Solutions to Homework Set No. 2 – Probability Theory (235A), Fall 2011
1.
(a) Let
X
be a random variable with distribution function
F
X
and piecewise continuous
density function
f
X
. Let [
a, b
]
⊂
R
be an interval (possibly infinite) such that
P
(
X
∈
[
a, b
]) = 1
,
and let
g
: [
a, b
]
→
R
be a monotone (strictly) increasing and differentiable function. Prove
that the random variable
Y
=
g
(
X
) (this is the function on Ω defined by
Y
(
ω
) =
g
(
X
(
ω
)),
in other words the composition of the two functions
g
and
X
) has density function
f
Y
(
x
) =
f
X
(
g

1
(
x
))
g
0
(
g

1
(
x
))
x
∈
(
g
(
a
)
, g
(
b
))
,
0
otherwise.
Solution.
It follows that
P
(
Y
∈
[
g
(
a
)
, g
(
b
)]) = 1 from
P
(
X
∈
[
a, b
]) = 1 and the mono
tonicity and continuity of
g
. Since
g
is strictly increasing and continuous,
F
Y
(
x
) =
P
(
Y
≤
x
) =
P
(
X
≤
g

1
(
x
)) =
F
X
(
g

1
(
x
))
and
f
Y
(
x
) =
d
dy
F
Y
(
x
) =
d
dy
F
X
(
g

1
(
x
))
for
x
∈
(
g
(
a
)
, g
(
b
)). Thus, by the chain rule,
f
Y
(
x
) =
f
X
(
g

1
(
x
))
g
0
(
g

1
(
x
))
(b) If
λ >
0, we say that a random variable has the exponential distribution with parameter
λ
if
F
X
(
x
) =
0
x <
0
,
1

e

λx
x
≥
0
,
and denote this
X
∼
Exp(
λ
)
.
Find an algorithm to produce a random variable with Exp(
λ
)
distribution using a random number generator that produces uniform random numbers in
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(0
,
1). In other words, if
U
∼
U
(0
,
1), find a function
g
: (0
,
1)
→
R
such that the random
variable
X
=
g
(
U
) has distribution Exp(
λ
).
Solution.
The density function for
X
is
f
X
=
λe

λx
. Use the formula in part (a), find
g
such that
1
g
0
(
g

1
(
x
))
=
λe

λx
.
You should obtain
g
(
x
) =

ln(1

x
)
/λ.
Now note that if
U
is uniform in (0
,
1) then so is
1

U
, so this
g
can be replaced by the slightly simpler
g
(
x
) =

1
λ
ln
x
.
(c) We say that a nonnegative random variable
X
≥
0 has the
lack of memory property
if it satisfies that
P
(
X
≥
t

X
≥
s
) =
P
(
X
≥
t

s
)
for all 0
< s < t.
Prove that exponential random variables have the lack of memory property.
Solution.
Let
X
∼
Exp(
λ
)
.
P
(
X
≥
t

X
≥
s
) =
P
(
X
≥
t
)
P
(
X
≥
s
)
=
e

λt
e

λs
=
e

λ
(
t

s
)
=
P
(
X
≥
t

s
)
(d) Prove that any nonnegative random variable that has the lack of memory property has
the exponential distribution with some parameter
λ >
0. (This is easier if one assumes that
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 DanRomik
 Probability, Probability theory

Click to edit the document details