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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?

all.over
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  • 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! – whuber Oct 30 '19 at 22:12
  • 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) $. – all.over Oct 30 '19 at 22:28
  • 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 $N(4)=N(0,4)=N(0,2)+N(2,4)=N(2)+N(2,4).$ – whuber Oct 30 '19 at 22:45
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    @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? – all.over Oct 30 '19 at 22:57
  • Good work: It looks like you're well on your way to a correct solution. – whuber Oct 31 '19 at 13:34

1 Answers1

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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}

Math1000
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