# Proof of Dependence of Random Variables

Is the following proof valid?

Let $X$ and $Y$ be jointly continuous random variables such that the joint density function is given by $$f(x,y) = \begin{cases} ye^{-(x+y)} & \text{ for } x>0, y>0 \\ 0 & \text{ otherwise } \end{cases}$$

Then $X$ and $Y$ are dependent.

Proof

Let $X$ and $Y$ be continuous random variables. Let the probability density function of $X$ be given by $f_X(x) = e^x$. Let the probability density function of $Y$ be given by $f_Y(y) = ye^y$. $$f_X(x)f_Y(y) = (e^{-x})(ye^{-y})= ye^{-(x+y)}$$ Jointly continuous random variables are independent if and only if $f(x,y) = f_X(x) f_Y(y) \hspace{8 px} \forall \hspace{7 px} x, y \in \mathbb{R}$. Thus, to prove that $X$ and $Y$ are dependent, it suffices to show that there exist some pair of real numbers $x, y$ such that $f(x, y) \neq f_X(x) f_Y(y)$. Suppose $x = -1$, and suppose $y = -1$. Then $f(x, y) = 0$. We also have that $f_X(-1)f_Y(-1) = -e^2 \neq 0$. So $X$ and $Y$ are dependent.

The marginal densities above are identified by integrating the joint pdf e.g. $$f_X(x) = \int_{-\infty}^\infty f(x,y) \; dy$$
• Is there another pair for which $f(x,y) \neq f_X(x) f_Y(y)$, or are the variables actually independent? – Wafflebaby Apr 14 '18 at 0:06