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Let $\lbrace N(t)\rbrace_{t\geq 0}$ be a Poisson process with intensity $\lambda = 3$. Compute $$P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right].$$

I understand when there is only one condition i.e. $P\left[N(6) = 2 | N(8) = 4\right]$. Since that is $$P\left[N(6) = 2 , N(8) = 4\right]/ P\left[N(8) = 4\right] = P\left[N(6) = 2 , N(8)-N(6) = 2\right]/ P\left[N(8) = 4\right].$$ But I am unsure on how to go about doing this one

Any tips on where to begin? Thanks

DMcMor
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V.L.
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    Hint: Compute $$P\left[N(6) = 2, N(8) = 4, N(3) = 1\right]=P\left[N(8)-N(6) = 2,N(6)-N(3)=1, N(3) = 1\right]$$ and $$P\left[N(8) = 4, N(3) = 1\right]=P\left[N(8)-N(3) = 3, N(3) = 1\right]$$ – Did Mar 29 '17 at 20:16

1 Answers1

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You can handle this using the definition of conditional probability, stationary increments and independent increments. $$ \begin{align*} P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right] &= \frac{P[N(6) = 2,N(8) = 4, N(3) = 1]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1, N(6) - N(3) = 1, N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1]P[N(6) - N(3) = 1]P[N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1]P[N(3)= 1]P[N(2)= 2]}{P[N(8) = 4, N(3) = 1]}\\ \end{align*} $$

I'll let you finish it up from there.

DMcMor
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