Let the random variables $X,Y$ have joint density function $f(x,y) = 3y(2-x)$ if $0 < y < 1$ and $y < x < 2-y$ and $f(x,y) = 0$ otherwise.
(a) Prove or disprove that $X$ and $Y$ are independent random variables.
(b) Find the probability of $X + Y \leq 1$.
I'm having a bit of trouble thinking about the bounds here, even after sketching the region on which I'm integrating the joint density function. Below is my solution -- can you tell me if I'm correct ?
(a) For $0 \leq x \leq 1$, $f_X(x) = \int_0^x 3y(2-x)dy = 3x^2 - \frac{3}{2}x^3$
For $1 < x \leq 2$, $f_X(x) = \int_0^{2-x} 3y(2-x)dy = \frac{3}{2}(2-x)^3$.
$f_X(x) = 0$ otherwise.
**The reason for the different cases here is because of how the sketched region looks when I drew it. But I feel odd about it.
For $0 < y < 1$, $f_Y(y) = \int_y^{2-y} 3y(2-x)dx = 6y - y^2$
$f_Y(y) = 0$ otherwise.
Thus, $X$ and $Y$ are not independent, since $f_{X,Y}(x,y) \neq f_X(x) \cdot f_Y(y)$.
(b) $P(X + Y \leq 1) = P(X \leq 1-Y) = \int_0^{\frac{1}{2}} \int_y^{1-y} 3y(2-x)dxdy = \frac{3}{16}$.
Thanks!
What you want for $(b)$ is the intersection of the red, blue, and green regions, so your bounds look good. However, in your integrals for $f_X(x)$, you wrote $dx$ where it should be $dy$. Doesn't affect your answers, but if you wrote it down that way on your homework then you should change it.