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Calculating Poisson Process probabilities

  1. Apr 8, 2006 #1
    I just want to check my answers/reasoning as I'm not sure if I assumed the right things to do these problems.

    N = {N(t), t>=0} ~ Poisson(1) and [tex]N_{(t,t+h]} = N(t+h)-N(t)[/tex]

    Determine P(N(4) =3|N(2) = 1)
    Here I presumed that since N(2) = 1, then there must be 2 more arrivals in the interval (2,4] so that N(4)=3 so I calculated
    [tex]P(N_{(0,2]} = 1 and N_{(2,4]} = 2) = P(N(2) = 1)P(N(2) = 2) = 4e^{-4}[/tex]
    since non-overlapping intervals are independent so I can multiply the probabilities (?). However I'm not too sure if I can break the intervals up like that?

    Determine [tex]P(N_{(4,7]} = 2 and N_{(3,6]} = 1)[/tex]
    Here I split up the intervals into (3,4], (4,6] and (6,7] although like I said before, I'm not sure if that should be done. Then I looked for the different combinations of events over those intervals so that the above would be true ie
    [tex]P(N_{(3,4]} = 0 and N_{(4,6]} = 1 and N_{(6,7]} = 1) + P(N_{(3,4]} = 1 and N_{(4,6]} = 0 and N_{(6,7]} = 2) = \frac{5e^{-4}}{2}[/tex]
    Again I assumed non-overlapping intervals were independent so I could multiply the probabilities together. It would be great if someone could double-check for me that those two are the only combinations (I did it three times already but I'm paranoid).

    Determine [tex]P(N_{(4,7]} = 2|N_{(1,5]} = 2)[/tex]
    I used the same sort of steps as I did before although it's a conditional but I thought that if I split up the intervals and made it independent, conditionals would not matter. I used intervals (1,4], (4,5] and (5,6] and found the combinations
    [tex]P(N_{(1,4]} = 2 and N_{(4,5]} = 0 and N_{(5,7]} = 2) + P(N_{(1,4]} = 1 and N_{(4,5]} = 1 and N_{(5,7]} = 1) + P(N_{(1,4]} = 0 and N_{(4,5]} = 2 and N_{(5,7]} = 0) = \frac{31e^{-6}}{2}[/tex]
    Last edited: Apr 8, 2006
  2. jcsd
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