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Poisson process that degrades uniformly

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1

$begingroup$

Given that people arrive according to a poisson process with rate 4 per minute and stay for $X$~Unif[0,10] number of minutes independently of the arrival times. What is the mean and variance of the number of people after 15 minutes.

I'm attempting this problem. First I noticed that anything before t=5 doesnt matter because anyone would have left who arrived before that. Then the expectation of anyone who arrives in that time period would be $lambda10 = (4)(10) = 40$ and they all stay with for expected time of 5 minutes. So I would expect that that the epectation would be 20.

I am the same sort of of shotty reasoning for variance which would just be var(possion)*var(X) = $(4^2)(frac10012) = frac4003$.

However I'm having alot of trouble trying to show any of this formally

probability poisson-process

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asked yesterday

Papa_fernPapa_fern

214

$endgroup$

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1

$begingroup$

Given that people arrive according to a poisson process with rate 4 per minute and stay for $X$~Unif[0,10] number of minutes independently of the arrival times. What is the mean and variance of the number of people after 15 minutes.

I'm attempting this problem. First I noticed that anything before t=5 doesnt matter because anyone would have left who arrived before that. Then the expectation of anyone who arrives in that time period would be $lambda10 = (4)(10) = 40$ and they all stay with for expected time of 5 minutes. So I would expect that that the epectation would be 20.

I am the same sort of of shotty reasoning for variance which would just be var(possion)*var(X) = $(4^2)(frac10012) = frac4003$.

However I'm having alot of trouble trying to show any of this formally

probability poisson-process

share|cite|improve this question

asked yesterday

Papa_fernPapa_fern

214

$endgroup$

add a comment | 

$begingroup$

Given that people arrive according to a poisson process with rate 4 per minute and stay for $X$~Unif[0,10] number of minutes independently of the arrival times. What is the mean and variance of the number of people after 15 minutes.

I'm attempting this problem. First I noticed that anything before t=5 doesnt matter because anyone would have left who arrived before that. Then the expectation of anyone who arrives in that time period would be $lambda10 = (4)(10) = 40$ and they all stay with for expected time of 5 minutes. So I would expect that that the epectation would be 20.

I am the same sort of of shotty reasoning for variance which would just be var(possion)*var(X) = $(4^2)(frac10012) = frac4003$.

However I'm having alot of trouble trying to show any of this formally

probability poisson-process

share|cite|improve this question

asked yesterday

Papa_fernPapa_fern

214

$endgroup$

share|cite|improve this question

asked yesterday

Papa_fernPapa_fern

214

share|cite|improve this question

asked yesterday

Papa_fernPapa_fern

214

asked yesterday

Papa_fernPapa_fern

214

asked yesterday

Papa_fernPapa_fern

214

1 Answer
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1

$begingroup$

Consulting this answer: Poisson process, number of people in store by time $t$

We are only interested in the distribution of $M(t)$, so we compute this from the joint distribution of $M(t)$ and $N(t)$ by summing over all possible values of $N(t)$:

beginalign
mathbb P(M(t)=m) &= sum_n=0^infty mathbb P(M(t)=m, N(t)=n)\
&= sum _n=0^infty frace^-lambda t (lambda (1-p) t)^m (lambda p t)^nm! n!\
&= frac ((1-p)lambda t)^mm!e^-(1-p)lambda t.
endalign

More generally, let $Xsim mathrmUnif(0,theta)$, then we compute

$$
p = int_0^t frac utheta mathsf du = frac t2theta,
$$
so that
$$
mathbb P(M(t)=m) = frac((1-t/2theta)lambda t)^mm!e^-(1-t/2theta)lambda t.
$$
This is a Poisson distribution with parameter $mu:=(1-t/2theta)lambda t$. The mean and variance are both given by $mu$, so plugging in $lambda=4$, $theta=10$, and $t=15$ we have $15$.

