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Past exam of the mathematics course of the University of Cambridge
/
2022
/
ia
/
Paper 2
/
4F
/
Solution
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2022
ia
Paper 2
4F
Created
2026-09-23
Updated
2026-09-25
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Expanding around the
expected value
μ
,
G
(
a
)
=
E
[(
X
−
μ
+
μ
−
a
)
2
]
=
E
[(
X
−
μ
)
2
]
+
(
μ
−
a
)
2
,
(1)
because
E
[
X
−
μ
]
=
0
. Therefore
G
(
a
)
=
σ
2
+
(
μ
−
a
)
2
≥
σ
2
,
(2)
with equality exactly when
a
=
μ
.
For the absolute loss,
H
(
a
)
=
∫
−
∞
a
(
a
−
x
)
f
(
x
)
d
x
+
∫
a
∞
(
x
−
a
)
f
(
x
)
d
x
.
(3)
Leibniz differentiation gives, with
F
(
a
)
=
∫
−
∞
a
f
(
x
)
d
x
,
H
′
(
a
)
=
F
(
a
)
−
(
1
−
F
(
a
))
=
2
F
(
a
)
−
1.
(4)
Thus
H
decreases while
F
(
a
)
<
1/2
and increases while
F
(
a
)
>
1/2
. It is minimized at any
median
, characterized in the continuous case by
∫
−
∞
a
f
(
x
)
d
x
=
2
1
.
(5)
Ancestors
(10)
4F
Paper 2
ia
2022
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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