Distribution of $X_{N(t)+1}$ in poisson process

Question

Assume $\{N(t)\}_{t\geq 0}$ is a poisson process with parameter $\lambda$, $X_n$ is the $n^{th}$ interarrival time, $n \in \{1, 2, 3, ...\}$, which means $X_n$ is exponential distribution with parameter $\lambda$.

Then how to compute the distribution of $X_{N(t)+1}$ ?

(This is a problem from Stochastic Process (Ross), chapter 3: renewal process)

It is the interarrival time which is exponential distribution with parameter $\lambda$ — Icy, Jun 15 '20 at 20:43
$N(t)$ takes the value $0$ so $X_{N(t)}$ is not defined in general. Either start the indexation from $0$ or consider $X_{N(t)+1}$ instead. — Michh, Jun 15 '20 at 21:29

score 1 · Accepted Answer · answered Jun 15 '20 at 22:14

To compute the cdf, simply write $$\mathbb{P}(X_{N(t)+1}\leq a) =\sum_{n=0}^\infty \mathbb{P}(X_{n+1} \leq a, \, N(t) = n).$$ Now observe that $\{N(t) =0\} = \{X_1 >t \}$ and $$\left\{N(t) = n\right\} = \left\{X_1+ \cdots + X_n \leq t < X_1 + \cdots + X_{n+1}\right\}, \quad \forall n \geq 1.$$ Thus it remains to compute $$\begin{align}\mathbb{P}(X_{n+1} \leq a, \, N(t) = n) &= \mathbb{P}(X_{n+1}\leq a, \,X_1+ \cdots + X_n \leq t < X_1 + \cdots + X_{n+1} )\\ &= \mathbb{P}(X_{n+1} \leq a, \, S_n \leq t < S_n + X_{n+1})\end{align}$$ where $S_n \sim \mathrm{Gamma}(n,\lambda)$ and $X_{n+1} \sim \mathrm{Exp}(\lambda)$ are independent. This is equal to $$\int_{[0,\infty)^2} \mathbf{1}_{x \leq a, \, y\leq t < x+y}\,\lambda e^{-\lambda x} \frac{\lambda^n}{(n-1)!}y^{n-1}e^{-\lambda y}\, dxdy.$$ Using Fubini's theorem we get $$\begin{align}\mathbb{P}(X_{N(t)+1}\leq a) &= \mathbb{P}(t < X_1 \leq a)+\sum_{n=1}^\infty \int_{[0,\infty)^2} \mathbf{1}_{x \leq a, \, y\leq t < x+y}\,\lambda e^{-\lambda x} \frac{\lambda^n}{(n-1)!}y^{n-1}e^{-\lambda y}\, dxdy\\ &=\mathbb{P}(t < X_1 \leq a)+\int_{[0,\infty)^2} \mathbf{1}_{x\leq a, \, y \leq t < x+y}\, \lambda^2 e^{-\lambda x}\, dx dy, \end{align}$$ which I now leave you to calculate.

The answer is $P(X_{N(t) + 1} > x) = (1+\frac{x}{\lambda})e^{-\lambda x} - e^{-\lambda t}$ . But I got a different answer using your method — Icy, Jun 15 '20 at 22:44
@Icy That cannot be the cdf as it does not go to $0$ as $x \to \infty$. As for the end of the computation, you need to distinguish the cases $a\leq t$ and $a>t$ (for the integral as well as $\mathbb{P}(t<X_1 \leq a)$. At least this was the problem that I faced. In the end I get $\mathbb{P}(X_{N(t)+1} \leq a ) = 1-(1+\lambda (a \wedge t)) e^{-\lambda a}$. — Michh, Jun 15 '20 at 23:33

Distribution of $X_{N(t)+1}$ in poisson process

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