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

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.

• A simple explanation would be to say that while a Poisson distribution only takes integer values, the average can take rational values Commented Aug 27, 2020 at 11:13
• Thank you, that helps me understand a bit more Commented Aug 27, 2020 at 14:33
• stats.stackexchange.com/questions/35042 is related: it shows that even when you round the average of independent Poisson variables you still don't get a Poisson distribution.
– whuber
Commented Aug 27, 2020 at 18:42
• The average will approximately follow a Gaussian distribution as a consequence of the central limit theorem. Commented Aug 27, 2020 at 21:35
• @oliversm - that Gaussian approximation is also true in the same sense (convergence in distribution of a standardised version) of a Poisson distribution as its parameter increases, and so of the sum of independent Poisson distributed random variables. Commented Aug 28, 2020 at 15:43

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


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

• This is useful, because it points out that the problem here isn't merely a discrete/non-discrete problem, but that average changes the shape of the curve.
– Yakk
Commented Aug 27, 2020 at 19:37
• @BruceET: Sorry, I don't follow. 35/3=11.6666... which is quite close to the mean(a) and 35/9=3.8888... which is quite close to var(a). This is not a response to the original question: I was not able to comment [Bruce's response because I don't have enough privileges] Commented Aug 28, 2020 at 11:06
• @IuriGavronski: BruceET's point is that the average has its expected value $n$ times its variance, while a Poisson distribution has the mean and varaince the same. I made a similar point in an answer to the linked question Commented Aug 28, 2020 at 15:49

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.

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.

• Can you clarify what ‘support’ means? Support of the average and N is the support of the Poisson distribution? Your post has really helped me to understand, I’m satisfied that I know the why. Commented Aug 28, 2020 at 7:15
• The support of a distribution is the set of values it can take. That is, a poisson distribution can only take "whole-number" values, but the mean might not be a whole number (it might be a fraction). That is how we know the mean can't follow a poisson distribution, because fractions are impossible under that distribution.
– JDL
Commented Aug 28, 2020 at 8:24