share|cite|improve this answer

answered 22 hours ago

Math1000Math1000

19.3k31745

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1

$begingroup$

Consulting this answer: Poisson process, number of people in store by time $t$

We are only interested in the distribution of $M(t)$, so we compute this from the joint distribution of $M(t)$ and $N(t)$ by summing over all possible values of $N(t)$:

beginalign
mathbb P(M(t)=m) &= sum_n=0^infty mathbb P(M(t)=m, N(t)=n)\
&= sum _n=0^infty frace^-lambda t (lambda (1-p) t)^m (lambda p t)^nm! n!\
&= frac ((1-p)lambda t)^mm!e^-(1-p)lambda t.
endalign

More generally, let $Xsim mathrmUnif(0,theta)$, then we compute

$$
p = int_0^t frac utheta mathsf du = frac t2theta,
$$
so that
$$
mathbb P(M(t)=m) = frac((1-t/2theta)lambda t)^mm!e^-(1-t/2theta)lambda t.
$$
This is a Poisson distribution with parameter $mu:=(1-t/2theta)lambda t$. The mean and variance are both given by $mu$, so plugging in $lambda=4$, $theta=10$, and $t=15$ we have $15$.

share|cite|improve this answer

answered 22 hours ago

Math1000Math1000

19.3k31745

$endgroup$

add a comment | 

1

$begingroup$

Consulting this answer: Poisson process, number of people in store by time $t$

We are only interested in the distribution of $M(t)$, so we compute this from the joint distribution of $M(t)$ and $N(t)$ by summing over all possible values of $N(t)$:

beginalign
mathbb P(M(t)=m) &= sum_n=0^infty mathbb P(M(t)=m, N(t)=n)\
&= sum _n=0^infty frace^-lambda t (lambda (1-p) t)^m (lambda p t)^nm! n!\
&= frac ((1-p)lambda t)^mm!e^-(1-p)lambda t.
endalign

More generally, let $Xsim mathrmUnif(0,theta)$, then we compute

$$
p = int_0^t frac utheta mathsf du = frac t2theta,
$$
so that
$$
mathbb P(M(t)=m) = frac((1-t/2theta)lambda t)^mm!e^-(1-t/2theta)lambda t.
$$
This is a Poisson distribution with parameter $mu:=(1-t/2theta)lambda t$. The mean and variance are both given by $mu$, so plugging in $lambda=4$, $theta=10$, and $t=15$ we have $15$.

share|cite|improve this answer

answered 22 hours ago

Math1000Math1000

19.3k31745

$endgroup$

add a comment | 

$begingroup$

Consulting this answer: Poisson process, number of people in store by time $t$

We are only interested in the distribution of $M(t)$, so we compute this from the joint distribution of $M(t)$ and $N(t)$ by summing over all possible values of $N(t)$:

beginalign
mathbb P(M(t)=m) &= sum_n=0^infty mathbb P(M(t)=m, N(t)=n)\
&= sum _n=0^infty frace^-lambda t (lambda (1-p) t)^m (lambda p t)^nm! n!\
&= frac ((1-p)lambda t)^mm!e^-(1-p)lambda t.
endalign

More generally, let $Xsim mathrmUnif(0,theta)$, then we compute

$$
p = int_0^t frac utheta mathsf du = frac t2theta,
$$
so that
$$
mathbb P(M(t)=m) = frac((1-t/2theta)lambda t)^mm!e^-(1-t/2theta)lambda t.
$$
This is a Poisson distribution with parameter $mu:=(1-t/2theta)lambda t$. The mean and variance are both given by $mu$, so plugging in $lambda=4$, $theta=10$, and $t=15$ we have $15$.

share|cite|improve this answer

answered 22 hours ago

Math1000Math1000

19.3k31745

$endgroup$

share|cite|improve this answer

answered 22 hours ago

Math1000Math1000

19.3k31745

answered 22 hours ago

Math1000Math1000

19.3k31745

answered 22 hours ago

Math1000Math1000

19.3k31745