The joint density function of $(X,Y)$ is $$f(x,y)=\begin{cases}6(1-x), & 0<y<x,0<x<1\\ 0, & \text{otherwise}\end{cases}$$ Then which of the following are correct?
- $X$ and $Y$ are not independent
- $f_Y(y)=\begin{cases}3(y-1)^2, & 0<y<1\\ 0, & \text{otherwise}.\end{cases}$
- $X$ and $Y$ are independent
- $f_Y(y)=\begin{cases}3(y-\frac{1}{2}y^2), & 0<y<1\\ 0, & \text{otherwise}.\end{cases}$
My work:
I have calculated $f_X(x)=\int_{0}^{x}6(1-x)\,dy=6(1-x)x$ and $f_Y(y)=\int_0^1 6(1-x)\,dx=3$.
So $f_{X,Y}(x,y)\not=f_X(x)f_Y(y)$, hence not independent. But I am unable to conclude (2) and (4). Please help me to solve. Thanks.
Draw a picture. It will show you that the pair $(X,Y)$ of random variables "lives" on the triangle with vertices $(0,0)$, $(1,0)$, $(1,1)$. For since the joint density is $6(1-x)$ when $y\lt x$ and $0\lt x\lt 1$, the density function is non-zero only in the part of the first quadrant below the line $y=x$ and to the left of the line $x=1$.
For the (marginal) density of $Y$, "integrate out" $x$. In principle we are calculating $\int_{-\infty}^\infty f(x,y)\,dx$.
But the joint density is $0$ up to $x=y$, and $0$ from $x=1$ on. So we calculate $\int_{x=y}^1 6(1-x)\,dx$. After a short while we get a version of 2.)