There are two service line. The random variables X and Y are the proportions of time that line 1 and line 2 are in use respectively. The joint probability density function for (X, Y)is given by
$$f(x,y) = \begin{cases} \frac{3}{2} (x^2+y^2), &\text 0\le x, y\le1\\ 0 &\text elsewhere \end{cases}$$ here is the problem.
a) Detemine whether or not X and Y are independent.
b) It is of interest to know something about the proportion of Z = X + Y, the sum of the two proportions. Find E(X+Y). Also fune E(XY)
c) Find Var(X), Var(Y), and Cov(X,Y)
d) Find Var(X+Y)
my answer is follow as
a) If X and Y are independent, $f_X(x)*f_Y(y)=f(x,y)$ $$f_X(x) = \frac{3}{2}\int_{0}^{1}x^2+y^2\,dy =\frac{3}{2}x^2+\frac{1}{2},$$ $$ f_Y(y)= \frac{3}{2}y^2+\frac{1}{2} $$ Consequently, X and Y are not independent.
b) $$E(X+Y)=\frac{3}{2}\int_{0}^{1}\int_{0}^{1}(x+y)(x^2+y^2)\,dydx=\frac{5}{4}$$ $$E(XY)=\frac{3}{2}\int_{0}^{1}\int_{0}^{1}xy(x^2+y^2)\,dydx=\frac{3}{8}$$
c) $$E(X^2)=\frac{3}{2}\int_{0}^{1}\int_{0}^{1}x^2(x^2+y^2)\,dydx=\frac{3}{2}*\frac{14}{45}=\frac{7}{15}$$ $$E(X)=\int_{0}^{1}\int_{0}^{1}x^2(x^2+y^2)\,dydx=\frac{3}{2}*\frac{5}{12}=\frac{5}{8}$$ $$Cov(X, Y)=E(XY)-E(X)E(Y)=\frac{3}{8}-(\frac{5}8{})^2?$$
From my calculation, I have negative covariance.
where is my fault? Is there anything I am misunderstanding?
I have not checked all your calculations but if the rv's are not independent (as it you correctly calculated) the covariance can be negative, positive or zero.
The value of your covariance is consistent as
$$|Cov(X,Y) |\leq \sigma_X \sigma_Y$$
Using your results you have that
$$|Cov(X,Y)|\approx 0.016$$
While
$$\sigma_X \sigma_Y \approx 0.076$$
...your calculation looks consistent...