# Proof Verification: Joint variance of the product of a random matrix with a random vector

BACKGROUND

QUESTIONS

Is the proof of my claim correct?

How might my proof be improved?

Claim: (1) The joint-covariance matrix of the product of a real random matrix $$X$$ of dimension $$v\times m$$ and a real random matrix $$Y$$ of dimension $$m\times 1$$ is a real matrix of dimension $$v\times v$$. (2) The element on the $$k^\textrm{th}$$ row and $$l^\textrm{th}$$ column of the joint-covariance matrix, which I denote as $$\operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l}$$, is given as $$\sum\limits_{i=1}^m\sum\limits_{j=1}^m \Bigl( \operatorname {cov}_X( X_{ki}, X_{lj}) + \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \Bigr)\Bigl( \operatorname {cov}_Y( Y_{i}, Y_{j} ) + \operatorname {E}_Y \left[ Y_{i} \right] \operatorname {E}_Y \left[ Y_{j} \right] \Bigr) -\operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \operatorname {E}_Y \left[ Y_{i} \right] \operatorname {E}_Y \left[ Y_{j} \right]$$

PROOF

PART I

By $$\mathbf{X}$$ I denote a real random matrix of dimension $$v\times m$$. By $$\mathbf{Y}$$ I denote a real random matrix of dimension $$m\times 1$$. I write these matrices explicitilty as \begin{align*} \mathbf{X} &= \begin{bmatrix} X_{11} & \cdots & X_{1m} \\ \vdots & \vdots & \vdots \\ X_{v1} & \cdots & X_{vm} \end{bmatrix},~\textrm{and} \\ \mathbf{Y} &= \begin{bmatrix} Y_{1} \\ \vdots \\ Y_{m} \end{bmatrix}~\textrm{respectively.} \end{align*} Apriori, I state that $$X_{ij}$$ and $$Y_k$$ are statistically independent for any and all $$i$$ in $$1,\ldots, v$$; any and all $$j$$ in $$1,\ldots, m$$; and any and all $$k$$ in $$1,\ldots, m$$.

The product $$\mathbf{X} \,\mathbf{Y}$$ can be written explicitly as \begin{align*} \mathbf{X} \,\mathbf{Y} &= \begin{bmatrix} X_{11} & \cdots & X_{1m} \\ \vdots & \vdots & \vdots \\ X_{v1} & \cdots & X_{vm} \end{bmatrix} \begin{bmatrix} Y_{1} \\ \vdots \\ Y_{m} \end{bmatrix} \\ &= \begin{bmatrix} \sum\limits_{i=1}^m X_{1i}\,Y_{i} \\ \vdots \\ \sum\limits_{i=1}^m X_{vi}\,Y_{i} \end{bmatrix} \end{align*} Adapting from [1], since $$\mathbf{X} \,\mathbf{Y}$$ is a vector-valued random vector, with values in $$\mathbb{R}^v$$, then a natural generalization of variance is $$\operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right].$$ From [2], since $$X_{ij}$$ and $$Y_k$$ are statistically independent, \begin{align*} \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] =& \operatorname {E}_X \left[\mathbf{X} \right] \, \operatorname {E}_Y \left[\mathbf{Y} \right] \end{align*} As a consequence, $$\operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right]$$ can be written explicitly as \begin{align*} \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] &= \begin{bmatrix} \sum\limits_{i=1}^m \operatorname {E}_X \left[ X_{1i} \right] \,\operatorname {E}_Y \left[ Y_i \right] \\ \vdots \\ \sum\limits_{i=1}^m \operatorname {E}_X \left[ X_{vi} \right] \,\operatorname {E}_Y \left[ Y_i \right] \end{bmatrix}. \end{align*} The covariance matrix is then written as the expected value of the product of $$v\times 1$$ vector with a $$1\times v$$ vector as \begin{align*} & \operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right] = %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad \operatorname {E} \left[ \begin{bmatrix} \sum\limits_{i=1}^m \left(X_{1i}\,Y_{i} - \operatorname {E}_X \left[ X_{1i} \right] \,\operatorname {E}_Y \left[ Y_i \right] \right) \\ \vdots \\ \sum\limits_{i=1}^m \left(X_{vi}\,Y_{i} - \operatorname {E}_X \left[ X_{vi} \right] \,\operatorname {E}_Y \left[ Y_i \right] \right) \end{bmatrix} \begin{bmatrix} \sum\limits_{i=1}^m \left(X_{1i}\,Y_{i} - \operatorname {E}_X \left[ X_{1i} \right] \,\operatorname {E}_Y \left[ Y_i \right] \right) \\ \vdots \\ \sum\limits_{i=1}^m \left(X_{vi}\,Y_{i} - \operatorname {E}_X \left[ X_{vi} \right] \,\operatorname {E}_Y \left[ Y_i \right] \right) \end{bmatrix} ^{\top }\right] . \end{align*}

