Existence of Multivariate normal distribution $(X_1,…,X_n)$ with $Corr(X_i,X_j) = rho;$












2












$begingroup$


Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 ;,; 1le i le n; ;} $$
$${Var(X_i) = 1;,; 1le i le n; ;} $$
$${Corr(X_i,X_j) = rho;;;1le i;neq j le n; ;}$$
for $ some; 0le rho < 1$.


Show that for any ${rhoin [0,1),(X_1,.....,X_n)}$ as
above exist, but not necessarily for $rhoin (-1,0]$.





I thought of doing it using density function of multivariate normal

That is,if density of
$(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) in mathbb{R^n}$
$${{g(x)=frac{1}{{(2pi)^frac n2}sqrt{detSigma}}e^{-frac12 x^TSigma^{-1}x},1(xin mathbb{R^n});;}}$$


where $Sigma =
begin{bmatrix}
1 & rho &cdots&cdots &rho \
rho & 1& ddots &&rho \
vdots&ddots&ddots&ddots &vdots\
rho&&&1&rho\
rho &cdots&&rho&1\
end{bmatrix}
_{ntimes n}$


${detSigma = (1+(n-1)rho){(1-rho)}^{n-1}}
$



I further thought that as $detSigma$ need to be $ge 0$ we can come up with some conditions that $rho$ need to satisfy.But I didn't get anything useful.
Maybe my thought process was entirely wrong.
Anyway I would be thankful for any help










share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps stop the craziness with enlarged fonts?
    $endgroup$
    – Did
    Dec 3 '18 at 6:54










  • $begingroup$
    I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:10












  • $begingroup$
    Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
    $endgroup$
    – Did
    Dec 3 '18 at 7:14












  • $begingroup$
    Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
    $endgroup$
    – Did
    Dec 3 '18 at 7:17










  • $begingroup$
    Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:22
















2












$begingroup$


Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 ;,; 1le i le n; ;} $$
$${Var(X_i) = 1;,; 1le i le n; ;} $$
$${Corr(X_i,X_j) = rho;;;1le i;neq j le n; ;}$$
for $ some; 0le rho < 1$.


Show that for any ${rhoin [0,1),(X_1,.....,X_n)}$ as
above exist, but not necessarily for $rhoin (-1,0]$.





I thought of doing it using density function of multivariate normal

That is,if density of
$(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) in mathbb{R^n}$
$${{g(x)=frac{1}{{(2pi)^frac n2}sqrt{detSigma}}e^{-frac12 x^TSigma^{-1}x},1(xin mathbb{R^n});;}}$$


where $Sigma =
begin{bmatrix}
1 & rho &cdots&cdots &rho \
rho & 1& ddots &&rho \
vdots&ddots&ddots&ddots &vdots\
rho&&&1&rho\
rho &cdots&&rho&1\
end{bmatrix}
_{ntimes n}$


${detSigma = (1+(n-1)rho){(1-rho)}^{n-1}}
$



I further thought that as $detSigma$ need to be $ge 0$ we can come up with some conditions that $rho$ need to satisfy.But I didn't get anything useful.
Maybe my thought process was entirely wrong.
Anyway I would be thankful for any help










share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps stop the craziness with enlarged fonts?
    $endgroup$
    – Did
    Dec 3 '18 at 6:54










  • $begingroup$
    I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:10












  • $begingroup$
    Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
    $endgroup$
    – Did
    Dec 3 '18 at 7:14












  • $begingroup$
    Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
    $endgroup$
    – Did
    Dec 3 '18 at 7:17










  • $begingroup$
    Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:22














2












2








2


0



$begingroup$


Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 ;,; 1le i le n; ;} $$
$${Var(X_i) = 1;,; 1le i le n; ;} $$
$${Corr(X_i,X_j) = rho;;;1le i;neq j le n; ;}$$
for $ some; 0le rho < 1$.


Show that for any ${rhoin [0,1),(X_1,.....,X_n)}$ as
above exist, but not necessarily for $rhoin (-1,0]$.





