X$ is a constant variable (not zero, not any other constant). This is the same as saying that the variables in $X$ together with the r.v. which is $1$ with probability $1$ are linearly independent.
For instance, if $Y$ has, say, a normal standard distribution, then $Y$ and $Y+1$ are linearly independent ($Y$ is not a linear combination of $Y+1$—that is, a scalar multiple—nor is the opposite the case). But the set ${Y,Y+1,1}$ is in fact linearly dependent, since, for instance, $Y=1cdot (Y+1)+(-1)cdot 1$.
In any case, if $vec X=(Y,Y+1)$, then $Var(vec X)$ would be positive semi-definite but not definite. Indeed,
$$var(Y)=var(Y+1)=sigma^2,$$
and also
$$cov(Y,Y+1)=cov(Y,Y)=var(Y).$$
So
$$Var(vec X)=left(begin{matrix}sigma^2 &sigma^2 \ sigma^2 &sigma^2 \end{matrix}right).$$
After all, what the property means is that definiteness of the covariance matrix is equivalent to the probability density or mass of probability of $vec X$ not to be entirely contained in any line, plane or hyperplane (passing through the origin or not).
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Actually not. The second statement says that no linear combination of the variables of $X$ is a constant variable (not zero, not any other constant). This is the same as saying that the variables in $X$ together with the r.v. which is $1$ with probability $1$ are linearly independent.
For instance, if $Y$ has, say, a normal standard distribution, then $Y$ and $Y+1$ are linearly independent ($Y$ is not a linear combination of $Y+1$—that is, a scalar multiple—nor is the opposite the case). But the set ${Y,Y+1,1}$ is in fact linearly dependent, since, for instance, $Y=1cdot (Y+1)+(-1)cdot 1$.
In any case, if $vec X=(Y,Y+1)$, then $Var(vec X)$ would be positive semi-definite but not definite. Indeed,
$$var(Y)=var(Y+1)=sigma^2,$$
and also
$$cov(Y,Y+1)=cov(Y,Y)=var(Y).$$
So
$$Var(vec X)=left(begin{matrix}sigma^2 &sigma^2 \ sigma^2 &sigma^2 \end{matrix}right).$$
After all, what the property means is that definiteness of the covariance matrix is equivalent to the probability density or mass of probability of $vec X$ not to be entirely contained in any line, plane or hyperplane (passing through the origin or not).
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Actually not. The second statement says that no linear combination of the variables of $X$ is a constant variable (not zero, not any other constant). This is the same as saying that the variables in $X$ together with the r.v. which is $1$ with probability $1$ are linearly independent.
For instance, if $Y$ has, say, a normal standard distribution, then $Y$ and $Y+1$ are linearly independent ($Y$ is not a linear combination of $Y+1$—that is, a scalar multiple—nor is the opposite the case). But the set ${Y,Y+1,1}$ is in fact linearly dependent, since, for instance, $Y=1cdot (Y+1)+(-1)cdot 1$.
In any case, if $vec X=(Y,Y+1)$, then $Var(vec X)$ would be positive semi-definite but not definite. Indeed,
$$var(Y)=var(Y+1)=sigma^2,$$
and also
$$cov(Y,Y+1)=cov(Y,Y)=var(Y).$$
So
$$Var(vec X)=left(begin{matrix}sigma^2 &sigma^2 \ sigma^2 &sigma^2 \end{matrix}right).$$
After all, what the property means is that definiteness of the covariance matrix is equivalent to the probability density or mass of probability of $vec X$ not to be entirely contained in any line, plane or hyperplane (passing through the origin or not).
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
Actually not. The second statement says that no linear combination of the variables of $X$ is a constant variable (not zero, not any other constant). This is the same as saying that the variables in $X$ together with the r.v. which is $1$ with probability $1$ are linearly independent.
For instance, if $Y$ has, say, a normal standard distribution, then $Y$ and $Y+1$ are linearly independent ($Y$ is not a linear combination of $Y+1$—that is, a scalar multiple—nor is the opposite the case). But the set ${Y,Y+1,1}$ is in fact linearly dependent, since, for instance, $Y=1cdot (Y+1)+(-1)cdot 1$.
In any case, if $vec X=(Y,Y+1)$, then $Var(vec X)$ would be positive semi-definite but not definite. Indeed,
$$var(Y)=var(Y+1)=sigma^2,$$
and also
$$cov(Y,Y+1)=cov(Y,Y)=var(Y).$$
So
$$Var(vec X)=left(begin{matrix}sigma^2 &sigma^2 \ sigma^2 &sigma^2 \end{matrix}right).$$
After all, what the property means is that definiteness of the covariance matrix is equivalent to the probability density or mass of probability of $vec X$ not to be entirely contained in any line, plane or hyperplane (passing through the origin or not).
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
edited Jan 10 at 2:40
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
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$begingroup$
Yes, they are equivalent because no component in $X$ belongs to the span of the remaining others
$endgroup$
– Legend Killer
Dec 28 '18 at 5:57