Discussion
\(Cov(X, Y)\) is
Answer: A
The covariance is -0.01.
First compute \(E(XY)\) over the unit square:
\[ \begin{aligned} E(XY) &= \tfrac{6}{5}\int\!\!\int xy(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{1}{4}\right) \\ &= \tfrac{6}{5}\cdot\tfrac{7}{24} = \tfrac{7}{20} = 0.35 \end{aligned} \]
With \(E(X) = E(Y) = 3/5\) from Q.52:
\[ Cov(X, Y) = 0.35 - (0.6)(0.6) = 0.35 - 0.36 = -0.01. \]
The covariance is not 0, because the density \(x + y^{2}\) does not factor into a function of \(x\) times a function of \(y\), so \(X\) and \(Y\) are not independent. The positive values 0.01 and 0.02 have the wrong sign.
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