If the joint density of $X$ and $Y$ is given by $f(x,\, y)=30xy^2$ if $x-1\leq y \leq 1-x,\, 0\leq x\leq 1$.

$(a)$ $P(X\leq 1/2,\, Y\leq 1/2)$

$(b)$ $P(X\leq 1/2,\, Y\leq 2)$

$(c)$ $P(X> Y)$

**Attempt**

$(a)$ $P(X\leq 1/2,\, Y\leq 1/2)=P(0\leq X\leq 1/2,\, x-1\leq Y\leq 1/2)=\int _0^{\frac{1}{2}}\int _{x-1}^{\frac{1}{2}}30xy^2\:dy\:dx=9/16$

$(b)$ $P(X\leq 1/2,\, Y\leq 2)=P(0\leq X\leq 1/2,\, x-1\leq Y\leq 2)=\int _0^{\frac{1}{2}}\int _{x-1}^{2}30xy^2\:dy\:dx=333/32$

$(c)$ $P(X> Y)=P(Y<X\leq 1,\, X-1\leq Y\leq 1-X )$

## Best Answer

Your answer to $(a)$ is correct.

Obviously, your answer to $(b)$ is incorrect, since probabilities cannot exceed $1$. So what happened? You have to observe that since $0 \le X \le 1$, the largest $Y$ can be is $1 - X \le 1 - 0 = 1$. So the condition $Y \le 2$ means nothing; i.e., $$\Pr[(X \le 1/2) \cap (Y \le 2)] = \Pr[X \le 1/2] = \int_{x=0}^{1/2} \int_{y=x-1}^{1-x} 30 xy^2 \, dy \, dx.$$ Selecting $2$ as the upper limit of integration for $Y$ will integrate $30xy^2$ over a region for which the probability is $0$.

For $(c)$, sketch the region of integration satisfied by the inequalities $$0 \le X \le 1, \\ X - 1 \le Y \le 1 - X, \\ X > Y.$$ Here is a hint: the region has vertices $$(0,0), (1/2, 1/2), (1,0), (0, -1).$$ Then split the region of integration into two disjoint sub-regions, and set up a double integral for each sub-region.