Systems 302
Tony E. Smith
SOLUTIONS TO PROBLEM SET 3
1.
(a) Let
i
x
= number of passengers in vehicle,
.
500
,...,
2
,
1
=
i
The
BLU estimator
for
the mean number of occupants,
)
(
X
E
=
µ
, is then given by:
500
1
1
1.778
500
i
i
xx
=
==
å
(b) Construct the Bernoulli random variable
y =
î
í
ì
1,
1
0,
x
otherwise
>
Then
BLU estimator
for
)
(
y
E
P
=
is given by
500
1
1
.36
500
i
i
yy
=
å
Partial Printout of JMPIN spreadsheet:
Rows
X
X_bar
Y
Y_bar
1
1
1.778
0
0.36
2
2
1.778
1
0.36
3
1
1.778
0
0.36
4
5
1.778
1
0.36
5
2
1.778
1
0.36
6
1
1.778
0
0.36
7
3
1.778
1
0.36
(c) If
j
X
= number of occupants in vehicle
j
= 1,2,&,12000 and
12000
1
,
j
j
TX
=
=
å
Then
( )
(
)
12000
( ),
j
j
E
TE
X
E
X
⋅
å
where
X
is occupancy of a randomly
sampled vehicle. Hence
12000
=⋅
is a linear unbiased estimator of ()
E T
.
Moreover, since
±
var(
)
var( )
X
<
µ
for any other linear unbiased estimator, ±
µ
, of
()
E X
implies that
±
var(12000
)
var(12000
)
X
⋅<
⋅
µ
, it follows by definition that
that
T
is the unique
BLU estimate
of ()
E T
. In the present case, the desired
estimate is
12000
21,366
tx
=
vehicles.
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For any estimator,
11
2 2
L
aX
=+
with
1
2
1
=
+
a
a
,
112 2
1
2
()
( )
( ) (
)
EL aEX
aEX
a a
=
+
µ = µ
So all three estimators are
unbiased
. This means that we need only compare their
variances to determine which is
most efficient
. But by the independence of the
random variables
1
X
and
2
X
:
22
1
1
2
2
var( )
var(
)
var(
)
L
L
a
X
a
X
Þ
12
(
/35)
(
/105)
aa
=σ
+
σ
So we must have:
(i)
2
1
var( )
105
L
σ
=
(ii)
2
2
2
11 9 1
var(
)
(
)
16 35
16 105
140
L
σ
⋅
+
⋅
=
(iii)
2
2
3
11 1 1
var(
)
(
)
4 35
4 105
105
L
σ
⋅
+ ⋅
=
Þ
2
L
is
best
(most
efficient
)
3.
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 Fall '08
 ese302
 Regression Analysis, BLU, WI, Mean squared error, Linear Unbiased Estimator

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