The joint density of $X$ and $Y$ is given by $$f_{X,Y}(x,y)= \left\{\begin{matrix}(x+y), \mbox{ } 0\leq x \leq1 , 0\leq y \leq1 \\ 0, \mbox{ otherwise}\end{matrix}\right.$$
a) Evaluate $f_{X}(X)$ and $f_{Y}(y)$.
b) Are $X$ and $Y$ independent?
c) Evaluate $Var[X]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solutions. Thanks for reading!!
My solutions:
a) $$f_{X}(x) = \int_{0}^{1}(x+y)dy = \left [ xy+\frac{y^{2}}{2} \right ]\Big|_0^1=x+\frac{1}{2}.$$
$$f_{Y}(y) = \int_{0}^{1}(x+y)dx = \left [ \frac{x^{2}}{2}+yx \right ]\Big|_0^1=y+\frac{1}{2}.$$
b) $X$ and $Y$ are independent if $f_{X,Y} = f_{X}(x)f_{Y}(y)$.
From a, $$f_{X}(x)f_{Y}(y) = (x+\frac{1}{2})(y+\frac{1}{2}) = xy + \frac{1}{2}x+ \frac{1}{2}y +\frac{1}{4}.$$
From given, $$f_{X,Y}= x+y.$$
So, since $xy + \frac{1}{2}x+ \frac{1}{2}y +\frac{1}{4} \neq x+y$, $X$ and $Y$ are NOT independent.
c)$$ Var[X] = E[X^{2}] - E[X]^{2}.$$ $$E[X^{2}] = \int_{a}^{b}x^{2}f_{X}(x)dx = \int_{0}^{1}(x^{3}+ \frac{1}{2}x^{2})dx = \left [ \frac{x^{4}}{4} +\frac{x^{3}}{6} \right ]\Big|_0^1$$ $$= \frac{1}{4} +\frac{1}{6}= \frac{3+2}{12}= \frac{5}{12}.$$ $$E[X]^{2} = (\int_{a}^{b}xf_{X}(x)dx)^{2} = (\int_{0}^{1}(x^{2}+ \frac{1}{2}x)dx)^{2} = )\left [ \frac{x^{3}}{3} +\frac{x^{2}}{4} \right ]\Big|_0^1)^{2}.(\frac{7}{12})^{2}=\frac{49}{144}.$$ $$\mbox{So, } Var[X] = \frac{5}{12} -\frac{49}{144}= \frac{60-49}{144}= \frac{11}{144}.$$