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Consider a nonhomogeneous Poisson process on $[0, T]$ with mean value function $m(t)$ for $t \in [0, T]$.

Let $X_1$ denote the time of the first arrival. Show that $(X_1\,|\,N(T) = 1)$ has the following cdf:

$$F(x) = \frac{m(x)}{m(T)},\quad x \in [0, T]$$

Can someone help me out here? In this case does x refer to the event $(X_1\,|\,N(T) = 1)$? Why is it not just $F(x)= m(T)$?

Ian
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user9999
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1 Answers1

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A cdf refers to a random variable, not an event.

By definition,

$$F(x)=\mathbb{P}(X_1\le x\,|\,N(T)=1).$$

Now note that $\{X_1\le x,N(T)=1\}=\{N(x)=1,N(T)-N(x)=0\}$. Therefore,

$$F(x)=\frac{\mathbb{P}(N(x)=1,N(T)-N(x)=0)}{\mathbb{P}(N(T)=1)}$$

and since $N(x)$ and $N(T)-N(x)$ are independent for a Poisson process,

$$F(x)=\frac{\mathbb{P}(N(x)=1)\mathbb{P}(N(T)-N(x)=0)}{\mathbb{P}(N(T)=1)}=\frac{m(x)e^{-m(x)}e^{-(m(T)-m(x))}}{m(T)e^{-m(T)}}=\frac{m(x)}{m(T)}.$$

Ian
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  • Why is it that you use the mean value function instead λT and 1 or λx and 1 for substitution into the pmf of the exponential distribution? – user9999 May 03 '14 at 08:53
  • I'm not sure I understand your question correctly. Anyway, when considering a nonhomogeneous Poisson process, then $X_1$ does not necessarily follow an exponential distribution. (What would you choose for its parameter? There is no $\lambda$ here.) All that you know is that the number of points in an interval $[a,b]$ follows a Poisson distribution of parameter $m(b)-m(a)$. Does this answer your question? – Ian May 03 '14 at 12:53
  • oh yes, it does. Thank you! – user9999 May 03 '14 at 14:11