8.1. The transformation r(X) = cX + d is strictly increasing. Its inverse
is s(Y ) = (Y − d)/c. Derivative of s(y) is 1/c. Consequently, density of Y is
f ((y − d)/c)/c.
n−1
8.2. Inverse of r(x) = xn is s(y) = y 1/n . Its derivative is y − n , hence
n−1
density of Y is y − n for y ∈ (0, 1) and zero otherwise.
8.3. Let us find the cumulative distribution
function of |X|.
Ra
P (|X| ≤ a) = P (−a ≤ X ≤ a) = −a f (x) dx for a ∈ [0, 1]. Therefore, the
density function of |X| is
Z x
d
f (t) dt = f (x) + f (−x).
dx −x
Consequently, density ofRR|X| is f (x) + f (−x) for x ∈ [0, 1] and zero otherwise.
8.4. We must have R2 f (x, y) dx dy = 1, hence
Z
1
Z
1
1
Z
c(x+y) dx dy = c
0
0
0
hence c = 1. Then
Z
P (X < 1/2) =
1/2
Z
x=1
x2 /2 + xy x=0 dy = c
1
Z
x + y dy dx =
0
0
Z
1
1/2+y dy = 1/2+1/2 = c,
0
1/2
x + 1/2 dx = 1/8 + 1/4 = 3/8.
0
8.5. P (XY ≤ z) is equal to the area of the set {(x, y) : 0 < x < 1, 0 <
y < 1, xy < z}. It is easier to find the area of the complement P (XY ≥ z) =
{(x, y) : 0 < x < 1, 0 < y < 1, xy ≥ z} in the unit square. It is equal to
Z
z
1
1
(1 − z/x) dx = (x − z ln x)|z = 1 − z + z ln z.
Consequently, P (XY < z) = z − z ln z = z(1 − ln z).
8.6. Joint distribution function is, by definition, P (X ≤ x, Y ≤ y). In this
case
P (X ≤ x, Y ≤ y) = P (X ≤ x, X ≤ y) = P (X ≤ min(x, y)) = min(x, y),
if min(x, y) ∈ [0, 1]. If min(x, y) > 1, then the probability is equal to 1, and if
min(x, y) < 0, it is equal to 0.
The function min(x, y) is not differentiable, hence there is no joint density.
1