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1
Massachusetts
Institute
of
Technology
Logistical
and
Transportation
Planning
Methods
Fall
2003
Solutions
to
quiz
1
Prepared
by
Margr´
et
Vilborg
Bjarnad´
ottir
(Bjarnad´
ottir,
2003,
(Outline
Kang,
2001))
X
1
, X
2
are
uniformly
distributed
between
0
and
a.
Let
G
(
a
)
∗
E
[
max
(
x
1
, x
2
)
3
]
and
consider
G
(
a
+
)
when
X
1
, X
2
are
uniformly
distributed
between
0
and
a
+
,
where
is
very
small.
Suppose
a
< X
2
≈
a
+
and
0
≈
X
1
≈
a
.
Then
we
know
that
max
(
x
1
, x
2
) =
x
2
.
There
fore
E
[
max
(
x
1
, x
2
)
3
]
∗
E
[
x
2
3
].
Since
X
1
and
X
2
are
independent,
G
(
a
+
)
for
this
case
can
be
computed
as
follows:
a
+
G
(
a
+
) =
E
[
max
(
x
1
, x
2
)
3
] =
E
[
x
2
3
] =
(
x
2
)
3
f
X
2
(
x
2
)
dx
2
,
a
where
f
X
2
(
x
2
)
is
the
probability
density
function
of
X
2
.
Because
X
2
is
uniformly
distributed
over
(
a,
a
+
],
f
X
2
(
x
2
) =
a
1
.
Thus,
1
a
+
G
(
a
+
) =
(
x
2
)
3
dx
2
a
±
²
a
+
1
1
4
=
x
4
2
a
=
1
·
1
³
(
a
+
)
4
−
a
4
´
4
=
1
·
1
³
4
a
3
+ 6
a
2
2
+ 4
a
3
+
4
´
4
1
1
³
3
´
=
·
(4
a
+
o
( )
,
4
where
o
( )
represents
higher
order
terms
of
satisfying
lim
0
o
( )
=
0
(“pathetic
terms”).
There
fore,
G
(
a
+
)
a
3
as
±
0.
By
symmetry
we
have
G
(
a
+
)
a
3
as
±
0
when
0
≈
X
2
≈
a
and
a < X
1
≈
a
+
.
Finally,
we
do
not
have
to
compute
G
(
a
+
)
for
the
case
where
a < X
1
≈
a
+
and
a < X
2
≈
a
+
because
the
associated
probability
is
negligible.
Page
1
of
5
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Massachusetts
Institute
of
Technology
Logistical
and
Transportation
Planning
Methods
Fall
2003
The
following
table
summarizes
G
(
a
+
)’s.
Case
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This note was uploaded on 11/08/2011 for the course AERO 16.72 taught by Professor Hansman during the Fall '06 term at MIT.
 Fall '06
 hansman

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