I thought of doing it using density function of multivariate normal

That is,if density of
$(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) in mathbb{R^n}$
$${{g(x)=frac{1}{{(2pi)^frac n2}sqrt{detSigma}}e^{-frac12 x^TSigma^{-1}x},1(xin mathbb{R^n});;}}$$


where $Sigma =
begin{bmatrix}
1 & rho &cdots&cdots &rho \
rho & 1& ddots &&rho \
vdots&ddots&ddots&ddots &vdots\
rho&&&1&rho\
rho &cdots&&rho&1\
end{bmatrix}
_{ntimes n}$


${detSigma = (1+(n-1)rho){(1-rho)}^{n-1}}
$



I further thought that as $detSigma$ need to be $ge 0$ we can come up with some conditions that $rho$ need to satisfy.But I didn't get anything useful.
Maybe my thought process was entirely wrong.
Anyway I would be thankful for any help










share|cite|improve this question











$endgroup$




Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 ;,; 1le i le n; ;} $$
$${Var(X_i) = 1;,; 1le i le n; ;} $$
$${Corr(X_i,X_j) = rho;;;1le i;neq j le n; ;}$$
for $ some; 0le rho < 1$.


Show that for any ${rhoin [0,1),(X_1,.....,X_n)}$ as
above exist, but not necessarily for $rhoin (-1,0]$.





I thought of doing it using density function of multivariate normal

That is,if density of
$(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) in mathbb{R^n}$
$${{g(x)=frac{1}{{(2pi)^frac n2}sqrt{detSigma}}e^{-frac12 x^TSigma^{-1}x},1(xin mathbb{R^n});;}}$$


where $Sigma =
begin{bmatrix}
1 & rho &cdots&cdots &rho \
rho & 1& ddots &&rho \
vdots&ddots&ddots&ddots &vdots\
rho&&&1&rho\
rho &cdots&&rho&1\
end{bmatrix}
_{ntimes n}$


${detSigma = (1+(n-1)rho){(1-rho)}^{n-1}}
$



I further thought that as $detSigma$ need to be $ge 0$ we can come up with some conditions that $rho$ need to satisfy.But I didn't get anything useful.
Maybe my thought process was entirely wrong.
Anyway I would be thankful for any help







probability normal-distribution






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share|cite|improve this question













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share|cite|improve this question








edited Dec 3 '18 at 6:54









Did

247k23223459




247k23223459










asked Dec 3 '18 at 5:36









AmeliaAmelia

329




329












  • $begingroup$
    Perhaps stop the craziness with enlarged fonts?
    $endgroup$
    – Did
    Dec 3 '18 at 6:54










  • $begingroup$
    I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:10












  • $begingroup$
    Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
    $endgroup$
    – Did
    Dec 3 '18 at 7:14












  • $begingroup$
    Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
    $endgroup$
    – Did
    Dec 3 '18 at 7:17










  • $begingroup$
    Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:22


















  • $begingroup$
    Perhaps stop the craziness with enlarged fonts?
    $endgroup$
    – Did
    Dec 3 '18 at 6:54










  • $begingroup$
    I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:10












  • $begingroup$
    Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
    $endgroup$
    – Did
    Dec 3 '18 at 7:14












  • $begingroup$
    Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
    $endgroup$
    – Did
    Dec 3 '18 at 7:17










  • $begingroup$
    Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:22
















$begingroup$
Perhaps stop the craziness with enlarged fonts?
$endgroup$
– Did
Dec 3 '18 at 6:54




$begingroup$
Perhaps stop the craziness with enlarged fonts?
$endgroup$
– Did
Dec 3 '18 at 6:54












$begingroup$
I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
$endgroup$
– Amelia
Dec 3 '18 at 7:10






$begingroup$
I wrote the density in enlarged fonts,so that it is more legible.the problem with your edit is that we have to strain our eyes to read the density especially the exponential part.I prefer not to strain thats all
$endgroup$
– Amelia
Dec 3 '18 at 7:10














$begingroup$
Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
$endgroup$
– Did
Dec 3 '18 at 7:14






$begingroup$
Quite bad idea. If you have doubts, look around you on the site and wonder why nobody does it.
$endgroup$
– Did
Dec 3 '18 at 7:14














$begingroup$
Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
$endgroup$
– Did
Dec 3 '18 at 7:17




