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What is the (referenced) exercice 6.18 about ? – Jean Marie Apr 04 '16 at 17:25
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About the intuitive way of considering the covariance, you can predict it is (slightly) negative because the general form of the support of the distribution has an approximate axis of symmetry (say $y=-(1/4)x+3/4$) with a negative slope. Otherwise said, if you make a certain number of simulations, you will get a "cloud of points" that will display a (slight) antinomy between $X$ and $Y$. – Jean Marie Apr 04 '16 at 17:49
1 Answers
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.
