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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.

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No it is not valid. You are failing to account for the support.

\begin{align*} f(x,y) &= ye^{-(x+y)} \mathbf{1}[x \ge 0 \text{ and } y \ge 0] \\ &= ye^{-(x+y)} \mathbf{1}[x \ge 0] \cdot \mathbf{1}[y \ge 0] \\ &= \underbrace{\left( e^{-x} \cdot \mathbf{1}[x \ge 0] \right)}_{f_X(x)} \underbrace{\left( y e^{-y} \cdot \mathbf{1}[y \ge 0] \right)}_{f_Y(y)} \end{align*}

The marginal densities above are identified by integrating the joint pdf e.g. $$f_X(x) = \int_{-\infty}^\infty f(x,y) \; dy$$

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  • $\begingroup$ Is there another pair for which $f(x,y) \neq f_X(x) f_Y(y)$, or are the variables actually independent? $\endgroup$ – Wafflebaby Apr 14 '18 at 0:06
  • $\begingroup$ @Wafflebaby They are independent. $\endgroup$ – Mike Earnest Apr 14 '18 at 0:07
  • $\begingroup$ I see how this works where the product is nonnegative. I don't quite see how it works when the product is negative. $\endgroup$ – Wafflebaby Apr 14 '18 at 0:20

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