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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.D
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notamathwiz
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3 Answers3

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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.

GeoffDS
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    +1 It is worth emphasizing, and especially to beginning students, that the factorization $f_{X,Y}(x,y)=g(x)h(y)$ must hold for all $x, y$, and not just in the expression specified. For example, the density $$f_{W,Z}(x,y) = \begin{cases}2\lambda^2e^{-\lambda(x+y)},&0 < x < y <\infty,\0,&\text{elsewhere,}\end{cases}$$ can be expressed as $g(x)h(y)$ but only for $0 < x < y < \infty$, not everywhere, and so $W$ and $Z$, which can be taken to be $\min(X,Y)$ and $\max(X,Y)$ in this instance, are not independent random variables. – Dilip Sarwate May 19 '13 at 13:21
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$$f_{xy}(x,y)\neq f_X(x)f_Y(y)$$

Seyhmus Güngören
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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.

response
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