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This continuous-time process with state space N starts in state 0 and repeats in the following way. It remains in the same state for some random waiting time at which point the state number increases by some random increment. The waiting times are independent and all follow the same fixed probability distribution taking values in R≥o. The increments are independent (and also independent of the waiting times) and all follow another fixed probability distribution taking values in {1, 2, 3, ...}. So, if the waiting times are S₁, S2, ... and the increments are n₁, n2,... then we start with X(0) = 0 and remain with X(t) = 0 for t € [0, S₁). Then we jump up by n₁ so X(t) = n₁ for t € [S₁, S₁ + S₂). Then, at time S₁ + S₂, we jump up to n₁ + n₂ and so on. More formally, for t € [S₁ + · + Sk, S1+ + Sk + Sk+1) we have X(t) = n₁ +. + nk. (a) Which choices of distribution for waiting time and increment show that th Poisson Process is a special case of this kind of process?

This continuous-time process with state space N starts in state 0 and repeats in the following way. It remains in the same state for some random waiting time at which point the state number increases by some random increment. The waiting times are independent and all follow the same fixed probability distribution taking values in R≥o. The increments are independent (and also independent of the waiting times) and all follow another fixed probability distribution taking values in {1, 2, 3, ...}. So, if the waiting times are S₁, S2, ... and the increments are n₁, n2,... then we start with X(0) = 0 and remain with X(t) = 0 for t € [0, S₁). Then we jump up by n₁ so X(t) = n₁ for t € [S₁, S₁ + S₂). Then, at time S₁ + S₂, we jump up to n₁ + n₂ and so on. More formally, for t € [S₁ + · + Sk, S1+ + Sk + Sk+1) we have X(t) = n₁ +. + nk. (a) Which choices of distribution for waiting time and increment show that th Poisson Process is a special case of this kind of process?

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.4: Expected Value
Problem 23E
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This continuous-time process with state space N starts in state 0 and repeats in the following way. It remains in the same state for some random waiting time at which point the state number increases by some random increment. The waiting times are independent and all follow the same fixed probability distribution taking values in Ro. The increments are independent (and also independent of the waiting times) and all follow another fixed probability distribution taking values in {1, 2, 3, ... ..}. So, if the waiting times are S₁, S2, ... and the increments are n₁, n₂,... then we start with X (0) = 0 and remain with X(t) = 0 for t = [0, S₁). Then we jump up by n₁ so X(t) = n₁ for t € [S₁, S₁ + S₂). Then, at time S₁ + S2, we jump up to n₁ + n₂ and so on. More formally, for t € [S₁ + … + Sk, S₁ + + Sk+ Sk+1) we have X(t) = n₁ + + nk. (a) Which choices of distribution for waiting time and increment show that the Poisson Process is a special case of this kind of process?
Transcribed Image Text:This continuous-time process with state space N starts in state 0 and repeats in the following way. It remains in the same state for some random waiting time at which point the state number increases by some random increment. The waiting times are independent and all follow the same fixed probability distribution taking values in Ro. The increments are independent (and also independent of the waiting times) and all follow another fixed probability distribution taking values in {1, 2, 3, ... ..}. So, if the waiting times are S₁, S2, ... and the increments are n₁, n₂,... then we start with X (0) = 0 and remain with X(t) = 0 for t = [0, S₁). Then we jump up by n₁ so X(t) = n₁ for t € [S₁, S₁ + S₂). Then, at time S₁ + S2, we jump up to n₁ + n₂ and so on. More formally, for t € [S₁ + … + Sk, S₁ + + Sk+ Sk+1) we have X(t) = n₁ + + nk. (a) Which choices of distribution for waiting time and increment show that the Poisson Process is a special case of this kind of process?
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You want to model customers arriving in a shop using this process instead of a Poisson Process. Using a random variable to determine the increments allows us to include a feature which the Poisson Process cannot. Write down this feature in non-mathematical language.

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