Problem 1
Suppose
X ∼ U nif (0, 1), Y ∼ U nif (0, 1) are independent
Z = X − Y . Be sure to
density function for the dierence,
uniform random variables in
(0, 1).
Find the probability
clearly explain ALL your work, in particular the domain
for all functions.
Solution:
First start with the cumulative distribution function:
F (z) = P (Z ≤ z) = P (X − Y ≤ z)
X −Y ≤ z . The region of integration is either
z < 0. (We also notice that the probability is P (X − Y ≤ z) = 0
and P (Z − Y ≤ z) = 1 if z > 1 since the entire square is covered)
You can think of this probability as the area of the square which has
a triangle or a pentagon depending on if
is
z < −1,
z>0
since none of the square is covered
or
See the diagram:
X-Y < z when z > 0
X-Y < z when z < 0
From the diagram one can compute the areas above by using the formula
should check that the width and height of the triangle are both
z+1
when
1
2
× Base × Height for a triangle. You
−1 < z < 0. The area of the pentagon
region is found by subtracting the area of the triangle from the area of the square.
1 − z when 0 < z < 1. Putting this all

0
z ≤ −1



 1 (z + 1)(z + 1)
−1 ≤ z ≤ 0
2
P (X − Y ≤ z) =
1
1 − 2 (1 − z) (1 − z) 0 ≤ z ≤ 1



1
z≥1

0
z ≤ −1



 1 (z + 1)2
−1 ≤ z ≤ 0
2
=
2
1

(1
−
z)
0≤z≤1
1
−

2


1
z≥1
triangular region missing from the pentagon is
The probability density function is the derivative of this:

0
z ≤ −1




(z + 1) −1 ≤ z ≤ 0
d
f (z) =
F (z) =

dz
(1 − z) 0 ≤ z ≤ 1



0
z≥1
This is a tent shaped function.
1
The base and height of the
together gives:
Problem 2
Suppose that
W ∼ Exp(λ) is an
U is chosen
the random variable
exponential random variable with parameter
to be uniform in the range
a) What is the joint density function
b) Find
f (w, u)
for
W
and
(0, W )
U ? Be sure
λ.
Conditionally on the value of
W,
to clearly indicate the domain.
E[U ].
Solution:
We interpret the following information from the problem statement (you needed to know the density function for a
uniform and an expoential random variable)
i) the density function for
W
is (this is because
fW (w)
ii) the conditional density of
U
given
W
W ∼ Exp(λ))
(
0
w≤0
=
−λw
λe
w≥0
is (this is because
(
fU |W (u |w ) =
1
w
0
U
is uniform in
[0, W ])
0<u<w
u ≥ w or u ≤ 0
The solution to part a) follows by the formula that the joint density function is the product of these
fW,U (w, u) = fU |W (u |w ) fW (w)
fU |W (u |w ) = fW,U (w, u)/fW (w)...these are just
(
λ −λw
e
w ≥ 0 and 0 < u < w
fW,U (w, u) = w
0
otherwise
(sometimes this is presented as
Thus we have:
rearrangements of each other)
To do part b) directly we can just integrate against the pdf:
ˆ ˆ
E [U ]
u · fW,U (w, u)dwdu
=
ˆ∞ ˆw
u·
=
0
λ −λw
e
dudw
w
0
ˆ∞ =
1 2 λ −λw
u e
2 w
0
w
dw
0
ˆ∞
=
=
=
λ −λw
we
dw
2
0
∞
λ 1 −λw
e
2 λ2
0
1
2λ
Here is another sneaky way to do this problem using the fact that
E [U ]
= E [E [U |W ]]
1
= E W
2
1
1
=
E [W ] =
2
2λ
2
E [U |W ] = 12 W
: