Covariance of X and Y on a Quadrilateral

Question

How should I go about determining the covariance? Also, how can I use intuition to determine if it should be positive or negative?

Answer 1

If $Y$ on average gets bigger as $X$ gets bigger, then the covariance is positive; if $Y$ on average gets smaller as $X$ gets bigger, then the covariance is negative. You're just picking a random point in that quadrilateral and asking that question about the horizontal and vertical coordinates of that point.

The covariance is $\operatorname{E}((X-\mu_X)(Y-\mu_Y))$ where $\mu_X=\operatorname{E}(X)$ and $\mu_Y=\operatorname{E}(Y)$. To find $\mu_X$ and $\mu_Y$, just integrate: \begin{align} \mu_X & = \iint_D x \cdot \text{constant} \, dy \, dx, \\[10pt] \mu_Y & = \iint_D y \cdot \text{constant} \, dy \, dx. \end{align} The density is constant because that's what it means to say the distribution is uniform. To find the "constant", you need this: $$ \iint_D \text{constant} \, dy \, dx = 1. $$ So the covariance is $$ \iint_D (x-\mu_X)(y-\mu_Y)\cdot\text{constant} \, dy\,dx. $$ However, there is a slightly quicker way, if you know this identity: $$ \operatorname{cov}(X,Y) = \operatorname{E}(XY) - \operatorname{E}(X)\operatorname{E}(Y) = \operatorname{E}(XY) - \mu_X \mu_Y. $$ Then you have $$ \operatorname{cov}(X,Y) = \iint_D xy\cdot\text{constant} \, dy\,dx - \mu_X\mu_Y. $$ Finding the "constant" is easy because the integral of a constant over a region is just that constant times the area of the region, and in this case the area is $3/2$, so you have $(3/2)\cdot\text{constant} = 1$.

Notice I am being careful about the difference between capital $X$ and $Y$ on the one hand, and on the other hand lower-case $x$ and $y$.

Since you say so little in your question, I'm omitting anything about how to find the integrals.