Recovering components of a Poisson process

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Consider a scenario with two independent input Poisson processes
with rates $P_1 sim Pois(lambda_1)$ and $P_2 sim Pois(lambda_2)$.
These enter a queueing system to form a combined process of rate
$lambda_1 + lambda_2$ and then incur some additional exponentially-distributed service delay (say with mean $1/mu$).

If the queue is M/M/1 and $mu > lambda_1 + lambda_2$,
the output process should also be Poisson
with rate $lambda_1 + lambda_2$. If an observer of the output process observes just the samples corresponding to $P_1$ (resp. $P_2$), does she see a Poisson process with rate $lambda_1$ (resp. $lambda_2$)?
If not, what does this process look like?

stochastic-processes poisson-process queueing-theory

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asked Jul 15 at 7:31

user219923

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Consider a scenario with two independent input Poisson processes
with rates $P_1 sim Pois(lambda_1)$ and $P_2 sim Pois(lambda_2)$.
These enter a queueing system to form a combined process of rate
$lambda_1 + lambda_2$ and then incur some additional exponentially-distributed service delay (say with mean $1/mu$).

If the queue is M/M/1 and $mu > lambda_1 + lambda_2$,
the output process should also be Poisson
with rate $lambda_1 + lambda_2$. If an observer of the output process observes just the samples corresponding to $P_1$ (resp. $P_2$), does she see a Poisson process with rate $lambda_1$ (resp. $lambda_2$)?
If not, what does this process look like?

stochastic-processes poisson-process queueing-theory

share|cite|improve this question

asked Jul 15 at 7:31

user219923

507

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Consider a scenario with two independent input Poisson processes
with rates $P_1 sim Pois(lambda_1)$ and $P_2 sim Pois(lambda_2)$.
These enter a queueing system to form a combined process of rate
$lambda_1 + lambda_2$ and then incur some additional exponentially-distributed service delay (say with mean $1/mu$).

If the queue is M/M/1 and $mu > lambda_1 + lambda_2$,
the output process should also be Poisson
with rate $lambda_1 + lambda_2$. If an observer of the output process observes just the samples corresponding to $P_1$ (resp. $P_2$), does she see a Poisson process with rate $lambda_1$ (resp. $lambda_2$)?
If not, what does this process look like?

stochastic-processes poisson-process queueing-theory

share|cite|improve this question

asked Jul 15 at 7:31

user219923

507

Consider a scenario with two independent input Poisson processes
with rates $P_1 sim Pois(lambda_1)$ and $P_2 sim Pois(lambda_2)$.
These enter a queueing system to form a combined process of rate
$lambda_1 + lambda_2$ and then incur some additional exponentially-distributed service delay (say with mean $1/mu$).

If the queue is M/M/1 and $mu > lambda_1 + lambda_2$,
the output process should also be Poisson
with rate $lambda_1 + lambda_2$. If an observer of the output process observes just the samples corresponding to $P_1$ (resp. $P_2$), does she see a Poisson process with rate $lambda_1$ (resp. $lambda_2$)?
If not, what does this process look like?

stochastic-processes poisson-process queueing-theory

share|cite|improve this question

asked Jul 15 at 7:31

user219923

507

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 7:31

user219923

507

asked Jul 15 at 7:31

user219923

507

asked Jul 15 at 7:31

user219923

507

507

add a comment | 

add a comment | 

1 Answer
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accepted

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $lambda_1+lambda_2$ in which each customer is independently marked as type $i$ with probability $fraclambda_isum_ilambda_i$. The customers leave in the same order, so the output process is again a process with rate $lambda_1+lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).

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answered Jul 15 at 9:45

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
  – user219923
  Jul 15 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
  – joriki
  Jul 15 at 17:54
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

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votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $lambda_1+lambda_2$ in which each customer is independently marked as type $i$ with probability $fraclambda_isum_ilambda_i$. The customers leave in the same order, so the output process is again a process with rate $lambda_1+lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).

share|cite|improve this answer

answered Jul 15 at 9:45

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
  – user219923
  Jul 15 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
  – joriki
  Jul 15 at 17:54
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $lambda_1+lambda_2$ in which each customer is independently marked as type $i$ with probability $fraclambda_isum_ilambda_i$. The customers leave in the same order, so the output process is again a process with rate $lambda_1+lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).

share|cite|improve this answer

answered Jul 15 at 9:45

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
  – user219923
  Jul 15 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
  – joriki
  Jul 15 at 17:54
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $lambda_1+lambda_2$ in which each customer is independently marked as type $i$ with probability $fraclambda_isum_ilambda_i$. The customers leave in the same order, so the output process is again a process with rate $lambda_1+lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).

share|cite|improve this answer

answered Jul 15 at 9:45

joriki

165k10180328

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $lambda_1+lambda_2$ in which each customer is independently marked as type $i$ with probability $fraclambda_isum_ilambda_i$. The customers leave in the same order, so the output process is again a process with rate $lambda_1+lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).

share|cite|improve this answer

answered Jul 15 at 9:45

joriki

165k10180328

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 9:45

joriki

165k10180328

answered Jul 15 at 9:45

joriki

165k10180328

answered Jul 15 at 9:45

joriki

165k10180328

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
  – user219923
  Jul 15 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
  – joriki
  Jul 15 at 17:54
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
  – user219923
  Jul 15 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
  – joriki
  Jul 15 at 17:54
  
  

  
  
  
  

Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
– user219923
Jul 15 at 17:38

Thanks! What about the case where $mu leq lambda_1 + lambda_2$? Ignoring the fact that the queue explodes, can we say that the output sub-processes will have rate $mu(lambda_1/(lambda_1 + lambda_2))$, etc.?
– user219923
Jul 15 at 17:38

1

1

@user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
– joriki
Jul 15 at 17:54

@user219923: Yes -- I think the same argument applies -- they're still in the same order, so they still have independent markings with probabilities $fraclambda_isum_ilambda_i$, so if the overloaded queue emits them at rate $mu$, they constitute subprocesses with rates $fracmulambda_isum_ilambda_i$.
– joriki
Jul 15 at 17:54

add a comment | 

 

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