Consider $Z_{1}(\omega)=\max\{0,\frac{1}{2}-\omega\}, Z_{2}(\omega)=\max\{0,\omega-\frac{1}{2}\}$.
Compute the covariance $\mbox{Cov}(Z_{1},Z_{2})$
I started by using the formula, $Cov(Z_{1},Z_{2})=E(Z_{1}Z_{2})-E(Z_{1}).E(Z_{2})$
Please click here for a figure of the given functions
$E(Z_{1})=\int^\frac12_0(\frac12-\omega).d\omega=\frac18$
$E(Z_{2})=\int^1_\frac12(\omega-\frac12\omega).d\omega=\frac18$
Then calculating, $E(Z_{1}Z_{2})=\int^1_0(\frac12-\omega).(\omega-\frac12\omega).d\omega$
However, the expectation I am getting for this is negative. This could be due to the constraint that function cannot be less than zero.
It is clear that the functions are not independent, but is the method above and the calculations correct? If not, could the community contributors please point out the correct method. Best Regards
I assume, from what I see that $\Omega=[0,1]$ and I assume that $\omega\sim U[0,1]$ . If so, notice that if $\omega\in\Omega$ is such that $\omega\geq 1/2$ then $Z_1(\omega)=0$ and if $\omega\leq 1/2$, $Z_2(\omega)=0$ so $\forall \omega\in\Omega, Z_1(\omega)Z_2(\omega)=0$. With this observation, we know that $\mathbb{E}(Z_1Z_2)=0$. To calculate the covariance, it is left to calculate $\mathbb{E}(Z_1)$ and $\mathbb{E}(Z_1)$. Notice that $\mathbb{E}(Z_1)=\int_0^1Z_1(\omega)d\omega=\int_0^{1/2}(\frac{1}{2}-\omega)d\omega=\frac{1}{8}=\mathbb{E}(Z_2)$. As a result $Cov(Z_1,Z_2)=-\frac{1}{64}$. So the computations are correct.
Now what is not quite, is your interpretation. When you look at the covariance, you don't care about the sign of the variables, but about how they move together. Think of the relationship between $Z_1$ and $Z_2$. Before $\omega=\frac{1}{2}$, $Z_1$ decreases, while $Z_2$ is not moving, but after $\omega=\frac{1}{2}$, $Z_1$ stops growing, while $Z_2$ starts growing. So you can see that they move in oposite direction, in the sense that they never increase or decrease together, and thats why the covariance is negative even if the RV's are positive.
Edit: As for independence, they of course won't be independent. There are different ways of seeing it, but the intuition is that $Z_1>0 \iff Z_2=0$, so by knowing the outcome of one variable you can infer something about the outcome of the other. Formally, a way of seeing it is the following:
$Z_1, Z_2$ are independent then $\forall f,g:\mathbb{R}\rightarrow \mathbb{R}$ measurable, we have $\mathbb{E}(f(Z_1)g(Z_2))=\mathbb{E}(f(Z_1))\mathbb{E}(g(Z_2))$ so, in particular for $f(x)=g(x)=x$, if $Z_1$ and $Z_2$ are independent then $\mathbb{E}(Z_1Z_2)=\mathbb{E}(Z_1)\mathbb{E}(Z_2) \iff Cov(Z_1,Z_2)=0$ but we already found that $Cov(Z_1,Z_2)<0$.