Poisson process and queuing system In a queuing system, the inter-arrival times are known to be exponentially distributed. 
My textbook states: "It can be shown that, if the underlying distribution of inter-arrival times { T1, T2, ..., Tn } is exponential, the arrival times are uniformly distributed on the interval (0, T). The arrival times- T1, (T1 + T2), (T1 + T2 + T3), ... , (T1 + ... + Tn) are obtained by adding inter-arrival times." 
But isn't the sum of exponential random variables distributed as Erlang? Implying that arrival times should be Erlang distributed?
How are arrival times distributed? Is the distribution of arrival times Uniform or Erlang? Why so? I'd like some clarity on the topic.
 A: If interarrival times ($IA$) are i.i.d. $IA\sim \text{Exponential}(\lambda)$ then the $n$th arrival time is $S_n \sim \text{Erlang}(n,\lambda)$. as a quick check, you can see that $S_1 \sim \text{Erlang}(1,\lambda)$ which is the same as $\text{Exponential}(\lambda)$.
However, the textbook is giving you more info. Given that the arrival times are on the interval $[0,T]$, they are distributed uniformly on that interval as a consequence of the interarrival distribution (or alternatively, a consequence of a stationary Poisson Process).
Notice the difference between arrival time $S_n$ and this conditional arrival time $(S_n|S_n\in [0, T])$
$$(S_n|S_n\in [0, T])\sim \text{Uniform}(0, T)$$. 
This extends to the case $(S_n|S_n\in [t_1, t_2])\sim \text{Uniform}(t_1, t_2)$ with $t_1<t_2$.
See the proof/derivation here on Math.SE.

With Exponential interarrivals, $(S_n|S_n\in [t_1, t_2])\sim \text{Uniform}(t_1, t_2)$ for $t_1<t_2$ doesn't invalidate that $S_n \sim \text{Erlang}(n,\lambda)$.  For a visual example, see two graphs below generated with the same Exponential interarrival times.



MATLAB code:
Rate = .85;
ArrivalNumber = 5;
SampleSize = 50000;
% Add Interarrival Times
S5 = sum((-1/Rate)*log(1-rand(SampleSize,ArrivalNumber)),2); 

fh=@(x) ((Rate^ArrivalNumber)*x.^(ArrivalNumber-1).*exp(-Rate.*x))...
    ./factorial(ArrivalNumber-1);  % Erlang PDF
Times = cumsum((-1/Rate)*log(1-rand(20000,50)),2);
IntervalTimes = [];
Interval = [15 35];
for i = 1:size(Times,1)
    Times_i = Times(i,:);
    IntervalTimes_i = Times_i(Times_i >= Interval(1) & Times_i <= Interval(2));
    IntervalTimes = [IntervalTimes; IntervalTimes_i(:)];
end

figure, hold on, box on
title('Event Times on Interval [15, 35]')
h = histogram(IntervalTimes,'Normalization','pdf','DisplayName','Simulation')
xlim([0 40])
p = plot([15 35],ones(1,2)./(35-15),'k-','LineWidth',2.2,'DisplayName','Uniform(15,35)')
xlabel('Arrival Times on Interval [15, 35]')
ylabel('PDF')
legend('show')

figure, hold on, box on
title('S_5 \sim Erlang(5,\lambda = 0.85)')
h = histogram(S5,'Normalization','pdf','DisplayName','Sum of Exponential(\lambda)')
Xrange = 0:.01:25;
p = plot(Xrange,fh(Xrange),'k-','LineWidth',2.2,'DisplayName','Erlang(5,\lambda)')
xlim([0 20])
legend('show')
xlabel('5th Arrival Time, S_5')
ylabel('PDF')

