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,...}.

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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.

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 n₁1, №2,…. 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 € [S1₁, 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.
·
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 n₁1, №2,…. 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 € [S1₁, 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. ·
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