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Neagative binomial

The mean-var parameterization of Negative Binomial distribution is mean=\(m=rp/(1-p)\), and var=\(m + \frac{1}{r}m^2 = (1+\frac{1}{r}m)m\). So if we want to generate some count data with mean \(\lambda\) and variance being say \(\phi\) times the mean, we can set \(r = \lambda/(\phi-1)\).

Poisson Log-normal

The mean-var relationship in a Poisson log normal model is mean=\(m=e^{\mu+\sigma^2/2}\) and var=\(m+(e^{\sigma^2}-1)e^{2\mu+\sigma^2} = (1+(e^{\sigma^2}-1)m)m\). So given \(\mu\), if we want to generate some count data with mean \(\lambda\) and variance being say \(\phi\) times the mean, we can set \(\sigma^2 = \log(1+\frac{\phi-1}{\lambda})\). (\(\phi>1\)). However \(\lambda\) itself is a function of \(\sigma^2\), so we can solve \((e^{\sigma^2}-1)e^{0.5\sigma^2} = (\phi-1)e^{-\mu}\) for \(\sigma^2\). Further assume \(\sigma^2 = \log(1+\alpha),\alpha>0\), then we solve \(\alpha^3+\alpha^2 - d^2 = 0\) where \(d = (\phi-1)e^{-\mu}\). The wolframalpha gives the following solution:

cubsolver = function(a){
   1/3*(((27*a)/2 + 3/2*sqrt(3)*sqrt(a*(27*a - 4)) - 1)^(1/3) + 1/((27*a)/2 + 3/2*sqrt(3)*sqrt(a*(27*a - 4)) - 1)^(1/3) - 1)
}

Lets verify it using polyroot when \(d^2=1\), corresponding to \(\phi=2,\mu=0\).

cubsolver(1)
[1] 0.7548777
polyroot(c(-1,0,1,1))
[1]  0.7548777+0.0000000i -0.8774388+0.7448618i -0.8774388-0.7448618i

So in this case the \(sigma^2\) is

log(1+cubsolver(1))
[1] 0.5623991

However the solution from wolframalpha seems to only support \(27*d^2-4 > 0\)?

So we can simply use polyroot then. The following function takes input of \(\mu,\phi\) and return the \(\sigma^2\):

calc_var_PLN = function(mu,phi){
  a = ((phi-1)*exp(-mu))^2
  return(log(1+Re(polyroot(c(-a,0,1,1))[1])))
}
calc_var_PLN(-5,2)
[1] 3.356975
calc_var_PLN(5,2)
[1] 0.006692988

Implications.

So for both cases, fixing \(\phi\), if the mean is large, the extra variance from the over-dispersion parameter(\(r\) in NB, \(\sigma^2\) in PLN) is small. This not too surprising for example in PLN, if \(\mu\) is small, then a lot of observations are 0 so in order to make variance larger, \(\sigma^2\) has to be larger.


sessionInfo()
R version 4.2.1 (2022-06-23)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 20.04.5 LTS

Matrix products: default
BLAS:   /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.9.0
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.9.0

locale:
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[10] LC_TELEPHONE=C         LC_MEASUREMENT=C.UTF-8 LC_IDENTIFICATION=C   

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] workflowr_1.7.0

loaded via a namespace (and not attached):
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