stopping time probability and expectation
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A broken clock moves the minute hand each minute randomly one minute ahead and one minute behind with equal probabilities. Each minute it does so independently of all other times it moved the minute hand.
The hour hand is on the other synchronized with the minute hand. The clock starts running at noon and let T denotes the first time that the clock shows either 1 o'clock or 11 o'clock, whatever happens first. Compute E(T).
I need some advice/suggestion on doing this problem.
probability
edited Dec 4 '18 at 18:04
amWhy
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
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add a comment |
$begingroup$
A broken clock moves the minute hand each minute randomly one minute ahead and one minute behind with equal probabilities. Each minute it does so independently of all other times it moved the minute hand.
The hour hand is on the other synchronized with the minute hand. The clock starts running at noon and let T denotes the first time that the clock shows either 1 o'clock or 11 o'clock, whatever happens first. Compute E(T).
I need some advice/suggestion on doing this problem.
probability
edited Dec 4 '18 at 18:04
amWhy
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
$endgroup$
1
$begingroup$
And we need your thoughts on it, to help you.
$endgroup$
– Did
Nov 20 '13 at 9:45
add a comment |
$begingroup$
A broken clock moves the minute hand each minute randomly one minute ahead and one minute behind with equal probabilities. Each minute it does so independently of all other times it moved the minute hand.
The hour hand is on the other synchronized with the minute hand. The clock starts running at noon and let T denotes the first time that the clock shows either 1 o'clock or 11 o'clock, whatever happens first. Compute E(T).
I need some advice/suggestion on doing this problem.
probability
edited Dec 4 '18 at 18:04
amWhy
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
$endgroup$
A broken clock moves the minute hand each minute randomly one minute ahead and one minute behind with equal probabilities. Each minute it does so independently of all other times it moved the minute hand.
The hour hand is on the other synchronized with the minute hand. The clock starts running at noon and let T denotes the first time that the clock shows either 1 o'clock or 11 o'clock, whatever happens first. Compute E(T).
I need some advice/suggestion on doing this problem.
probability
probability
edited Dec 4 '18 at 18:04
amWhy
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
edited Dec 4 '18 at 18:04
amWhy
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
edited Dec 4 '18 at 18:04
amWhy
1
edited Dec 4 '18 at 18:04
amWhy
1
edited Dec 4 '18 at 18:04
amWhy
1
1
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
asked Nov 20 '13 at 7:33
Bob ramerizBob rameriz
121
121
1
$begingroup$
And we need your thoughts on it, to help you.
$endgroup$
– Did
Nov 20 '13 at 9:45
add a comment |
1
$begingroup$
And we need your thoughts on it, to help you.
$endgroup$
– Did
Nov 20 '13 at 9:45
1
1
$begingroup$
And we need your thoughts on it, to help you.
$endgroup$
– Did
Nov 20 '13 at 9:45
$begingroup$
And we need your thoughts on it, to help you.
$endgroup$
– Did
Nov 20 '13 at 9:45
add a comment |
1 Answer
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This question is old, but it can be solved by treating the clock's advance or retreat as a random walk starting at $0$, and absorbing barriers at $-60$ and $60$. Then the expected time to absorption is $60 times 60 = 3600$. The analysis can be found in various resources; see, for example, http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
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$begingroup$
This question is old, but it can be solved by treating the clock's advance or retreat as a random walk starting at $0$, and absorbing barriers at $-60$ and $60$. Then the expected time to absorption is $60 times 60 = 3600$. The analysis can be found in various resources; see, for example, http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
$endgroup$
add a comment |
$begingroup$
This question is old, but it can be solved by treating the clock's advance or retreat as a random walk starting at $0$, and absorbing barriers at $-60$ and $60$. Then the expected time to absorption is $60 times 60 = 3600$. The analysis can be found in various resources; see, for example, http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
$endgroup$
add a comment |
$begingroup$
This question is old, but it can be solved by treating the clock's advance or retreat as a random walk starting at $0$, and absorbing barriers at $-60$ and $60$. Then the expected time to absorption is $60 times 60 = 3600$. The analysis can be found in various resources; see, for example, http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
$endgroup$
This question is old, but it can be solved by treating the clock's advance or retreat as a random walk starting at $0$, and absorbing barriers at $-60$ and $60$. Then the expected time to absorption is $60 times 60 = 3600$. The analysis can be found in various resources; see, for example, http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
answered Dec 4 '18 at 20:52
Brian TungBrian Tung
25.7k32554
25.7k32554
add a comment |
add a comment |
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And we need your thoughts on it, to help you.
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– Did
Nov 20 '13 at 9:45