• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Inter event times Q 4.1

N

Nimisha

Member
Hi
Please explain that in Q 4.1 ii) how will we denote the required probability in terms of T and N.
Is it P(Nt=1)?
Not too clear with this concept.
Thank you.
 
Hello

\( P(N_t = 1) = P(N_t - N_0 = 1) \) is a special case of the probability. More generally, the question is asking you to work out:

\( P(N_{s + t} - N_s = 1) \) for any s >= 0, t > 0

ie for any interval of length t (so from some time s to time s + t) the probably of seeing one arrival in that interval. Due to the nature of a Poisson process, this probability is the same for any s>=0 for some fixed t >0.

For a Poisson process, the distribution of \( N_{s + t} - N_s \) is \( Poi(\lambda t) \) for any s >= 0, t > 0.

Hope this helps

Andy
 
Back
Top