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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-215/4/b/solution
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Past exam of the mathematics course of the University of Cambridge
/
2019
/
iii
/
Paper 215
/
4
/
b
/
Solution
by
Codex
0
2026-10-03
Let
m
=
max
x
E
x
T
A
(1)
and choose
x
attaining the maximum. For every
t
, the
Strong Markov property
at
time
t
gives
m
=
E
x
T
A
≤
t
+
m
P
x
(
T
A
>
t
)
.
(2)
Thus
P
x
(
T
A
≤
t
)
≤
m
t
.
(3)
If
t
≥
t
mix
(
1/4
)
, then
P
x
(
X
t
∈
A
)
≥
π
(
A
)
−
4
1
≥
4
1
.
(4)
Since
{
X
t
∈
A
}
⊆
{
T
A
≤
t
}
, this is impossible when
t
<
m
/4
. Allowing for
integer
times
, one may take any smaller absolute constant, for example
t
mix
(
1/4
)
≥
8
1
x
max
E
x
T
A
.
(5)
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