# How to properly truncate exponential distribution to represent random memory-less arrivals?

Typically, Poisson and exponential distributions are used to represent random memory-less arrival processes. It has come to my attention, however, that a more realistic distribution is a truncated (or would it be shifted?) exponential distribution in order to not have physically impossible inter-arrival times (e.g. 0.1 seconds between two people/cars/etc).

I have been doing some reading, and from my understanding that would be achieved by shifting the exponential distribution to the right (i.e. if the minimum acceptable inter-arrival time is 2 seconds, then shift the distribution to the right by 2 seconds).

My confusion now is in regards to the properties of the typical exponential distribution that no longer hold, mainly mean = variance. Instead (and I could be wrong), it seems that the mean becomes 1/lambda + shift whereas the variance remains as 1/lambda. Not only that, but the most important input, being the mean inter-arrival time, no longer represent true conditions due to the shift (if the original mean was 5 seconds, now the new mean is 7 seconds).

Would this need correction? If so, how is this corrected?

• Exponential distributions are often used as a rough approximation to the truth because of their mathematical simplicity and because the no-memory property often avoids the inconvenience of having to model the past. You might find a distribution similar to an exponential with support $[2, \infty),$ mean 5 and standard deviation 5, but it would neither have the no-memory property nor the simple form of an exponential. (In the context of your question, it's the Poisson distribution that has numerically equal mean and variance.) – BruceET Apr 12 at 6:13
• Thank you so much for your response! I'll have some thinking to do. – Deuterium Apr 13 at 0:22
• @BruceET - That could be an answer. – Pere Apr 13 at 10:15