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Let $N(t)$ be a non-homogeneous Poisson process with rate function $\lambda(r),r\geq 0$. I have to find joint density function of arrival times $T_1$ and $T_2-T_1$ and to show that they are dependent. This is not a homework. Firstly I tried to calculate joint cdf and got $P(T_1\leq x,T_2-T_1\leq y)=$ $(1-e^{-\int^x_0 \lambda(r)dr})(1-e^{-\int^{x+y}_x \lambda(r)dr})$, I am not sure whether this is correct. Then I tried to find separate probabilities, $P(T_1\leq x)=P(N(x)>1)=$ $1-e^{-\int^x_0 \lambda(r)dr}$ and $P(T_2\leq y+T_1)=P(N(y+T_1)>2)=$ $1-e^{\int^{y+T_1}_0 \lambda(r)}$, the latter does not seem very "legal" so another idea was $P(T_2\leq y+T_1)=\int_0^{\infty} P(N(y+T_1)>2|T_1=s)P(N(s)=1)ds$. This question seems similar, but unfortunately did not really help me. Is my solution correct? The main problem is $T_2-T_1$ since it is non-homogeneous process, I am not sure how to deal with it.

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You're fine on $T_1$. $P(T_2 - T_1 > y \vert T_1) = e^{-\int_{T_1}^{T_1 + y} \lambda(x) dx}$ –  mike Apr 23 '12 at 19:00

1 Answer 1

up vote 3 down vote accepted

The answer is already on the other page, basically, but here we go.

For every $t\geqslant0$, let $\Lambda(t)=\int\limits_0^t\lambda(s)\mathrm ds$, hence $\frac{\mathrm d}{\mathrm dt}\Lambda(t)=\lambda(t)$.

  • The probability of the event $[T_1\geqslant t]$ that there is no arrival in $(0,t)$ is $\mathrm e^{-\Lambda(t)}$. Hence the density $f_1$ of the distribution of $T_1$ is minus the derivative of this, that is, $f_1(t)=\lambda(t)\mathrm e^{-\Lambda(t)}$.
  • Conditionally on $[T_1=t]$, the probability of the event $[T_2-T_1\geqslant s]$ that there is no arrival in $(t,t+s)$ is $\mathrm e^{\Lambda(t)-\Lambda(t+s)}$. Hence the conditional density $f_2(\ \mid t)$ of the distribution of $T_2-T_1$ is minus the derivative of this, that is, $f_2(s\mid t)=\lambda(t+s)\mathrm e^{\Lambda(t)-\Lambda(t+s)}$.

Putting everything together, one sees that the joint density $f$ of $(T_1,T_2-T_1)$ is given by, for every nonnegative $t$ and $s$, $$ f(t,s)=f_1(t)f_2(s\mid t)=\lambda(t)\lambda(t+s)\mathrm e^{-\Lambda(t+s)}. $$ Sanity check: In the homogenous case, $\lambda(t)=\lambda$ for every $t$ hence $\Lambda(t)=\lambda t$, $f_1(t)=\lambda\mathrm e^{-\lambda t}$ and $f(t,s)=\lambda^2\mathrm e^{-\lambda(t+s)}=f_1(t)f_1(s)$. Thus $T_1$ and $T_2-T_1$ are independent and exponentially distributed with parameter $\lambda$, as they should be.

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