Why does the sum of Poisson distributed random variables have a Poisson distribution but the average of the variables do not?

Question

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 1

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

$E((X_1+X_2+X_3)/ 3) = 1/3(4 + 10 + 20) = 35/3,$ $Var((X_1+X_2+X_3)/3) = 1/9(5 + 10 + 20) = 35/9\ne 35/3.$

Answer 2

The Poisson distribution is a probability distribution defined on the set $\mathbb N$ of natural numbers $0,1,2,\dots$.
We also say that $\mathbb N$ is the support of the Poisson distribution. This distribution is often used to model experiments whose outcomes represent counts.

If $X$ is a random variable following a Poisson distribution with parameter $\lambda$ then for a natural number $k \in \mathbb N$, $$ \mathbb P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}. $$

It can be shown that the sum $X+Y$ of two independent Poisson-distributed variables $X,Y$ still follows a Poisson distribution.

Now, assume that you have $N$ independent random variables $X_1, \dots, X_N$ each of them following a Poisson distribution.
Their sum $X_1+ \dots + X_N$ will be a natural number and by an induction reasonment we can show that $X_1+ \dots + X_N$ also follows a Poisson-distribution.

However their average, $\frac{X_1 + \dots + X_N}{N}$, does not need to be a natural number.
For example if $N=3$ and $X_1 = 1, X_2 = 0, X_3 = 7$ then $\frac{X_1 +X_2 + X_3}{3} = \frac{8}{3} \approx 2.67.$
Thus the average of Poisson random variables can take non-integer values (but it also can take integer values) which is against the definition of a Poisson distribution.
More precisely, the support of the average is not $\mathbb N$ but rather belongs to $\mathbb Q$ the set of rational numbers (which contains $\mathbb N$).
This means that the average can't (by definition) follow a Poisson distribution.

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In the same spirit, the statement above "It can be shown that the sum $X+Y$ of two independent Poisson-distributed variables $X,Y$ still follows a Poisson distribution" is not true if $X$ and $Y$ are not independent anymore.
Take for example $Y=X$ (thus $X$ and $Y$ are not independent) then the sum $X+Y=2X$ only takes even values and thus $\mathbb P(2X=1) = \mathbb P(2X=3) = \dots = 0$ which is not in agreement with the definition of a Poisson distribution since the quantity $e^{-\lambda} \frac{\lambda^k}{k!}$ is strictly greater than $0$ for all natural numbers $k$.

I hope this is clear enough to help.