TITLE: Suppose X and Y have constant density on the region in Figure 26.2
QUESTION [1 upvotes]: Figure 26.2 is above
a) Are $X$ and $Y$ independent?
b) Find $P(X+Y <2)$
c) Find $P(X + Y < 2.5)$
So I know the area of the region of figure 26.2) is $3$ so $f(x,y) = \cfrac{1}{3}$ for the region of figure 26.2)
a)
I'm not sure how I would do this
would marginal density $f(y)$ be piecewise with
$f(y) = \int_0^2\cfrac{1}{3}dx=2/3$ for $0\leq y \leq 1$
and $f(y) = \int_0^2\cfrac{1}{3} = 1/3$ for $1\leq y\leq 2$
and would marginal density $f(x)$ be piecewise with
$f(x) = \int_0^2 \cfrac{1}{3}dy = 2/3$ for $0 \leq x \leq 1$
$f(x) = \int_0^1 \cfrac{1}{3}dy = 1/3$ for $1 \leq x \leq 2$
so would $f(x)f(y)$ also be piecewise with
$f(x)f(y) = \cfrac{4}{9}$ for $0\leq y\leq 1$ and $0\leq x \leq 1$
$f(x)f(y) = \cfrac{1}{9}$ for $1\leq y\leq 2$ and $1\leq x \leq 2$
neither of which equal $f(x,y) = 1/3$? so we can say that X and Y are not independent?
How would you do this question?
b) $P(X + Y < 2)$ is just multiplying the joint pdf by the area of the triangle with vertices at $(0,0), (0,2) (2,0)$ so it's equal to $2/3$ yes?
c)$P(X + Y < 2)$. Imagine we have triangles with vertices at $(0.5,2), (1,2), (1,1.5)$ and at $(1.5,1),(2,1),(2,.5)$ then $P(X + Y < 2) = 1 - f(x,y)\cdot$ area of the those triangles combined = $1 - \cfrac{1}{3}(\cfrac{1}{8} + \cfrac{1}{8}) = 11/12$
So can someone help me with part a) and tell me if my answers to b and c are fine?
Thank You

REPLY [0 votes]: For part 1, to show not independent, it is sufficient to provide some event where you show that independence is violated.
Here, one can check that both X and Y have some positive probability of being above 1. ie $P(X \in (1,2])>0$ and $P(Y \in (1,2])>0$.
But $P(X \in (1,2]\cap Y \in (1,2]) = 0 \neq P(X \in (1,2])\cdot P(Y \in (1,2])$. Therefore they are not independent.

Part 2 and 3 look correct.
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