The result of Example 2.7.5 can be interpreted as follows. A stream of entities arrive at a junction in such a way that the time between successive arrivals (interarrival time) has an exponential distribution with mean $1 / \lambda$. If the entities (customers) are alternately routed along two separate paths such that the first customer takes the first path, the second the second path, the third the first path, the fourth the second path etc., so that the odd-numbered arrivals take the first path and the even-numbered ones take the second path, then the interarrival time on each path has an Erlang-2 distribution with average value $2 / \lambda$. Generalize the above result to show that if a stream of customers having an exponential interarrival time is split deterministically into $k$ streams, then the interarrival times along each new stream have an Erlang- $k$ distribution with mean $k / \lambda$.