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Given a Poisson Stochastic Process with parameter $\lambda$, determine the the distribution of the inter-arrival time given that it contains $\tau > 0$ (it is not exponential).

My approach

I want to calculate $$\Pr [X(t) = 1 | X(\tau) = 0], \qquad t > \tau$$ From Bayes Theorem it follows that:

$$\Pr [X(t) = 1 | X(\tau) = 0] = \frac {\Pr [X(\tau) = 0 | X(t) = 1] \Pr [X(t) = 1]} {\Pr [X(\tau) = 0]} = \frac {(1 - \tau / t)t \lambda e^{-\lambda t}} { e^{-\lambda \tau} } = (t - \tau) \lambda e^{-\lambda (t - \tau)}$$

bolzano
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1 Answers1

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Other route for calculating $P(X(t)=1\mid X(\tau)=0)$.

$$P(X(t)=1\mid X(\tau)=0)=P(X(t)-X(\tau)=1\mid X(\tau)=0)=P(X(t)-X(\tau)=1)=e^{-\lambda(t-\tau)}\lambda(t-\tau)$$where the second equality is based on independence.

P.S.

I don't understand what the real question is.

edit:

If $Y$ denotes the first inter-arrival time with $Y>\tau$ then:$$P(Y-\tau>t)=P(X(t+\tau)-X(\tau))=0)=e^{-\lambda t}$$showing that $Y-\tau$ has exponential distribution with parameter $\lambda$.

Then $Y=\tau+(Y-\tau)=\tau+Z$ where $Z$ has exponential distribution with parameter $\lambda$.

drhab
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