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A question has been asked regarding the Variance of product of dependent variables. I am interested in the case in which $X$ is a vector and $Y$ is a scalar and the two variables are independent.

What is $\mbox{var}(\mathbf{X}Y)$, where $\mathbf{X}\perp Y$?

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Vivek Subramanian
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    The only way to interpret this notation consistently is that "$\mbox{var}(\mathbf{X}Y)$" has to be the vector of variances of the $X_iY$, rather than full variance-covariance matrix. But then the last term on the right side (which is either a matrix or a scalar, depending on whether you think of vectors as being columns or rows) does not match its dimensions. The second question is similarly problematic. Please, then, explain what your notation means. – whuber Feb 13 '15 at 16:22
  • Thanks, @whuber. My vectors are column vectors. The notation $\mbox{var}(\mathbf{X}Y)$ refers to the matrix describing the covariances between the rows of $\mathbf{X}Y$. If you call this matrix $\mathbf{C}$, then $\mathbf{C}_{st}$ gives the covariance between $X_sY$ and $X_tY$. Does this make sense? – Vivek Subramanian Feb 13 '15 at 16:26
  • It does, but now the terms in your second question have dimensions (from left to right) $1\times 1$, $n\times n$, $n\times n$, and $n\times n$ where $X$ and $Y$ each are of dimension $n$. If instead you wanted to ask about $\text{var}(XY^\prime)$, then $XY^\prime$ would be a matrix with $n^2$ coefficients and its variance-covariance matrix would have $n^4$ entries. There's no way you can make the left and right sides match. – whuber Feb 13 '15 at 16:30
  • Right, okay, that makes sense. Let's ignore the second question - I didn't put too much thought into it and just realized that $\mbox{var}(\mathbf{X}^\top\mathbf{Y})$ is simply $\sum{\mbox{var}(X_sY_s)}$, and each entry of the sum can be computed using the formula in the question I link to in my first statement. – Vivek Subramanian Feb 13 '15 at 16:39
  • The first question isn't much different from that: the matrix on the left has two kinds of terms of the forms $\text{var}(X_iY)$ (whose values you already know from the previous thread you reference) and $\text{cov}(X_iY, X_jY)$ (whose values can be obtained using the same method described in that thread). – whuber Feb 13 '15 at 16:42

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As @whuber pointed out, we can calculate the variance of each entry of $\mathbf{X}Y$ using the formula in the question linked to in my first statement. Thus, $$\mbox{cov}(X_iY, X_iY) = \mbox{var}(X_iY) = \mbox{var}(X_i)\mbox{var}(Y) + \mbox{var}(X_i)\mathbb{E}[Y]^2 + \mbox{var}(Y)\mathbb{E}[X_i]^2$$

Moreover, the covariance between $X_iY$ and $X_jY$ is given by: $$\begin{align*} \mbox{cov}(X_iY, X_jY) &= \mathbb{E}[X_iX_jY^2] - \mathbb{E}[X_iY]\mathbb{E}[X_jY]\\ &= \mathbb{E}[X_iX_j]\mathbb{E}[Y^2] - \mathbb{E}[X_i]\mathbb{E}[X_j]\mathbb{E}[Y]\mathbb{E}[Y]\\ &= (\mbox{cov}(X_i, X_j) + \mathbb{E}[X_i]\mathbb{E}[X_j])\mathbb{E}[Y^2] -\mathbb{E}[X_i]\mathbb{E}[X_j]\mathbb{E}[Y]^2\\ &= \mbox{cov}(X_i, X_j)\mathbb{E}[Y^2] + (\mathbb{E}[X_i]\mathbb{E}[X_j])(\mathbb{E}[Y^2] - \mathbb{E}[Y]^2)\\ &= \mbox{cov}(X_i, X_j)\mathbb{E}[Y^2] + \mathbb{E}[X_i]\mathbb{E}[X_j]\mbox{var}(Y)\\ &= \mbox{cov}(X_i, X_j)(\mbox{var}(Y) + \mathbb{E}[Y]^2) + \mathbb{E}[X_i]\mathbb{E}[X_j]\mbox{var}(Y) \end{align*}$$

Note that when we replace $X_jY$ with $X_iY$ in the formula for $\mbox{cov}(X_iY, X_jY)$, we recover the formula for $\mbox{var}(X_iY)$. Thus, we can construct a covariance matrix $\mathbf{C}$ for $\mathbf{X}Y$ where $C_{ij} = \mbox{cov}(X_iY, X_jY)$.

Vivek Subramanian
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  • I can't figure out how to change the commit title of my revision to this answer, so if someone could please help me with that, that'd be great. I'd like it to read, "Old answer had mistake - used E[Y]^2 instead of E[Y^2]. Fixed." – Vivek Subramanian Feb 16 '15 at 23:06