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STP 421 - Problem Set 5 Solutions
1.) Let X and Y be jointly continuous random variables with joint density function
f (x, y) = c(y 2 − x2 )e−y ,
−y ≤ x ≤ y, 0 < y < ∞.
(a) Find c so that f is a density function.
(b) Find the marginal densities of X and Y .
(c) Find the expected value of X.
Solutions:
(a) The normalizing constant c is determined by the condition
Z
1 =
f (x, y)dxdy
R2
Z y
Z ∞
= c·
e−y
(y 2 − x2 )dx dy
0
−y
Z ∞
2 3
−y
3
= c·
e
2y − y dy
3
0
Z ∞
4
= c·
e−y y 3 dy
3 0
4
= c · Γ(4).
3
Since Γ(4) = 3! = 6, it follows that c = 18 .
(b) The marginal density of X at x ∈ R is
Z
fX (x) =
f (x, y)dy
R
Z
1 ∞ 2
=
y − x2 e−y dy
8 |x|
Z ∞
2 −z−|x|
1
2 −|x|
=
z + |x| e
dz − x e
8 0
1
=
1 + |x| e−|x| ,
4
while the marginal density of Y at y ∈ (0, ∞) is
Z
fY (y) =
f (x, y)dx
R
Z y
1
y 2 − x2 e−y dx
=
8 −y
1 −y
2 3
3
=
e
2y − y
8
3
1 3 −y
=
y e ,
6
1
i.e., the marginal distribution of Y is that of a Gamma distributed random variable with parameters α = 3 and λ = 1.
(c) The expected value of X is
Z
∞
xfX (x)dx = 0,
E[X] =
−∞
since x · fX (x) is an odd function.
2.) Let X and Y be independent standard uniform random variables and let a, b and c be
positive real numbers. Find the probability that aX + bY ≤ c.
Solution:
Because (X, Y ) is uniformly distributed on the unit square [0, 1] × [0, 1], the probability of an
event {(X, Y ) ⊂ A}, where A is a subset of the unit square, is equal to the area of A. In this
problem, the set A is the intersection of the region beneath the line y = cb − ab x and the unit
square. There are five cases:


1
if c ≥ a + b



c−b
c−a
1


1− b
if a + b ≥ c ≥ max{a, b}
1 − 2 1 − a
c−b
1
b
P(aX + bY ≤ c) =
if a ≥ c ≥ b
a + 2a


1
a
c−a

if b ≥ c ≥ a


b + 2b

 c2
if min{a, b} ≥ c
2ab
3.) Show that if X and Y are jointly continuous, then X + Y is a continuous random variable
while X, Y and X + Y are not jointly continuous.
Solution:
Suppose that X and Y are jointly continuous with joint density fX,Y (x, y). Then the cumulative
distribution function of X + Y is equal to
Z ∞
Z t−x
P{X + Y ≤ t} =
dx
dyfX,Y (x, y).
−∞
−∞
Since this function is differentiable with respect to t, it follows that X + Y is a continuous
random variable with density
Z ∞
pX+Y (t) =
dxfX,Y (x, t − x).
−∞
2
To show that X, Y , and X + Y are not jointly continuous, let A = {(x, y, z) : z − x − y = 0}
and note that P((X, Y, X + Y ) ∈ A) = 1. If these variables were jointly continuous, then there
would exists a density function p : R3 → [0, ∞] such that
Z
p(x, y, z)dxdydz
P(E) =
E
for every measurable subset E ⊂
have
R3 .
However, since A is a two-dimensional subspace of R3 , we
Z
f (x, y, z)dxdydz = 0
A
for every measurable function f : R → [0, ∞]. This shows that X, Y , and X + Y do not have a
joint density and therefore are not jointly continuous.
4.) Suppose that X1 , · · · , Xn are independent exponential random variables with parameters
λ1 , · · · , λn . Find the distribution of Y = min{X1 , · · · , Xn }. Hint: Calculate the probability
P(Y > t).
Solution:
Since Y > t if and only if Xi > t for all i = 1, · · · , n, it follows that
P(Y > t) = P(X1 > t, · · · , Xn > t)
n
Y
=
P(Xi > t)
(by independence)
i=1
=
n
Y
e−λi t
i=1
(
= exp −t
n
X
)
λi
,
i=1
which shows that Y is exponentially distributed with rate parameter
3
Pn
i=1 λi .