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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/28k/b/solution
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Past exam of the mathematics course of the University of Cambridge
/
2022
/
ii
/
Paper 3
/
28K
/
b
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-09-25
Since
X
n
∼
N
(
μ
,
1/
n
)
,
E
T
n
=
E
X
n
2
=
μ
2
+
n
1
,
(1)
so
Bias
(
T
n
)
=
n
1
.
(2)
Every leave-one-out
mean
has
variance
1/
(
n
−
1
)
, and hence
E
B
n
=
(
n
−
1
)
(
n
−
1
1
−
n
1
)
=
n
1
.
(3)
Therefore
Bias
(
T
JACK
)
=
0
.
(4)
Total
articles
:
1
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