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Under what conditions is an arbitrary matrix $A$ the covariance matrix of a column vector of random variables?

0

$begingroup$

Let $X = beginbmatrix X_1 & X_2 & dots& X_n endbmatrix ^T $ be an arbitrary column vector of random variables.

Given an $n times n$ matrix $A$, how do I determine whether there exists an $X$ such that $K_XX = A$ ? I'm asking about both the case where the elements of $X$ are constrained to be independent and the case where the elments of $X$ are not constrained to be independent.

Paraphrasing a bit from Wikipedia, let $X' = X - textE[X]$ and $Y' = Y - textE[Y]$ .

$$ K_XY ;; stackreltextdef= ;; textE[X'(Y'^T)] $$

which means:

$$ K_XX ;; stackreltextdef= ;; textE[X'(X'^T)] $$

Which I think means that the $ij$th entry looks like this:

$$ (K_XX)_ij ;;stackreltextdef=;; textE[X'X']_ij $$

And this is where I get stuck. Since multiplication is commutative, in order for $A$ to be the covariance matrix of some column vector of random variables, it has to be symmetric.

However, as for actually working backwards and picking individual random variables inside $X'$ I'm stumped.

If I constrain each random variable in $X'$ to be independent, then I'm still picking a càdlàg CDF for each variable, up to the restriction that the mean is zero. That's a huge amount of freedom, so it seems like it should be possible to hit an arbitrary symmetric matrix $A$.

But that's just a vague intuition. I'm not sure how to break the problem down and prove it one way or the other.

linear-algebra covariance

share|cite|improve this question

edited Apr 9 at 17:30

Gregory Nisbet

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

$endgroup$

add a comment | 

0

$begingroup$

Let $X = beginbmatrix X_1 & X_2 & dots& X_n endbmatrix ^T $ be an arbitrary column vector of random variables.

Given an $n times n$ matrix $A$, how do I determine whether there exists an $X$ such that $K_XX = A$ ? I'm asking about both the case where the elements of $X$ are constrained to be independent and the case where the elments of $X$ are not constrained to be independent.

Paraphrasing a bit from Wikipedia, let $X' = X - textE[X]$ and $Y' = Y - textE[Y]$ .

$$ K_XY ;; stackreltextdef= ;; textE[X'(Y'^T)] $$

which means:

$$ K_XX ;; stackreltextdef= ;; textE[X'(X'^T)] $$

Which I think means that the $ij$th entry looks like this:

$$ (K_XX)_ij ;;stackreltextdef=;; textE[X'X']_ij $$

And this is where I get stuck. Since multiplication is commutative, in order for $A$ to be the covariance matrix of some column vector of random variables, it has to be symmetric.

However, as for actually working backwards and picking individual random variables inside $X'$ I'm stumped.

If I constrain each random variable in $X'$ to be independent, then I'm still picking a càdlàg CDF for each variable, up to the restriction that the mean is zero. That's a huge amount of freedom, so it seems like it should be possible to hit an arbitrary symmetric matrix $A$.

But that's just a vague intuition. I'm not sure how to break the problem down and prove it one way or the other.

linear-algebra covariance

share|cite|improve this question

edited Apr 9 at 17:30

Gregory Nisbet

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

$endgroup$

add a comment | 

$begingroup$

Let $X = beginbmatrix X_1 & X_2 & dots& X_n endbmatrix ^T $ be an arbitrary column vector of random variables.

Given an $n times n$ matrix $A$, how do I determine whether there exists an $X$ such that $K_XX = A$ ? I'm asking about both the case where the elements of $X$ are constrained to be independent and the case where the elments of $X$ are not constrained to be independent.

Paraphrasing a bit from Wikipedia, let $X' = X - textE[X]$ and $Y' = Y - textE[Y]$ .

$$ K_XY ;; stackreltextdef= ;; textE[X'(Y'^T)] $$

which means:

$$ K_XX ;; stackreltextdef= ;; textE[X'(X'^T)] $$

Which I think means that the $ij$th entry looks like this:

$$ (K_XX)_ij ;;stackreltextdef=;; textE[X'X']_ij $$

And this is where I get stuck. Since multiplication is commutative, in order for $A$ to be the covariance matrix of some column vector of random variables, it has to be symmetric.

However, as for actually working backwards and picking individual random variables inside $X'$ I'm stumped.

If I constrain each random variable in $X'$ to be independent, then I'm still picking a càdlàg CDF for each variable, up to the restriction that the mean is zero. That's a huge amount of freedom, so it seems like it should be possible to hit an arbitrary symmetric matrix $A$.

But that's just a vague intuition. I'm not sure how to break the problem down and prove it one way or the other.

linear-algebra covariance

share|cite|improve this question

edited Apr 9 at 17:30

Gregory Nisbet

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

$endgroup$

share|cite|improve this question

edited Apr 9 at 17:30

Gregory Nisbet

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

share|cite|improve this question

edited Apr 9 at 17:30

Gregory Nisbet

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712

asked Apr 8 at 20:13

Gregory NisbetGregory Nisbet

872712