how to tell whether x and y are independent or not Suppose that $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$ Find $\operatorname{Var(X+Y)}$. 
I'm having trouble with this problem the way to find $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+2\operatorname{Cov(X,Y)}$, however if $X$ and $Y$ are independent, then $\operatorname{Cov(X, Y)}=0$, the answers indicated that $X$ and $Y$ are independent since they just used $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+0$, my question is how do I tell whether $X$ and $Y$ are independent or not, based on looking only at $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$
 A: If you can factor the joint density into a product that is a function of x times a function of y, then $X$ and $Y$ are independent and their marginal densities are a constant multiple of the two functions in the product.  That is, if 
$$f_{X, Y}(x, y) = g(x) h(y)$$
for some functions $g(x)$ and $h(y)$, then $X$ and $Y$ are independent and
$$f_X(x) = c g(x)$$
and
$$f_Y(y) = \frac{1}{c} h(y),$$
for some constant $c$.  In fact, this statement is an if and only if.  And, in your specific problem, you can write
$$f_{X, Y}(x, y) = (\lambda e^{-\lambda x}) \cdot (\lambda e^{-\lambda y}) = g(x) h(y)$$
so $X$ and $Y$ are independent.  That doesn't tell us what $f_X(x)$ or $f_Y(y)$ are except they are constant multiples of $g(x)$ and $h(y)$, respectively.  I broke it up like I did to be symmetrical and because each function would be the pdf of an exponential.
In this case, you can actually figure out the marginal densities with a bit of thought.  Since we know an exponential has density $\lambda e^{-\lambda x}$ we know that integrates to 1.  So, if we took away the $\lambda$, i.e., $e^{-\lambda x}$, it would integrate to $\frac{1}{\lambda}$.  Or, if we had any other constant $c$ in front, $c e^{-\lambda x}$, it would integrate to $\frac{c}{\lambda}$.  So, we must have $c = \lambda$ in front to ensure the PDF integrates to 1.  So, the $g(x)$ and $h(y)$ I picked above must in fact be the marginal PDFs for $X$ and $Y$.  Thus, if you are very familiar with some of the major densities, you can figure these out without ever integrating to get the marginal densities.  It's just factoring.
A: $$f_{xy}(x,y)\neq f_X(x)f_Y(y)$$
A: You need to find the marginal densities and show that the joint is the product of the marginals in order to show that they are independent.
