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Let $N(t)$ be a Poisson Process with rate $\lambda$

Find $\displaystyle P(N(4) \le 2N(2) \mid N(2) = 1) = \frac{\sum_{i = 0 }^2P(N(4) = i, N(2) = 1)}{P(N(2) = 1)} = ?$

Can I split this up using the independent increments?

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  • $\begingroup$ Presuming "$N(t)$" means the number of events in the interval $[0,t],$ the assumption of independent increments is essential for making any progress at all. But why haven't you proceeded to evaluate the answer to see whether it makes sense? There's less work involved than writing the question! $\endgroup$
    – whuber
    Oct 30, 2019 at 22:12
  • $\begingroup$ As I wasn't entirely sure in which way to split it up: Namely, $P(N(4) - N(2) = i - N(2), N(2) = 1)$ which seems incorrect but so does $P(N(4) - N(2) = i - 1, N(2) = 1) $. $\endgroup$
    – all.over
    Oct 30, 2019 at 22:28
  • $\begingroup$ You might find it helpful and revealing to adopt a more precise notation: let $N(s,t)$ be the number of events in the interval $(s,t]$ and note that $N(4)=N(0,4)=N(0,2)+N(2,4)=N(2)+N(2,4).$ $\endgroup$
    – whuber
    Oct 30, 2019 at 22:45
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    $\begingroup$ @whuber this leads me to $P(N(0,4) \le 2N(0,2) \mid N(0,2) = 1) = P(N(2,4) = 0 ) + P(N(2,4) = 1) + P(N(2,4) = 2)$ and $N(2,4) \sim$ Poisson$((4-2)\lambda)$. Is this the correct logic? $\endgroup$
    – all.over
    Oct 30, 2019 at 22:57
  • $\begingroup$ Good work: It looks like you're well on your way to a correct solution. $\endgroup$
    – whuber
    Oct 31, 2019 at 13:34

1 Answer 1

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\begin{align} \mathbb P(N(4)\leqslant 2N(2)\mid N(2)=1) &= \frac{\mathbb P(N(4)\leqslant 2N(2), N(2)=1)}{\mathbb P(N(2) = 1}\\ &= \frac{\sum_{i=0}^2 \mathbb P(N(4)=i, N(2)=1)}{\mathbb P(N(2)=1)}\\ &= \frac{\sum_{i=0}^2 \mathbb P (N(4) - N(2) = i,N(2)=1)}{\mathbb P(N(2)=1)}\\ &=\frac{\sum_{i=0}^2 \mathbb P (N(4) - N(2) = i)\mathbb P(N(2)=1)}{\mathbb P(N(2)=1)}\\ &= \sum_{i=0}^2 \mathbb P(N(4) - N(2) = i)\\ &= e^{-2\lambda}(1 + 2\lambda +2\lambda^2). \end{align}

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