A question about variance-covariance matrix.












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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







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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
















0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






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asked Dec 28 '18 at 5:39









Tao XTao X

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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


















  • $begingroup$
    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










1 Answer
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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).






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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).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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).






        share|cite|improve this answer











        $endgroup$



        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).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 2:40

























        answered Dec 28 '18 at 5:58









        Alejandro Nasif SalumAlejandro Nasif Salum

        4,765118




        4,765118






























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