What part of the Covariance formula determines its sign?

Question

We know that

$$\text{Cov}(x,z) = E(xz) - E(x)E(z)$$

Can it be the case that $\text{Cov}(x,z)>0$ but $E(xz)<0$$?

Or viceversa: $\text{Cov}(x,z)<0$ but $E(xz)>0$?

In other words, is the sign of a covariance determined with certainty by the first element of its decomposition?

Answer 1

Since $E(x)E(z)$ can have any sign, of course you can have either of those.

For example, with $x$ and $z$ taking values $\pm 1$, consider the joint probabilities given by the following tables:

$$ \left[ \begin {array}{c|cc} x \backslash z &-1& 1 \\\hline -1& 0.6 & 0.2 \\ 1 & 0.2 & 0 \end {array} \right] $$

where $E(x) = E(z) = -0.6$, $E(xz) = 0.2$, $\text{Cov}(x,z) = -0.16$, and

$$ \left[ \begin {array}{c|cc} x \backslash z &-1& 1 \\\hline -1& 0.2 & 0.6 \\ 1 & 0 & 0.2 \end {array} \right] $$ where $E(x) = -0.6$, $E(z) = 0.6$, $E(xy) = -0.2$, $\text{Cov}(x,z) = +0.16$.

Answer 2

We know that $$ \operatorname{cov}(X,Y) = \operatorname{cov}(X-a,Y-b), $$ so the covariance is unchanged if we shift $X$ and $Y$ by constants. So $$E[(X-a)(Y-a)] = \operatorname{cov}(X,Y) + E[X-a]E[Y-a]. $$ Suppose $\operatorname{cov}(X,Y) = \operatorname{cov}(X-a,Y-a)$ is negative. If we choose $a$ really large, the left-hand side, $E[(X-a)(Y-a)] = E[XY]-a(E[X+Y])+a^2$ is dominated by $a^2$, and so is positive. Equally, if $\operatorname{cov}(X,Y)>0$, we can consider $\operatorname{cov}(X-a,Y+a)$. So in both cases, we can find $Z$ and $W$ so that $\operatorname{cov}(Z,W)$ has different sign to $E[ZW]$.