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.

**Answer**

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)
[1] 35.0542 # mean & var both aprx 35 w/in margin of sim err
[1] 35.14318
a = t/3; mean(a); var(a)
[1] 11.68473 # obviously unequal for average of three
[1] 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.

**Attribution***Source : Link , Question Author : Microbiota , Answer Author : BruceET*