Q: Let $(X,Y)$ be a two dimensional random variable with the following density function:
$f_{_{X,Y}}(x,y)=\begin{cases}8xy&, 0\le x\le y \le 1 \\ 0&, o.w.\end{cases}$
Are $X,Y$ independent?
My attempts: in the lecture notes I have, it says that two random variables $X,Y$ are independent $\iff \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$.
I found $f_{_X}, f_{_Y}$:
- $ f_{_X}(x)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dy= \begin{cases} \int_\limits x^1 8xy\,dy&, 0\le x \le 1 \\ 0&, o.w. \end{cases} = \begin{cases} 4x-4x^3 &, 0\le x \le 1 \\ 0&, o.w. \end{cases} $
- $ f_{_Y}(y)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dx= \begin{cases} \int_\limits 0^y 8xy \,dx &, 0\le y \le 1 \\ 0&, o.w. \end{cases} = \begin{cases} 4y^3 &, 0\le y \le 1 \\ 0&, o.w. \end{cases} $
this is the part I'm not sure about:
$
f_{_X}(x)\cdot f_{_Y}(y) =
\begin{cases}
(4x-4x^3)\cdot 4y^3 & ,0 \le x, y \le 1 \\
0 &, o.w.
\end{cases}
$
so $f_{_{X,Y}}(x,y) \neq f_{_X}(x)\cdot f_{_Y}(y)$.
It's doesn't seem to be the right way to show the inequality. Also, what if the equality was true? it's possible that the same function has two ways to be written.
So my question is what should I do to prove or disprove that $ \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$, not only in this case but in the more general case.
Please keep in mind that I'm an undergraduate taking my first probability course, so please try to keep it simple.
Thanks
Your approach is right. $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ a.e. is necessary and sufficient for independence of $X$ and $Y$ provided the joint dendisty $f_{X,Y}$ exists.