# Counting samples seem to not be Poisson distributed, need sanity check

I have an exercise where I have to use Poisson one-way classification / Regression of some data. The data I have is a set of 120 samples grouped the following labels A, B, C, D, E, and F. For each group, there are 20 samples (or 20 repetitions) with a count value. Now that is all good and from what I can tell is well suited for the assumption that it may fit a Poisson distribution.

However, as I understand it one of the properties of a random variable $$Y$$ that follows $$Y \sim Po(\lambda)$$ Then it follows that $$E(Y) = V(Y)$$ But when I calculate the means (expected) and the variance for the data according to the grouping: Then I get

|          |         A |         B |         C |         D |          E |         F |
|----------+-----------+-----------+-----------+-----------+------------+-----------|
|----------+-----------+-----------+-----------+-----------+------------+-----------|
| Mean     |      4.90 |      9.45 |      8.65 |      1.45 |      18.35 |      0.80 |
| Variance | 9.8842105 | 6.4710526 | 7.3973684 | 1.3131579 | 15.6078947 | 0.5894737 |
|----------+-----------+-----------+-----------+-----------+------------+-----------|


So from what I can tell $$E(Y) \neq V(Y)$$ and just to show how I computed this using R:

( Means <- tapply(D$NumberPGrains, D$Era, mean) )
( Variances <- tapply(D$NumberPGrains, D$Era, var) )


This means, from my understanding that the data is not Poisson distributed. So my question is: Am I wrong can this still be Poisson distributed, is there something I have missed?

And just to clarify, the exercise literally states to follow Poisson one-way classification (the title of the exercise: "Question 3 -Poisson one-way classification model"), but right now I have hard time seeing the purpose of that.

A true Poisson distribution will have its mean exactly equal to its variance. For a sampling of a Poisson distribution, however, there will be some deviation - with only 20 samples, it's unlikely that you'd see the mean and variance of the sample be exactly equal. For the most part, you seem to have a strong correlation between the mean and variance, which is good. You could also find the confidence intervals around your parameter estimates, to take a hypothesis testing approach to determine whether your mean and variance estimates are really statistically different from one another. With a very large sample size, you'll have very good estimates which should be very nearly equal if the data is indeed Poisson distributed, but for lower sample sizes, your estimates won't be as good, so some numerical differences between the mean and variance are expected.

• Thanks a lot :) Yes I will follow with some hypothesis testing as well. I was just worried thanks for calming me down :) Thanks for a good answer :) – Lars Nielsen Mar 26 at 13:49

It's always good to simulate a little to get a feeling for the randomness involved in a situation like this. Here is a little something I hacked together, which may be enlightening.

I'll take the means you observed as the true means of your underlying data generating process. Then I'll assume that the data are actually Poisson distributed, and generate simulations using these assumptions. Then I'll calculate the variance for such simulations. I'll do this, say, fifty times, so I end up with fifty different variances we might observe under the null hypothesis of Poisson distributions with the means you observed. Finally, I'll plot these variances (as dot coulds, jittered horizontally), and also the variances you actually observed (as horizontal red lines):

In panels B-F, the observed variances are pretty much in the middle of the point clouds of simulated variances, so they seem to be quite consistent with a Poisson assumption. In panel A, the variance you observe is way higher than all the variances we observed, so this one does not appear consistent with a Poisson assumption. (We could call this an informal hypothesis test and say that we got $$p=\frac{1}{50}=0.02$$, though this doesn't account for the uncertainty in the mean.)

R code:

n_sims <- 50
n_sample <- 20
means <- c(4.90, 9.45, 8.65, 1.45, 18.35, 0.80)
vars_obs <- c(9.8842105, 6.4710526, 7.3973684, 1.3131579, 15.6078947, 0.5894737)

vars <- matrix(NA,nrow=n_sims,ncol=length(means),
dimnames=list(NULL,LETTERS[seq_along(means)]))
for ( ii in 1:n_sims ) {
set.seed(ii)    # for reproducibility
vars[ii,] <- sapply(means,function(mm)var(rpois(n_sample,mm)))
}

opar <- par(mfrow=c(2,3))
for ( jj in seq_along(means) ) {
plot(rnorm(n_sims),vars[,jj],pch=19,main=colnames(vars)[jj],
ylim=range(c(vars[,jj],vars_obs[jj])),xaxt="n",las=1,ylab="",xlab="")
abline(h=vars_obs[jj],col="red",lwd=2)
}
par(opar)

• Thanks a lot :D (edit: Damn DK keyboard was enabled XD) – Lars Nielsen Mar 26 at 14:25
• Good back-of-the-envelope example to show that the divergence is largest for group A. This hypothesis test shows that it's unlikely to find a variance as high as 9.88 when sampling a Poisson distribution with mean 4.9. As you allude to, we should also account for uncertainty in the mean estimate, so the true hypothesis test would be to check if the difference in mean and variance is significantly different from 0, while accounting for a multiple hypothesis correction for the fact that there are 6 groups. – Nuclear Wang Mar 26 at 16:29

One can formally test the fit of the data to the theoretical poisson distribution using a chi-squared goodness of fit test, via the goodfit() function included in the vcd package.

• Null hypothesis significance test: If p < .05 or whatever
threshold we might choose in advance, we will reject the hypothesis
that x comes from a Poisson process.
• More flexible approach: we simply assess the degree of evidence
against the null based on the size of the p-value, where smaller
p = stronger evidence.

require(vcd)

x=0:5

freqx=c(7,12,7,3,1,1) # Counts at each level of x
gf = goodfit (x, type= "poisson", method= "ML")


summary(gf)

    Goodness-of-fit test for poisson distribution

X^2 Likelihood Ratio = 1.91, df = 4, P(> X^2) = 0.75

plot (gf, main= "x vs Poisson distribution")


In plots such as these from vcd, bars lifted above the x-axis reflect negative residuals (too few occurrences to fit the theoretical distribution); bars that extend below, positive residuals (too many). This may seem less than intuitive, and the plot can't be modified in some of the usual ways (text size relative to bar size, etc), and so the following approach may be more helpful.

require(fitdistrplus)
poisson = fitdist(x, 'pois', method = 'mle')
print(poisson)  # Identify the best value of lambda (which indicates both mean and variance) to insert below
dist = dpois(0:5, lambda = 2.5) * sum(freqx) # assign lambda from the fit above
df = as.data.frame(dist)
windows()
# Then we plot first the observed distribution, as vertical lines, and then the theoretical distribution as the curve
plot(x, freqx,  type='h', lwd=2, main = 'Curve = Fitted Poisson Values',
cex.axis=1.2, cex.lab=1.2)
lines(x, df\$dist, col = 'red', lwd=3)


• freqx=c(7,12,7,3,1,1) # Counts at each level of x x is the individual groups right? Just to be sure :) – Lars Nielsen Mar 27 at 10:00
• @LarsNielsen - x is the value at which we observe a certain count: in these plots, x = "Number of Occurences" while "Freqx" = the count of the number of times we observe such a number of occurrences. 0 occurrences happened 7 times, etc. I left vcd's default labels as they are, though I'm sure you could improve on them. Cheers ~ – rolando2 Mar 27 at 10:17
• Okay, thanks for the explanation – Lars Nielsen Mar 27 at 10:44