$begingroup$
Why is what so? Why is it a terrible idea to add LARGE and large everywhere randomly? If you are truly interested in understanding why, perhaps follow the suggestion in my previous comment. Re the exponential part, simply use exp(...) instead of e^{...}.
$endgroup$
– Did
Dec 3 '18 at 7:17












$begingroup$
Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
$endgroup$
– Amelia
Dec 3 '18 at 7:22




$begingroup$
Ok thanks.As I mentioned before this was my first time .So sorry for the inconvenience if any
$endgroup$
– Amelia
Dec 3 '18 at 7:22










2 Answers
2






active

oldest

votes


















1












$begingroup$

The $Sigma$ you desire can be written as $(1-rho)I + rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $Sigma$ now, and state a condition for $Sigmasucceq 0$?




The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $Sigma$ are $1-rho + nrho$ and $1 - rho$. So if $rho in [0, 1)$, these eigenvalues are nonnegative so $Sigma succeq 0$. If $rho in (-1, 0]$, then you can find an $n$ such that $1 - rho + n rho < 0$ so $Sigma not succeq 0$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
    $endgroup$
    – Vim
    Dec 3 '18 at 6:07












  • $begingroup$
    I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32










  • $begingroup$
    after that can u please check if my reasoning is correct
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32






  • 1




    $begingroup$
    so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:35












  • $begingroup$
    @Amelia See my edit.
    $endgroup$
    – angryavian
    Dec 3 '18 at 6:36



















1












$begingroup$

Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly
$AA^T$. Now what $A$ can satisfy your requirement?



To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as
$$C=rho 11^T + (1-rho)I$$
And therefore the required covariance matrix is just
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.



For $rhoin(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.



Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.



For $rho<0$, when does $C$ fail to be PSD? When there exists a non zero $xinBbb R^n$ such that
$$x^TCx = rho (1^T x)^2 + (1-rho)|x|^2 < 0$$
Or
$$|1^Tx| / |x|> left(frac{1 + |rho|}{|rho|}right)^{1/2}.$$
But since $xmapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
    $endgroup$
    – Amelia
    Dec 3 '18 at 5:56












  • $begingroup$
    Yes please see edit.
    $endgroup$
    – Vim
    Dec 3 '18 at 5:59










  • $begingroup$
    thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:47










  • $begingroup$
    @Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
    $endgroup$
    – Vim
    Dec 3 '18 at 8:31











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The $Sigma$ you desire can be written as $(1-rho)I + rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $Sigma$ now, and state a condition for $Sigmasucceq 0$?




The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $Sigma$ are $1-rho + nrho$ and $1 - rho$. So if $rho in [0, 1)$, these eigenvalues are nonnegative so $Sigma succeq 0$. If $rho in (-1, 0]$, then you can find an $n$ such that $1 - rho + n rho < 0$ so $Sigma not succeq 0$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
    $endgroup$
    – Vim
    Dec 3 '18 at 6:07












  • $begingroup$
    I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32










  • $begingroup$
    after that can u please check if my reasoning is correct
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32






  • 1




    $begingroup$
    so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:35












  • $begingroup$
    @Amelia See my edit.
    $endgroup$
    – angryavian
    Dec 3 '18 at 6:36
















1












$begingroup$

The $Sigma$ you desire can be written as $(1-rho)I + rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $Sigma$ now, and state a condition for $Sigmasucceq 0$?




The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $Sigma$ are $1-rho + nrho$ and $1 - rho$. So if $rho in [0, 1)$, these eigenvalues are nonnegative so $Sigma succeq 0$. If $rho in (-1, 0]$, then you can find an $n$ such that $1 - rho + n rho < 0$ so $Sigma not succeq 0$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
    $endgroup$
    – Vim
    Dec 3 '18 at 6:07












  • $begingroup$
    I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32










  • $begingroup$
    after that can u please check if my reasoning is correct
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32






  • 1




    $begingroup$
    so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:35












  • $begingroup$
    @Amelia See my edit.
    $endgroup$
    – angryavian
    Dec 3 '18 at 6:36














1












1








1





$begingroup$

The $Sigma$ you desire can be written as $(1-rho)I + rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $Sigma$ now, and state a condition for $Sigmasucceq 0$?




