Let f(x,y)=k, if x>0, y>0, and x+y<3 and 0 otherwise.

Question

Let f(x,y)=k, if x>0, y>0, and x+y<3 and 0 otherwise.

Find k. Find P(X+Y<=1) Find P(X^2+Y^2<=1) Find P(Y>X)

I've found that k= 2/9 and P(X+Y<=1)=1/9 but I'm unsure how to set up the second two. I'm also curious whether or not x and y are independent.

Answer 1

You got the second one wrong: $P\{X+Y \leq 1\}=1/9$. For the third one you can observe that $x>0,y>0,x^{2}+y^{2} \leq 1$ imply that $x+y<3$: $(x-y)^{2} \geq 0$ gives $(x+y)^{2} \leq 2(x^{2}+y^{2}) \leq 2 <3$. Hence $P\{X^{2}+Y^{2} \leq 1 \}$ is simply k times the area of one fouth of the circle of radius 1 which is $(2/9)(\pi /4)=\frac {\pi} {18}$. Next, you can observe that intrechanging X and Y does not change $f(x,y)$. This implies that $P\{X>Y\} = P\{Y>X\}$ and hence $P\{Y>X\}=1/2$. Finally, you can draw a picture to see that $P\{Y<1\}$ is k times the sum of the areas of a rectangle with sides 1 and 2 and a right-angled triangle with sides 1 each which works out to 11/9. Similarly, $P\{X<1\}$=11/9. But $P\{X<1,Y<1\}=2/9$ which is not equal to $P\{X<1\}P\{Y<1\}$. Hence the random variables are not independent.

Answer 2

The r.v.'s $X,Y$ have joint pdf $f(x,y) = k1_{\Omega}$, where $\Omega = \{(x,y) \in \Bbb{R}^2 \mid x>0, y>0, x+y < 3\}$. OP has found $k = \frac29$ and $P(X+Y\le1) = \frac29$, but I've found

$$\begin{aligned} P(X+Y \le 1) &= \frac{ \text{Area of } \{(x,y) \in \Bbb{R}^2 \mid x>0, y>0, x+y \le 1\} }{ \text{Area of } \Omega} \\ &= \frac{1/2}{9/2} = \frac19. \end{aligned}$$

Similarly, $$\begin{aligned} P(X^2+Y^2\le1) &= \frac{ \text{Area of } \{(x,y) \in \Bbb{R}^2 \mid x>0, y>0, x^2+y^2 \le 1\} }{ \text{Area of } \Omega} \\ &= \frac{\pi/4}{9/2} = \frac{\pi}{18}. \end{aligned}$$

By symmetry, $P(X<Y) = P(X>Y) = \frac12$.

$X$ and $Y$ aren't independent because \begin{align} P(0<X<\sqrt3, 0<Y < \sqrt3) &= \frac{3}{9/2} = \frac23 \text{, but} \\ P(0<X<\sqrt3) \, P(0<Y < \sqrt3) &= \left(\frac{(3+\sqrt3)/2}{9/2}\right)^2 \notin \Bbb{Q} \\ \therefore P(0<X<\sqrt3, 0<Y < \sqrt3) &\ne P(0<X<\sqrt3) \, P(0<Y < \sqrt3) \end{align}

Remark: By a careful choice of $L$ in $P(0<X<L)$, we don't even need to calculate the probability.