Poisson process: Arrival times

Let's see the Poisson process $\{N(t)\}$ as the phone call problem. That is, $N(t)$ is the random variable representing the number of phone calls received by time $t$. Then, the arrival time $T_n$ of the $n$-th call is defined as the earliest time at which the random variable $N(t)=n$. The inter-arrival time $Q_n=T_n-T_{n-1}$ is the time between the successive calls (i.e., between $n-1$ and $n$). In particular, $Q_1=T_1-0=T_1$.

We have $$\begin{array}{}\text{(Eq:QnTn)}& T_n=Q_1+Q_2+\cdots +Q_n.\end{array}$$

Note that $\{Q_i\}$ are independent and identically distributed (i.i.d.) random variables. Therefore, we first determine the distribution of $Q_n$ and then combine them to find the distribution of $T_n$.

Inter-arrival times $Q_n$

Now, observe the following.

The probability of having no call at time $t$ is $$p_0(t)=e^{-\lambda t}.$$

Thus, the probability of having the first call by time $t$ is$$\begin{array}{rcl}Pr(T_1\le t)& =& Pr(\text{``1 or more calls arrived by time }t\text{''})\\ & =& Pr(N(t)\ge 1)\\ & =& p_1(t)+p_2(t)+\cdots =\sum _{n=1}^{\mathrm{\infty }}{p}_n(t)\\ & =& 1-p_0(t)\\ & =& 1-e^{-\lambda t}.\end{array}$$

Since each call is independent of other calls and all $Q_n$'s are identically distributed, we may assume that $$Pr(Q_n\le t)=Pr(Q_1\le t)=Pr(T_1\le t).$$ In other words, each call is the first call since the last call. 

Thus, the cumulative distribution function of $Q_n$ is $$Pr(Q_n\le t)=1-e^{-\lambda t}.$$ Accordingly, the density function of $Q_n$ is $${\rho }_{Q_n}(t)=\frac{dPr(Q_n\le t)}{dt}=\lambda e^{-\lambda t}.$$

Therefore, the inter-arrival time $Q_n$ follows the exponential distribution with parameter $\lambda$. The mean and variance of $Q_n$ are $$\begin{array}{}\text{(eq:mean)}& \mathbb{E}(Q_n)& =& \frac{1}{\lambda },\text{(eq:var)}& \mathbb{V}(Q_n)& =& \frac{1}{{\lambda }^2}.\end{array}$$

Arrival times $T_n$

Since we have (Eq:QnTn) above, and $Q_n$'s are i.i.d., it is convenient to use the moment generating function to find the distribution of $T_n$. 

Actually, we have the following well-known theorem.

Theorem [The sum of exponential variates is a gamma variate]

If i.i.d. random variables $X_1,X_2,\cdots ,X_n$ follow the exponential distribution Exp($\lambda$), then their sum $Y=X_1+X_2+\cdots +X_n$ follows the gamma distribution Gamma($n,\lambda$) (with $n$ and $\lambda$ being the shape and rate parameters, respectively). Accordingly, the density function of $Y$ is $${\rho }_Y(y)=\frac{\lambda (\lambda y{)}^{n-1}{e}^{-\lambda y}}{(n-1)!}.$$

Proof. Exercise. ■

Using this theorem, we find that $T_n$ follows a gamma distribution and its density function is $${\rho }_{T_n}(t)=\frac{\lambda (\lambda t{)}^{n-1}{e}^{-\lambda t}}{(n-1)!}.$$

The mean and variance of $T_n$ are $$\begin{array}{rcl}\mathbb{E}(T_n)& =& \frac{n}{\lambda }(=n\mathbb{E}(Q_n)),\\ \mathbb{V}(T_n)& =& \frac{n}{{\lambda }^2}(=n\mathbb{V}(Q_n)),\end{array}$$ as expected.