The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $Sigma$ are $1-rho + nrho$ and $1 - rho$. So if $rho in [0, 1)$, these eigenvalues are nonnegative so $Sigma succeq 0$. If $rho in (-1, 0]$, then you can find an $n$ such that $1 - rho + n rho < 0$ so $Sigma not succeq 0$.







share|cite|improve this answer











$endgroup$



The $Sigma$ you desire can be written as $(1-rho)I + rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $Sigma$ now, and state a condition for $Sigmasucceq 0$?




The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $Sigma$ are $1-rho + nrho$ and $1 - rho$. So if $rho in [0, 1)$, these eigenvalues are nonnegative so $Sigma succeq 0$. If $rho in (-1, 0]$, then you can find an $n$ such that $1 - rho + n rho < 0$ so $Sigma not succeq 0$.








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share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 6:35

























answered Dec 3 '18 at 5:56









angryavianangryavian

40.6k23280




40.6k23280












  • $begingroup$
    Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
    $endgroup$
    – Vim
    Dec 3 '18 at 6:07












  • $begingroup$
    I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32










  • $begingroup$
    after that can u please check if my reasoning is correct
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32






  • 1




    $begingroup$
    so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:35












  • $begingroup$
    @Amelia See my edit.
    $endgroup$
    – angryavian
    Dec 3 '18 at 6:36


















  • $begingroup$
    Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
    $endgroup$
    – Vim
    Dec 3 '18 at 6:07












  • $begingroup$
    I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32










  • $begingroup$
    after that can u please check if my reasoning is correct
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:32






  • 1




    $begingroup$
    so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:35












  • $begingroup$
    @Amelia See my edit.
    $endgroup$
    – angryavian
    Dec 3 '18 at 6:36
















$begingroup$
Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
$endgroup$
– Vim
Dec 3 '18 at 6:07






$begingroup$
Actually, unless each Xi has unit variance, the correlation matrix isn't the same as the covariance matrix. (They are, up to a scaling, though)
$endgroup$
– Vim
Dec 3 '18 at 6:07














$begingroup$
I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
$endgroup$
– Amelia
Dec 3 '18 at 6:32




$begingroup$
I get the point that eigen values of $Sigma; are; (1-rho + nrho);and rho; $.
$endgroup$
– Amelia
Dec 3 '18 at 6:32












$begingroup$
after that can u please check if my reasoning is correct
$endgroup$
– Amelia
Dec 3 '18 at 6:32




$begingroup$
after that can u please check if my reasoning is correct
$endgroup$
– Amelia
Dec 3 '18 at 6:32




1




1




$begingroup$
so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
$endgroup$
– Amelia
Dec 3 '18 at 6:35






$begingroup$
so for $Sigma$ to be semi positive definite all its eigen values must be $ge$ 0.Thus $rho;le;1$ and $rho gefrac{-1}{n-1}$ for any value of n.Thus if $nto infty$ then$rhoge 0$
$endgroup$
– Amelia
Dec 3 '18 at 6:35














$begingroup$
@Amelia See my edit.
$endgroup$
– angryavian
Dec 3 '18 at 6:36




$begingroup$
@Amelia See my edit.
$endgroup$
– angryavian
Dec 3 '18 at 6:36











1












$begingroup$

Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly
$AA^T$. Now what $A$ can satisfy your requirement?



To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as
$$C=rho 11^T + (1-rho)I$$
And therefore the required covariance matrix is just
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.



For $rhoin(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.



Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.



For $rho<0$, when does $C$ fail to be PSD? When there exists a non zero $xinBbb R^n$ such that
$$x^TCx = rho (1^T x)^2 + (1-rho)|x|^2 < 0$$
Or
$$|1^Tx| / |x|> left(frac{1 + |rho|}{|rho|}right)^{1/2}.$$
But since $xmapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
    $endgroup$
    – Amelia
    Dec 3 '18 at 5:56












  • $begingroup$
    Yes please see edit.
    $endgroup$
    – Vim
    Dec 3 '18 at 5:59










  • $begingroup$
    thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:47










  • $begingroup$
    @Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
    $endgroup$
    – Vim
    Dec 3 '18 at 8:31
















1












$begingroup$

Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly
$AA^T$. Now what $A$ can satisfy your requirement?



