If the inter-arrival times of customers are i.i.d. exponential distribution, is it necessary that the number of customers is a Poisson process?

Question

Suppose customers arrive with time interval $U_i$ i.i.d. $Exp(\lambda)$, therefore,

$$F(U_i\le t)=1-e^{-\lambda t}$$

The arrival time of customer $i$ is

$$T_i=\sum^i_{j=1}{U_j}$$

The number of customers having arrived by time $t$ is therefore

$$N(t)=\sum^\infty_{i=1}{1_{\{T_i\le t\}}}$$

$N(t)$ is a counting process.

$N(t)$ is said to be a Poisson counting process if it has the three properties.

This condition seems to imply that i.i.d. exponentially-distributed inter-arrival time doesn't guarantee it is a Poisson process unless it satisfies the four properties.

1. Is i.i.d. exponentially-distributed inter-arrival time a necessary and sufficient condition that it is a Poisson process?
2. Does this mean the probability distribution of $N(t)$ is not unique or tractable without assumptions such as the three properties?
3. If 1 is true, is there any other possible process besides Poisson to formulate $N(t)$?

Answer 1

Alternative definition of Poisson$(\lambda)$ process is \begin{equation} N(t) = \begin{cases} 0, X_1>t\\ \sup\{n:X_1 + X_2 + ... +X_n \leq t, X_1\leq t\} \end{cases} . \end{equation} where $(X_i)_{i=1,2,...}$ is a sequence of i.i.d. random variables, each from $exp(\lambda)$.

This definition directly points out that the process that you defined is without a doubt a Poisson process.

If the process starts in $0$ and has i.i.d. exponentially-distributed inter-arrival time then it is a Poisson process.