Question related to Poisson process
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Suppose busses arrive at a bus stop either with an inter-arrival time of exactly 1 min or with an inter-arrival time of exactly 10 mins. Suppose the 1 min inter-arrival times occur with probability 2/3 and the 10 min inter-arrival times occur with probability 1/3. What is the average inter-arrival time between bus arrivals? .What is the probability that a person, arriving uniformly at random, lands in the 10 min inter-arrival time?
Answers to questions are 4 and 5/6 respectively.
For the first one I tried solving it this way: either it is 10 min interval or 1 min interval so multiplied each one with their probabilities, adding them up and then taking reciprocal (as the average in exponential is 1/lambda). But that seems to be wrong. And for secondly I have no clue
probability probability-theory poisson-process
asked Dec 14 '18 at 10:51
pufflespuffles
669
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add a comment |
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Suppose busses arrive at a bus stop either with an inter-arrival time of exactly 1 min or with an inter-arrival time of exactly 10 mins. Suppose the 1 min inter-arrival times occur with probability 2/3 and the 10 min inter-arrival times occur with probability 1/3. What is the average inter-arrival time between bus arrivals? .What is the probability that a person, arriving uniformly at random, lands in the 10 min inter-arrival time?
Answers to questions are 4 and 5/6 respectively.
For the first one I tried solving it this way: either it is 10 min interval or 1 min interval so multiplied each one with their probabilities, adding them up and then taking reciprocal (as the average in exponential is 1/lambda). But that seems to be wrong. And for secondly I have no clue
probability probability-theory poisson-process
asked Dec 14 '18 at 10:51
pufflespuffles
669
$endgroup$
$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
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– Did
Dec 15 '18 at 6:45
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I dont understand your intuition. Can you please rephrase it
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– puffles
Dec 15 '18 at 7:10
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@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
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I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05
add a comment |
$begingroup$
Suppose busses arrive at a bus stop either with an inter-arrival time of exactly 1 min or with an inter-arrival time of exactly 10 mins. Suppose the 1 min inter-arrival times occur with probability 2/3 and the 10 min inter-arrival times occur with probability 1/3. What is the average inter-arrival time between bus arrivals? .What is the probability that a person, arriving uniformly at random, lands in the 10 min inter-arrival time?
Answers to questions are 4 and 5/6 respectively.
For the first one I tried solving it this way: either it is 10 min interval or 1 min interval so multiplied each one with their probabilities, adding them up and then taking reciprocal (as the average in exponential is 1/lambda). But that seems to be wrong. And for secondly I have no clue
probability probability-theory poisson-process
asked Dec 14 '18 at 10:51
pufflespuffles
669
$endgroup$
Suppose busses arrive at a bus stop either with an inter-arrival time of exactly 1 min or with an inter-arrival time of exactly 10 mins. Suppose the 1 min inter-arrival times occur with probability 2/3 and the 10 min inter-arrival times occur with probability 1/3. What is the average inter-arrival time between bus arrivals? .What is the probability that a person, arriving uniformly at random, lands in the 10 min inter-arrival time?
Answers to questions are 4 and 5/6 respectively.
For the first one I tried solving it this way: either it is 10 min interval or 1 min interval so multiplied each one with their probabilities, adding them up and then taking reciprocal (as the average in exponential is 1/lambda). But that seems to be wrong. And for secondly I have no clue
probability probability-theory poisson-process
probability probability-theory poisson-process
asked Dec 14 '18 at 10:51
pufflespuffles
669
asked Dec 14 '18 at 10:51
pufflespuffles
669
edited Dec 15 '18 at 4:57
puffles
asked Dec 14 '18 at 10:51
pufflespuffles
669
asked Dec 14 '18 at 10:51
pufflespuffles
669
asked Dec 14 '18 at 10:51
pufflespuffles
669
669
$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
$endgroup$
– Did
Dec 15 '18 at 6:45
$begingroup$
I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:10
$begingroup$
@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
$begingroup$
I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05
add a comment |
$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
$endgroup$
– Did
Dec 15 '18 at 6:45
$begingroup$
I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:10
$begingroup$
@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
$begingroup$
I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05
$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
$endgroup$
– Did
Dec 15 '18 at 6:45
$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
$endgroup$
– Did
Dec 15 '18 at 6:45
$begingroup$
I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:10
$begingroup$
I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:10
$begingroup$
@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
$begingroup$
@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
$begingroup$
I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05
$begingroup$
I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05
add a comment |
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$begingroup$
After $3n$ busses passed, if $n$ is large, by the LLN, roughly $2n$ corresponded to 1 min inter-arrival times and $n$ to 10 min inter-arrival times, hence the mean inter-arrival time is roughly $(2ncdot1+ncdot10)/(3n)=4$. Likewise, after $3n$ busses passed, if $n$ is large, by the LLN, roughly $12n$ min passed, during which roughly $2n$ minutes corresponded to 1 min inter-arrival times, and roughly $10n$ minutes corresponded to 10 min inter-arrival times, thus the probability of a 10 min interval is roughly $10n/(12n)=5/6$.
$endgroup$
– Did
Dec 15 '18 at 6:45
$begingroup$
I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:10
$begingroup$
@Did I dont understand your intuition. Can you please rephrase it
$endgroup$
– puffles
Dec 15 '18 at 7:18
$begingroup$
I dont understand what you fail to get in the precise proof (not intuition, thanks) in my first comment. Can you please phrase what you are missing here?
$endgroup$
– Did
Dec 15 '18 at 10:05