Solutions

4.1

$\displaystyle f(x,y) = \begin{cases}\frac{1}{4} & -1 \le x, y \le 1 0 & \text{otherwise} \end{cases}$

a.

$\displaystyle P(X^{2}+Y^{2}<1)$	$\displaystyle = \frac{\text{area of circle}}{\text{total area}}$
	$\displaystyle = \frac{\pi}{4}$

b.

$\displaystyle P(Y < 2 X)$	$\displaystyle = \frac{1}{2}$ by symmetry, or
	$\displaystyle = \int_{-1}^{1}\int_{y/2}^{1}\frac{1}{4}dxdy = \int_{-1}^{1}\frac{1}{4}(1-y/2)dy$
	$\displaystyle =\left[\frac{y}{4}-\frac{y^{2}}{16}\right]_{-1}^{1} = \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

c.

$P(\vert X+Y\vert<2)=1$

4.4

a.

$\displaystyle 1$	$\displaystyle = \int_{0}^{1}\int_{0}^{2}C(x+2y)dx dy = \int_{0}^{1}C(2+4y)dy$
	$\displaystyle = C(2+2)=4C$

b.

$\displaystyle f_{X}(x)$	$\displaystyle = \int_{0}^{1}\frac{1}{4}(x+2y)dy 1_{[0,2]}(x)$
	$\displaystyle = \frac{1}{4}(x+1) 1_{[0,2]}(x)$

c.

$\displaystyle F(x,y)$	$\displaystyle = \int_{0}^{y}\int_{0}^{x}\frac{1}{4}(u+2v)du dv$
	$\displaystyle = \int_{0}^{y}\frac{1}{4}(\frac{x^{2}}{2}+2vx)dv$
	$\displaystyle = \frac{1}{4}(\frac{x^{2}}{2}y+y^{2}x) = \frac{1}{8}x^{2}y+\frac{1}{4}y^{2}x$

for

and

d.

$\displaystyle Z$	$\displaystyle = 9/(X+1)^{2}$
$\displaystyle X$	$\displaystyle = 3/\sqrt{Z}-1$
$\displaystyle dx$	$\displaystyle = \frac{3}{2}\frac{1}{z^{3/2}}dz$
$\displaystyle f_{Z}(z)$	$\displaystyle = \frac{1}{4} \frac{3}{\sqrt{z}} \frac{3}{2}\frac{1}{z^{3/2}} = \frac{9}{8z^{2}}$

for $z \in [1,9]$ .

4.5

a.

$\displaystyle P(X > \sqrt{Y})$	$\displaystyle = P(Y < X^2) = \int_0^1 \int_0^{x^2} x+y dy dx$
	$\displaystyle = \int_0^1 x^3 + \frac{x^4}{2} dx = \frac{1}{4} + \frac{1}{10} = \frac{7}{20} = 0.35$

b.

$\displaystyle P(X^{2} < Y < X)$	$\displaystyle = \int_{0}^{1}\int_{x^2}^{x}2x dy dx$
	$\displaystyle = \int_{0}^{1}2x(x-x^{2})dx = \int_{0}^{1}2x^{2}-2x^{3}dx$
	$\displaystyle = \frac{2}{3}-\frac{2}{4} = \frac{1}{6}$

4.14

a.

$\displaystyle P(X^{2}+Y^{2}<1) = \int\int_{x^{2}+y^{2}<1}\frac{1}{2\pi}e^{-\frac{x^{2}+y^{2}}{2}}dx dy$

Can do $x=-\sqrt{1-y^{2}}$ to $x=+\sqrt{1-y^{2}}$ , or

$\displaystyle x$	$\displaystyle = r \cos \theta$	$\displaystyle dxdy$	$\displaystyle =r dr d\theta$
$\displaystyle y$	$\displaystyle = r \sin \theta$

for $0 \le r < 1$ and $0 \le \theta < 2\pi$ :

$\displaystyle P(X^{2}+Y^{2})$	$\displaystyle = \int_{0}^{1}\int_{0}^{2\pi}\frac{1}{2\pi}r e^{-r^{2}/2}d\theta dr$
	$\displaystyle = \left.-e^{-r^{2}/2}\right\vert _{0}^{1} = 1-e^{-1/2}$

b.