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I have the Poisson process $\{N(t)\}_{t\geq 0}$ with rate $\lambda=2$. Given that four events occur during the time interval $[0,2]$, what is the probability that the first event occurs before time $t=1$?

From what I understand, I need to calculate $\mathbb{P}(N(1)\geq1\mid N(2)-N(0)=4).$

So I assume I must use the conditional probability formula \begin{equation}\frac{\mathbb{P}(N(1)\geq 1,N(2)-N(0)=4)}{\mathbb{P}(N(2)-N(0)=4)} \end{equation}

I struggle now to see the intersection between the two parts of my numerator. I also am not too confident my workings for the denominator are correct. \begin{equation} \mathbb{P}(N(2)-N(0)=4)=e^{-2}\frac{(2)^4}{4!}=e^{-2}\frac{2}{3} \end{equation} Could someone explain to me how to identify the intersection in the numerator and if my calculation for the denominator is correct?

Bernard
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TK99
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1 Answers1

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In the definition of Poisson Process it is assumed that $N(0)=0$. [Ref. https://en.wikipedia.org/wiki/Poisson_point_process ]

$N(2)-N(0)=N(2)$ is Poisson with parameter $4$. So the denominator is $e^{-4}\frac {4^{4}} {4!}$.

Hint for the numerator: Let $X=N(1)$ and $Y=N(2)-N(1)$. Then $X$ and $Y$ are independent with $Poiss(2)$ distribution. . Hence $P(X \geq 1, X+Y=4)= \sum\limits_{n=1}^{4} P(X=n) P(Y=4-n)=\sum\limits_{n=1}^{4}e^{-2} \frac {2^{n}} {n!} e^{-2}\frac {2^{4-n}} {(4-n)!}$. .

Kavi Rama Murthy
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