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(1) What is the definition of independence between two random vectors $\mathbf X$ and $\mathbf Y$? I think it's more than just pairwise independence between all the elements $X_i$ and $Y_j$.

(2) The interpretation of having the covariance matrix between two random vectors equal to zero is that the elements of the vectors have pairwise covariance zero, because the $ij^{th}$ element of the covariance matrix is the covariance between $X_i$ and $Y_j$. In the joint normal setting, this is the same as saying the elements of the vectors are pairwise independent. How come this is enough to conclude that the two vectors are completely independent, i.e. not just pairwise independent?

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(1) What is the definition of independence between two random vectors $\mathbf X$ and $\mathbf Y$?

The definition of independence between two random vectors is the same as that between two ordinary random variables: Random vectors $\mathbf{x}$ and $\mathbf{y}$ are independent if and only if their joint distribution is equal to the product of their marginal distributions. That is:

$$p(x, y) = p(x) p(y)$$

Or, to write things explicitly in terms of the individual elements of each vector, let $\mathbf{x} = [\mathbf{x}_1, \dots, \mathbf{x}_n]^T$ and $\mathbf{y} = [\mathbf{y}_1, \dots, \mathbf{y}_m]^T$. Then $\mathbf{x}$ and $\mathbf{y}$ are independent if and only if:

$$p(x_1, \dots, x_n, y_1, \dots, y_m) = p(x_1, \dots, x_n) p(y_1, \dots, y_m)$$

Note that the elements of $\mathbf{x}$ may depend on each other, and likewise for $\mathbf{y}$. But there's no dependence between the elements of $x$ and $y$.

(2) The interpretation of having the covariance matrix between two random vectors equal to zero is that the elements of the vectors have pairwise covariance zero, because the $ij^{th}$ element of the covariance matrix is the covariance between $X_i$ and $Y_j$. In the joint normal setting, this is the same as saying the elements of the vectors are pairwise independent. How come this is enough to conclude that the two vectors are completely independent, i.e. not just pairwise independent?

This follows from the particular form of the Gaussian distribution. Suppose random vectors $\mathbf{x} \sim \mathcal{N}(\mu_x, C_x)$ and $\mathbf{y} \sim \mathcal{N}(\mu_y, C_y)$ are jointly Gaussian, and $\text{cov}(\mathbf{x_i}, \mathbf{y_j}) = 0$ for all $i,j$. Then $\mathbf{x}$ and $\mathbf{y}$ are independent because their joint distribution factors into the product of their marginal distributions.

Proof

We can write the joint distribution of $\mathbf{x}$ and $\mathbf{y}$ by concatenating them to form random vector $\mathbf{z} = \left[ \begin{matrix} \mathbf{x} \\ \mathbf{y} \end{matrix} \right]$. $\mathbf{z}$ has a Gaussian distribution with mean $\mu = \left[ \begin{matrix} \mu_x \\ \mu_y \end{matrix} \right]$. Because the covariance between all entries of $\mathbf{x}$ and $\mathbf{y}$ is zero, $\mathbf{z}$ has a block diagonal covariance matrix $C = \left[ \begin{matrix} C_x & \mathbf{0} \\ \mathbf{0} & C_y \\ \end{matrix} \right ]$. The joint density of $\mathbf{x}$ and $\mathbf{y}$ is equal to the density of $\mathbf{z}$:

$$p(x, y \mid \mu_x, \mu_y, C_x, C_y) = p(z \mid \mu, C) = \text{det}(2 \pi C)^{-\frac{1}{2}} \exp \left[ -\frac{1}{2} (z-\mu)^T C^{-1} (z-\mu) \right]$$

Because $C$ is block diagonal, its inverse is $C^{-1} = \left[ \begin{matrix} C_x^{-1} & \mathbf{0} \\ \mathbf{0} & C_y^{-1} \\ \end{matrix} \right ]$ and we can write:

$$(z-\mu)^T C^{-1} (z-\mu) = (x-\mu_x)^T C_x^{-1} (x-\mu_x) + (y-\mu_y)^T C_y^{-1} (y-\mu_y)$$

As a further consequence of the block diagonal form of $C$, the determinant is:

$$\text{det}(C) = \text{det}(C_x) \ \text{det}(C_y)$$

Using these facts and a little algebra, we can re-write the joint distribution as:

$$p(x, y \mid \mu_x, \mu_y, C_x, C_y) =$$

$$\text{det}(2 \pi C_x)^{-\frac{1}{2}} \exp \left[ -\frac{1}{2} (x-\mu_x)^T C_x^{-1} (x-\mu_x) \right]$$

$$\text{det}(2 \pi C_y)^{-\frac{1}{2}} \exp \left[ -\frac{1}{2} (y-\mu_y)^T C_y^{-1} (y-\mu_y) \right]$$

Clearly, this is just the product of the Gaussian marginal distributions of $\mathbf{x}$ and $\mathbf{y}$.

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  • $\begingroup$ Thanks, this answer is exactly what I was looking for! As a quick follow up, is the joint normal distribution the only one with this property that covariance zero results in the joint pdf factoring (and hence independence)? $\endgroup$ – mikario Nov 11 '18 at 18:19
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(1) Regarding your first question, $X$ and $Y$ are independent if we can simply factorize the joint PDF as: $p_{X,Y}(x,y)=p_X(x)p_Y(y)$, irrespective of $X$ and $Y$ being vectors or not.

(2) I'm wondering where you saw that having pairwise independence of $X_i,Y_j$ for all $i,j$ pairs result in complete independence. In general, it is not true. In specific cases of RVs, I don't know if there exists such situation but, in joint normal setting, I can tweak this post of @Sarwate to conform to the situation here. Let's have two vectors as $X=[X_1]$, $Y=[Y_1,Y_2]$. Assume, these are pairwise independent normal RVs, and $Y_1$ and $Y_2$ are jointly normal therefore. But, as the post shows, their joint PDF is not the multiplication of their individual PDFs, in which we can say that they're not independent together.

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  • $\begingroup$ Sorry, maybe I was a bit unclear. My point regarding (2) is that for a jointly normal $\mathbf X$ and $\mathbf Y$, checking covariance zero is enough to verify independence. This is the same thing as checking pairwise covariance zero between the $X_i$ and $Y_j$, since that is exactly what the elements of the covariance matrix represents. And having covariance zero between any pair of jointly normal r.v.s implies independence, right? $\endgroup$ – mikario Nov 4 '18 at 17:37
  • $\begingroup$ @mikario The point of the example in my post that gunes cited above is that it is possible for three normal random variables that are not jointly normal nor mutually independent random variables to nonetheless be pairwise independent normal random variables. The covariance matrix of these random variables is indeed the identity matrix, but simply checking the covariance matrix without continuing to vociferously insist that all the random variables are jointly normal (not just pairwise joint normality) is inadequate. $\endgroup$ – Dilip Sarwate Nov 10 '18 at 17:04

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