From a biology background and not strong in statistics.
From what I have read the sum of Poisson distributed random independent variables have a Poisson distribution but the average of these variables do not have a Poisson distribution. Why is that, can someone show me the maths?
I thought the average would still have a Poisson distribution.
Some background: this concerns technical replicates in RNA-seq. Marioni et al found that technical replicates follow a Poisson distribution. Tools that accommodate technical replicates sum the values but do not average the values. I can accept this at face value but I would like to understand the maths/stats behind this.
Comment in answer format to show simulation:
@periwinkle’s Comment that the average takes non-interger values should be enough.
However, the mean and variance of a Poisson random variable are numerically equal,
and this is not true for the mean of independent Poisson random variables. Easy
to verify by standard formulas for means of variances of linear combinations.
Also illustrated by a simple simulation in R as below:
set.seed(827) x1 = rpois(10^4, 5); x2 = rpois(10^4, 10); x3 = rpois(10^4, 20) t = x1+x2+x3; mean(t); var(t)  35.0542 # mean & var both aprx 35 w/in margin of sim err  35.14318 a = t/3; mean(a); var(a)  11.68473 # obviously unequal for average of three  3.904797
E((X_1+X_2+X_3)/ 3) = 1/3(4 + 10 + 20) = 35/3,
Var((X_1+X_2+X_3)/3) = 1/9(5 + 10 + 20) = 35/9\ne 35/3.