sum of independant poisson processes help

#1
HI, I've a question that says accidents to people and animals are independant poisson processes with intensity lambda and mu respectively. I'm asked given that the total amount of accident last year was 4, what is the distribution of no. of accidents to people? I'm not quite sure how to approach this, I know the total no.of accidents has distribution poisson(Lambda + mu), how do I obtain the distribution of accidents to people? Would I use a mark, say Mi = 1 if people, 0 if animals, and probability of Mi=1 equals to 0.5 so accidents to people is poisson lambdax0.5? Thanks for any help
 

BGM

TS Contributor
#4
Would I use a mark, say Mi = 1 if people, 0 if animals, and probability of Mi=1 equals to 0.5 so accidents to people is poisson lambdax0.5?
Yes you need to calculate the probability but it is not 0.5. I can give you the idea to calculate this probability.

Note that you have 2 independent Poisson process, so you have the corresponding 2 independent exponentially distributed inter-arrival time too. Note that the arrival time of the next accident is the minimum of these two inter-arrival time (by memoryless property); i.e. it depends on which inter-arrival time is shorter. If the inter-arrival time of the animal accidents is smaller, then the next accident occur is an animal accident.
As a result, the probability that a particular accident is an accident to people is the probability that the inter-arrival time is smaller than the animal accident one.

After determine such probability you can immediately deduce the distribution.
 
#5
Yes you need to calculate the probability but it is not 0.5. I can give you the idea to calculate this probability.

Note that you have 2 independent Poisson process, so you have the corresponding 2 independent exponentially distributed inter-arrival time too. Note that the arrival time of the next accident is the minimum of these two inter-arrival time (by memoryless property); i.e. it depends on which inter-arrival time is shorter. If the inter-arrival time of the animal accidents is smaller, then the next accident occur is an animal accident.
As a result, the probability that a particular accident is an accident to people is the probability that the inter-arrival time is smaller than the animal accident one.

After determine such probability you can immediately deduce the distribution.
Oh right, but if the inter arrival time for both are exponentially distributed, I'm still not sure how to get probabilities because i'm not given specific values for lambda or mu, and both of these I would need to substitute into the exponential formula wouldn't I?
 

BGM

TS Contributor
#6
Of course the resulting probability is in terms of \( \mu \) and \( \lambda \). I suppose you know the continuous version of the law of total probability which is extremely useful and important to deal with the problem about the stochastic processes. And also you will need to do a little bit integration with the pdf/cdf of the exponential.
 
#7
Of course the resulting probability is in terms of \( \mu \) and \( \lambda \). I suppose you know the continuous version of the law of total probability which is extremely useful and important to deal with the problem about the stochastic processes. And also you will need to do a little bit integration with the pdf/cdf of the exponential.
No, I think I only know law of total probability in relation to discrete case where P(A) = expected value of probability of A given X. I'm really confused in this question, I never ask for specific answers but could you lay the foundation of the answer out for me and I will do the integration with the pdf/cdf of exponential, I have many questions like this one and once I know how to do one, I will hopefully be able to do them all, but I'm confused at the minute so having the basis of the answer would be a great help.
 

BGM

TS Contributor
#8
Suppose \( X \sim \exp(\lambda), Y \sim \exp(\mu), X,Y \) are independent. Please calculate \( \Pr\{X < Y\} \)