APPM 4/5520
Solutions to Problem Set Four
1.
(a)
E
[
S
2
2
] =
1
n
−
1
E
[
summationdisplay
(
X
i
−
X
)
2
]
Option One:
Square (
X
i
−
X
) and take the sum through:
summationdisplay
(
X
i
−
X
)
2
=
summationdisplay
(
X
2
i
−
2
X
+
X
2
)
=
∑
X
2
i
−
2
X
∑
X
i
+
n
X
2
=
∑
X
2
i
−
2
X
·
n
·
1
n
∑
X
i
+
n
X
2
=
∑
X
2
i
−
2
X
·
n
X
+
n
X
2
=
∑
X
2
i
−
2
n
X
2
+
n
X
2
=
∑
X
2
i
−
n
X
2
So
E
[
∑
(
X
i
−
X
)
2
]
=
E
[
∑
X
2
X
−
n
X
2
]
=
∑
E
[
X
2
i
]
−
n
E
[
X
2
]
=
n
·
E
[
X
2
1
]
−
n
E
[
X
2
]
Since
E
[
X
2
1
] =
V ar
[
X
1
] + (
E
[
X
1
])
2
=
σ
2
+
μ
2
and
E
[
X
2
] =
V ar
[
X
] +
parenleftBig
E
[
X
]
parenrightBig
2
=
σ
2
n
+
μ
2
we have
E
[
∑
(
X
i
−
X
)
2
]
=
n
(
σ
2
+
μ
2
)
−
n
parenleftBig
σ
2
n
+
μ
2
parenrightBig
=
nσ
2
−
σ
2
= (
n
−
1)
σ
2
.
Putting it all together:
E
[
S
2
2
] =
1
n
−
1
E
[
summationdisplay
(
X
i
−
X
)
2
] =
1
n
−
1
·
(
n
−
1)
σ
2
=
σ
2
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So
S
2
2
is an unbiased estimator of
σ
2
.
Option Two:
Square (
X
i
−
X
) and find the expectation but leave the sum outside of
the expected value:
E
[(
X
i
−
X
)
2
]
=
E
[
X
2
i
−
2
XX
i
+
X
2
]
=
E
[
X
2
i
]
−
2
E
[
XX
i
] +
E
[
X
2
]
As before, we have
E
[
X
2
i
] =
V ar
[
X
i
] + (
E
[
X
i
])
2
=
σ
2
+
μ
2
and
E
[
X
2
] =
V ar
[
X
] +
parenleftBig
E
[
X
]
parenrightBig
2
=
σ
2
n
+
μ
2
.
As for the middle term,
E
[
XX
i
]
=
E
[
X
i
X
]
E
bracketleftBig
X
i
1
n
∑
n
j
=1
X
j
bracketrightBig
=
1
n
∑
n
j
=1
E
[
X
i
X
j
]
When
j
negationslash
=
i
, this expectation is
E
[
X
i
X
j
]
indep
=
E
[
X
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 Fall '11
 Manuel
 Variance, Probability theory, Exponential distribution, CDF

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