I am tying to model a Poisson process in MATLAB using the exponentially decaying distribution which is the time distribution for the elapsed time for the next event after any given event.
In MATLAB, I sampled from an exponential distribution multiple times and created a histogram of the sampled values. With enough samples, this of course, gets you back the original exponential distribution. I also tried to find the time distribution of the second event (the one after the 'next') - I sampled the first time value and when I sampled again , I added this new value to the first time value. Here are the curves. How do I arrive at an analytical expression for the second consecutive event after any given event?
The pdf of a sum of two independent exponential(1) random variables is the convolution of the individual pdfs, meaning:
$$f_{X+Y}(x)=\int_{-\infty}^\infty e^{-(x-y)} e^{-y} 1_{x-y>0} 1_{y>0} dy = \begin{cases} 0 & x<0 \\ \int_0^x e^{-x} dy = xe^{-x} & x \geq 0 \end{cases}.$$
In general for a sum of $n+1$ iid exponential($\lambda$) random variables, you obtain a Gamma distribution (or more specifically an Erlang distribution) with pdf $\lambda^{n+1} x^n e^{-\lambda x}/n!$.