How Is E(S5) Calculated in a Poisson Process?

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In a Poisson process with rate λ, the expected time of the ith occurrence, E(Si), can be calculated using the formula E(Si) = i / λ. For the case where X(1) = 5, the expected time for the fifth occurrence, E(S5), is thus 5 / λ. The initial confusion revolved around whether this formula was sufficient or if additional factors needed consideration. Ultimately, the correct approach was confirmed, resolving the query. The discussion highlights the importance of understanding inter-arrival times in Poisson processes.
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Summary: inter-arrival times problem

Let {X(t) : t ≥ 0} be a Poisson process with rate λ.

a-) Let Si denote the time of the ith occurrence, i = 1, 2, . . . . Suppose it is known that X(1) = 5. Find ## E(S_{5}) ##.

My attempt: Is the answer simply ## n / \lambda = 5 / \lambda ##? Or is there more to it that I am missing?

Thanks for the help
 
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CTK said:
Summary: inter-arrival times problem

Let {X(t) : t ≥ 0} be a Poisson process with rate λ.

a-) Let Si denote the time of the ith occurrence, i = 1, 2, . . . . Suppose it is known that X(1) = 5. Find ## E(S_{5}) ##.

My attempt: Is the answer simply ## n / \lambda = 5 / \lambda ##?
Why would that be the answer?
 
PeroK said:
Why would that be the answer?
Hi PeroK. I guess I have just figured it out, so no worries. Thanks for your reply.
 
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