To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as
$$C=rho 11^T + (1-rho)I$$
And therefore the required covariance matrix is just
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.



For $rhoin(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.



Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.



For $rho<0$, when does $C$ fail to be PSD? When there exists a non zero $xinBbb R^n$ such that
$$x^TCx = rho (1^T x)^2 + (1-rho)|x|^2 < 0$$
Or
$$|1^Tx| / |x|> left(frac{1 + |rho|}{|rho|}right)^{1/2}.$$
But since $xmapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
    $endgroup$
    – Amelia
    Dec 3 '18 at 5:56












  • $begingroup$
    Yes please see edit.
    $endgroup$
    – Vim
    Dec 3 '18 at 5:59










  • $begingroup$
    thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:47










  • $begingroup$
    @Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
    $endgroup$
    – Vim
    Dec 3 '18 at 8:31














1












1








1





$begingroup$

Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly
$AA^T$. Now what $A$ can satisfy your requirement?



To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as
$$C=rho 11^T + (1-rho)I$$
And therefore the required covariance matrix is just
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.



For $rhoin(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.



Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.



For $rho<0$, when does $C$ fail to be PSD? When there exists a non zero $xinBbb R^n$ such that
$$x^TCx = rho (1^T x)^2 + (1-rho)|x|^2 < 0$$
Or
$$|1^Tx| / |x|> left(frac{1 + |rho|}{|rho|}right)^{1/2}.$$
But since $xmapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.






share|cite|improve this answer











$endgroup$



Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly
$AA^T$. Now what $A$ can satisfy your requirement?



To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as
$$C=rho 11^T + (1-rho)I$$
And therefore the required covariance matrix is just
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.



For $rhoin(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.



Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.



For $rho<0$, when does $C$ fail to be PSD? When there exists a non zero $xinBbb R^n$ such that
$$x^TCx = rho (1^T x)^2 + (1-rho)|x|^2 < 0$$
Or
$$|1^Tx| / |x|> left(frac{1 + |rho|}{|rho|}right)^{1/2}.$$
But since $xmapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 8:52

























answered Dec 3 '18 at 5:41









VimVim

8,07731348




8,07731348












  • $begingroup$
    so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
    $endgroup$
    – Amelia
    Dec 3 '18 at 5:56












  • $begingroup$
    Yes please see edit.
    $endgroup$
    – Vim
    Dec 3 '18 at 5:59










  • $begingroup$
    thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:47










  • $begingroup$
    @Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
    $endgroup$
    – Vim
    Dec 3 '18 at 8:31


















  • $begingroup$
    so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
    $endgroup$
    – Amelia
    Dec 3 '18 at 5:56












  • $begingroup$
    Yes please see edit.
    $endgroup$
    – Vim
    Dec 3 '18 at 5:59










  • $begingroup$
    thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
    $endgroup$
    – Amelia
    Dec 3 '18 at 6:47










  • $begingroup$
    @Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
    $endgroup$
    – Vim
    Dec 3 '18 at 8:31
















$begingroup$
so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
$endgroup$
– Amelia
Dec 3 '18 at 5:56






$begingroup$
so $Sigma=AA^T$.For this to happen $Sigma$ must be a positive definite matrix -By cholesky decomposition.Is that what u meant
$endgroup$
– Amelia
Dec 3 '18 at 5:56














$begingroup$
Yes please see edit.
$endgroup$
– Vim
Dec 3 '18 at 5:59




$begingroup$
Yes please see edit.
$endgroup$
– Vim
Dec 3 '18 at 5:59












$begingroup$
thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
$endgroup$
– Amelia
Dec 3 '18 at 6:47




$begingroup$
thanks @viv.Yours was a more general proof and I appreciate your help.If u find my question acceptable can u please up vote it.I am a newbie and I lack in reputation that's why.I am sorry to bother – Amelia 1 min ago edit
$endgroup$
– Amelia
Dec 3 '18 at 6:47












$begingroup$
@Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
$endgroup$
– Vim
Dec 3 '18 at 8:31




$begingroup$
@Jean-ClaudeArbaut sorry you're right. If rho is a small negative number, then it's still positive semidefinite because $11^T$ is a bounded linear operator.
$endgroup$
– Vim
Dec 3 '18 at 8:31


















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