How can I compute $P(X > 3Y)$ where $X, Y \sim U(-1, 1)$? I got $1/2$ since $\int_{-1}^{1} \int_{3y}^{1} \frac{1}{4} \mathop{dx dy} = 1/2$ but it seems wrong since $P(X > Y)$ should be $1/2$.
How about $P(X > 3Y \mid X > 0)?$ I think this is $(1/6)/2 = 1/12$ by drawing a picture but not completely sure.
Here's my picture for the first part:
On
How can I compute $P(X > 3Y)$ where $X, Y \sim U(-1, 1)$? I got $1/2$ since $\int_{-1}^{1} \int_{3y}^{1} \frac{1}{4} \mathop{dx dy} = 1/2$ but it seems wrong since $P(X > Y)$ should be $1/2$.
As your diagram shows, both $X$ and $Y$ are distributed symmetrically around $0$, and a line through the origin will bisect the $(-1,1)^2$ square whatever is its gradient. Thusly, you should actually anticipate that: $$\forall m\in\Bbb R~.\mathsf P(Y<mX)=1/2$$
How about $P(X > 3Y \mid X > 0)?$ I think this is $(1/6)/2 = 1/12$ by drawing a picture but not completely sure.
$$\mathsf P(X>3Y\mid X>0)=\dfrac{\displaystyle\int_0^1\int_{-1}^{x/3}f_{X,Y}(x,y)\, \mathrm d y\,\mathrm d x}{1/2}=\dfrac 7{12}$$
$[\small\text{Note: }\mathsf P(X>3Y\mid X>0, Y>0)=1/6, \mathsf P(X>3Y\mid X>0,Y\leq 0)=1]$
And, of course:
$$\mathsf P(X>3Y)=\int_{-1}^1\int_{-1}^{x/3} \dfrac 14\,\mathrm dy\,\mathrm d x = \dfrac 12$$
Independence of $X$ and $Y$ is essential for this. Otherwise you cannot compute $P(X>3Y)$.
Your integral is wrong. The integral from $3y$ to $1$ w.r.t. $x$ appears only when $3y<1$ or $y <\frac 1 3$. Similarly you have to make sure that $3y >-1$. So the correct integral is $\int_{-1/3}^{1/3} \int_{3y}^{1} \frac 1 4 dxdy+\int_{-1}^{-1/3} \int_{-1}^{1} \frac 1 4 dxdy$.
For the conditional probability compute $\int_{0}^{1/3} \int_{3y}^{1} \frac 1 4 dxdy+\int_{-1/3}^{0} \int_{0}^{1} \frac 1 4 dxdy+\int_{-1}^{-1/3} \int_{0}^{1} \frac 1 4 dxdy$. and divide by $P(X>0)$ which is $\frac 1 2$.