Are a sum and a product of two covariance matrices also a covariance matrix? Suppose I have covariance matrices $X$ and $Y$. Which of these options are then also covariance matrices?


*

*$X+Y$

*$X^2$

*$XY$



I have a bit of trouble understanding what exactly is needed for something to be  a covariance matrix. I suppose it is meant that for instance if $X=\operatorname{cov}(X_1,X_2)$, and $Y=\operatorname{cov}(Y_1,Y_2)$ that for 1 to hold true we should have that $\operatorname{cov}(X_1,X_2) + \operatorname{cov}(Y_1,Y_2) = \operatorname{cov}(Z_1, Z_2)$, where $Z_1$ and $Z_2$ are some other random variables. However, I can't see why that would hold true for any of the three options. Any insight would be apprciated.
 A: Background
A covariance matrix $\mathbb{A}$ for a vector of random variables $X=(X_1, X_2, \ldots, X_n)^\prime$ embodies a procedure to compute the variance of any linear combination of those random variables.  The rule is that for any vector of coefficients $\lambda = (\lambda_1, \ldots, \lambda_n)$,
$$\operatorname{Var}(\lambda X) = \lambda \mathbb{A} \lambda ^\prime.\tag{1}$$
In other words, the rules of matrix multiplication describe the rules of variances.
Two properties of $\mathbb{A}$ are immediate and obvious:

*

*Because variances are expectations of squared values, they can never be negative.  Thus, for all vectors $\lambda$, $$0 \le \operatorname{Var}(\lambda X) = \lambda \mathbb{A} \lambda ^\prime.$$ Covariance matrices must be non-negative-definite.


*Variances are just numbers--or, if you read the matrix formulas literally, they are $1\times 1$ matrices.  Thus, they do not change when you transpose them.  Transposing $(1)$ gives $$\lambda \mathbb{A} \lambda ^\prime = \operatorname{Var}(\lambda X)  = \operatorname{Var}(\lambda X) ^\prime = \left(\lambda \mathbb{A} \lambda ^\prime\right)^\prime = \lambda \mathbb{A}^\prime \lambda ^\prime.$$  Since this holds for all $\lambda$, $\mathbb{A}$ must equal its transpose $\mathbb{A}^\prime$: covariance matrices must be symmetric.
The deeper result is that any non-negative-definite symmetric matrix $\mathbb{A}$ is a covariance matrix.  This means there actually is some vector-valued random variable $X$ with $\mathbb{A}$ as its covariance.  We may demonstrate this by explicitly constructing $X$.  One way is to notice that the (multivariate) density function $f(x_1,\ldots, x_n)$ with the property $$\log(f) \propto -\frac{1}{2} (x_1,\ldots,x_n)\mathbb{A}^{-1}(x_1,\ldots,x_n)^\prime$$ has $\mathbb{A}$ for its covariance.  (Some delicacy is needed when $\mathbb{A}$ is not invertible--but that's just a technical detail.)
Solutions
Let $\mathbb{X}$ and $\mathbb{Y}$ be covariance matrices.  Obviously they are square; and if their sum is to make any sense they must have the same dimensions.  We need only check the two properties.

*

*The sum.

*

*Symmetry $$(\mathbb{X}+\mathbb{Y})^\prime = \mathbb{X}^\prime + \mathbb{Y}^\prime = (\mathbb{X} + \mathbb{Y})$$ shows the sum is symmetric.

*Non-negative definiteness.  Let $\lambda$ be any vector.  Then $$\lambda(\mathbb{X}+\mathbb{Y})\lambda^\prime = \lambda \mathbb{X}\lambda^\prime + \lambda \mathbb{Y}\lambda^\prime \ge 0 + 0 = 0$$ proves the point using basic properties of matrix multiplication.



*I leave this as an exercise.


*This one is tricky.  One method I use to think through challenging matrix problems is to do some calculations with $2\times 2$ matrices.  There are some common, familiar covariance matrices of this size, such as $$\pmatrix{a & b \\ b & a}$$ with $a^2 \ge b^2$ and $a \ge 0$.  The concern is that $\mathbb{XY}$ might not be definite: that is, could it produce a negative value when computing a variance?  If it will, then we had better have some negative coefficients in the matrix.  That suggests considering $$\mathbb{X} = \pmatrix{a & -1 \\ -1 & a}$$ for $a \ge 1$.  To get something interesting, we might gravitate initially to matrices $\mathbb{Y}$ with different-looking structures.  Diagonal matrices come to mind, such as $$\mathbb{Y} = \pmatrix{b & 0 \\ 0 & 1}$$ with $b\ge 0$.  (Notice how we may freely pick some of the coefficients, such as $-1$ and $1$, because we can rescale all the entries in any covariance matrix without changing its fundamental properties.  This simplifies the search for interesting examples.)
I leave it to you to compute $\mathbb{XY}$ and test whether it always is a covariance matrix for any allowable values of $a$ and $b$.
A: A real matrix is a covariance matrix if and only if it is symmetric positive semi-definite.
Hints:
1) If $X$ and $Y$ are symmetric, is $X+Y$ symmetric? If $z^TX z \ge 0$ for any $z$ and $z^TY z \ge 0$ for any $z$, what can you conclude about $z^T(X+Y)z$?
2) If $X$ is symmetric, is $X^2$ symmetric? If the eigenvalues of $X$ are non-negative, what can you conclude about the eigenvalues of $X^2$?
3) If $X$ and $Y$ are symmetric, can you conclude that $XY$ is symmetric, or can you find a counter-example?
