If each arrival is exponentially distributed, then the $k$th arrival time is Erlang distributed. The Erlang PDF is: $$ f_{Y_k}(y) = \lambda e^{-\lambda y} \frac{(\lambda y)^{k-1}}{(k-1)!} $$ How is this derived?
How is the Erlang pdf derived?
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2You need each arrival to be _exponentially_ distributed (and independent of past arrivals). If arrival times are discrete and geometrically distributed with parameter $p$, the $k$-th arrival time is a _negative binomial_ or _Pascal_ random variable with parameters $(k,p)$. Both derivations can be done with a $(k-1)$-fold convolution, via moment-generating functions or characteristic functions, etc. – 2012-12-04
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0Here's a derivation given by convolution: http://www.math.unl.edu/~scohn1/428s05/queue3.pdf Essentially the Erlang distribution is the result of convolving the exponential distribution with itself k-1 times. Note that convolution as the sum of random variables is particularly important to understanding what's going on here. – 2016-10-05