Columbia University
IEOR 4404: Simulation
Fall 2009
Solutions to Assignment 5
1. (Exercise 5.14)
(a) Consider
X
with distribution
G
. Since the denominator of the frac
tion is equal to
G
(
b
)

G
(
a
), we can easily guess that the infor
mation we have to use in order to write
F
as a conditional dis
tribution is
a
≤
X
≤
b
.
Let us prove it.
Assume therefore that
F
(
x
) =
P
(
X
≤
x

a
≤
X
≤
b
) and prove that
F
(
x
) =
G
(
x
)
−
G
(
a
)
G
(
b
)
−
G
(
a
)
:
F
(
x
)
=
P
(
X
≤
x

a
≤
X
≤
b
)
=
P
(
X
≤
x

a
≤
X
≤
b
)
P
(
a
≤
X
≤
b
)
P
(
a
≤
X
≤
b
)
=
P
(
X
≤
x, a
≤
X
≤
b
)
P
(
a
≤
X
≤
b
)
using Bayes rule
=
P
(
a
≤
X
≤
x
)
P
(
a
≤
X
≤
b
)
since
x
∈
[
a, b
]
=
G
(
x
)

G
(
a
)
G
(
b
)

G
(
a
)
Therefore, we proved that
F
can be written as
F
(
x
) =
P
(
X
≤
x

a
≤
X
≤
b
).
(b) It is important to notice that the distribution
F
is defined only on the
interval [
a, b
]. Therefore, by differentiating
F
, we obtain the following
density:
f
(
x
) =
F
′
(
x
) =
g
(
x
)
G
(
b
)

G
(
a
)
1
[
a,b
]
(
x
)
.
Now, we have to find an upper bound
c
for the ratio
f
(
y
)
g
(
y
)
for all
y
∈
[
a, b
].
Given the definition of
f
above, it is easy to see that
c
=
1
G
(
b
)
−
G
(
a
)
works. Therefore, we are able to apply the usual ac
ceptance/rejection algorithm to generate
Y
with distribution
F
:
STEP 1. Generate a random variable
X
with distribution
G
.
STEP 2. Generate
U
∼
Uniform[0
,
1]
STEP 3. If
U
≤
f
(
X
)
c g
(
X
)
, then set Y=X. Otherwise, go back to STEP 1.
By running the algorithm above, we know that
Y
will have the de
sired distribution
F
. Nonetheless, the question asks us to show that
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the following modified version of the usual acceptance/rejection al
gorithm also provides
Y
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 Spring '10
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 Normal Distribution, Poisson Distribution, Probability theory, Exponential distribution

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