The covariance matrix has a dimension of $$v\times v$$

PART II

By $$\operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l}$$ I denote the element on at the $$k^\textrm{th}$$ row and $$l^\textrm{th}$$ column of the covariance matrix. Since the expectation the covariance matrix is equal to the matrix of expecatations of the covarniance matrix' elements, and since the expectation of a sum is equal to the sum of expectations, I write $$\operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l}$$ as: \begin{align*} & \operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l} %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E} \left[\left(X_{ki}\,Y_{i} - \operatorname {E}_X \left[ X_{ki} \right] \,\operatorname {E}_Y \left[ Y_i \right] \right) \left(X_{lj}\,Y_{j} - \operatorname {E}_X \left[ X_{lj} \right] \,\operatorname {E}_Y \left[ Y_j \right] \right)\right] \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki}\, X_{lj}\, \right] \, \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] \\ &\quad - \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_j \right] \\ &\quad - \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_i \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_i \right] \, \operatorname {E}_Y \left[ Y_j \right] %%%%%%%%%%%55 %%%%%%%%%%%%% %%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki}\, X_{lj}\, \right] \, \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] \\ &\quad - \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_j \right] . \end{align*} I now attempt to separate the variables \begin{align*} & \operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l} %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki}\, X_{lj} \right] - \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] \\ &\quad - \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_j \right] %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki}\, X_{lj} \right] - \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \left( \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] - \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \right) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki}\, X_{lj} \right] - \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \left( \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \right) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] \\ &\quad - \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \, \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_j \right] . %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \left( \operatorname {E}_X \left[ X_{ki}\, X_{lj} \right] - \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \, \left( \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] - \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \right) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \, \left( \operatorname {E}_X \left[ X_{ki}\, X_{lj} \right] - \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \right) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \, \left( \operatorname {E}_Y \left[ Y_{i} \,Y_{j} \right] - \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \right) \end{align*} From the definition of covariance [3], I rewrite the above as follows. \begin{align*} & \operatorname {E} \left[(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )(\mathbf{X} \,\mathbf{Y}- \operatorname {E} \left[\mathbf{X} \,\mathbf{Y} \right] )^{\top }\right]_{k,l} %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% %%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {cov}_X( X_{ki}, X_{lj}) \, \operatorname {cov}_Y( Y_{i}, Y_{j}) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \, \operatorname {E}_Y \left[ Y_{i} \right] \, \operatorname {E}_Y \left[ Y_{j} \right] \, \operatorname {cov}_X( X_{ki}, X_{lj}) \\ &\quad + \sum\limits_{i=1}^m\sum\limits_{j=1}^m \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \, \operatorname {cov}_Y( X_{i}, Y_{j}) %%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% \\ &\quad = \sum\limits_{i=1}^m\sum\limits_{j=1}^m \Bigl( \operatorname {cov}_X( X_{ki}, X_{lj}) + \operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \Bigr)\Bigl( \operatorname {cov}_Y( Y_{i}, Y_{j} ) + \operatorname {E}_Y \left[ Y_{i} \right] \operatorname {E}_Y \left[ Y_{j} \right] \Bigr) -\operatorname {E}_X \left[ X_{ki} \right] \operatorname {E}_X \left[ X_{lj} \right] \operatorname {E}_Y \left[ Y_{i} \right] \operatorname {E}_Y \left[ Y_{j} \right] \end{align*}

BIBLIOGRAPHY

• Comments are not for extended discussion; this conversation has been moved to chat.
– whuber
Aug 5, 2019 at 14:00
• It looks correct, but you can get there in about 25 fewer steps by breaking the ideas down into simpler ones. The calculation does two things: (1) it applies the definition of matrix multiplication and bilinearity of covariance to express the result as a double sum and (2) it computes $\operatorname{Cov}(AB,CD)$ where random variables $(A,C)$ are independent of $(B,D).$ Step (1) can be accomplished in one line and step (2) in one or two lines, depending on how much algebraic detail you feel your readers need to see. Such brevity and simplicity lead to greater confidence in the result.
– whuber
Aug 5, 2019 at 14:08

I would suggest amending your work by simplifying the notation and the algebra, because a clear derivation is easier to check and more convincing than a long one and puts much less burden on your readers, as well as revealing the key ideas in the result.

Let $$(A,B,C,D)$$ be any random variables for which $$(A,C)$$ is independent of $$(B,D).$$ From that fact and the definition of covariance as $$\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y],$$ you may compute

\eqalign{ \operatorname{Cov}(AB,CD) &= E[ABCD]-E[AB]E[CD] \\&= E[AC]E[BD] - E[A]E[B]E[C]E[D] \\ &= \left(\operatorname{Cov}(A,C)+E[A]E[C]\right) \left(\operatorname{Cov}(B,D)+E[B]E[D]\right) - E[A]E[B]E[C]E[D]. }

Consequently, letting $$A=X_{ki}, B=y_i, C=X_{lj},$$ and $$D=y_j,$$ the definition of matrix multiplication and the bilinearity of covariance yield

\eqalign{ \operatorname{Cov}((Xy)_k, (Xy)_l) &= \operatorname{Cov}\left(\sum_i X_{ki}y_i\ \sum_j X_{lj}y_j\right) \\ &= \sum_{i,j}\operatorname{Cov}\left(X_{ki}y_i X_{lj}y_j\right) \\ &= \sum_{i,j}\left(\operatorname{Cov}(X_{ki},X_{lj})+E[X_{ki}]E[X_{lj}]\right) \left(\operatorname{Cov}(y_i,y_j)+E[y_i]E[y_j]\right) - E[X_{ki}]E[y_i]E[X_{lj}]E[y_j], }