Saturday, 2 August 2014

statistics - Variance stabilization for Poisson data

Intro

Let $Z > 0$ be a random variable with the mean and variance defined as $\mathbb{E}\{ Z \}$ and $\operatorname{Var}\{ Z \}$, respectively. The variance stabilization transform (VST) $f(z)$ turns heteroskedastic data $z$ to homoskedastic data $f(z)$ with constant variance, e.g., variance equals 1.



Poisson distribution
For Poisson distributed data with $\mathbb{E}\{ Z \} = \operatorname{Var}\{ Z \}=\lambda$ this VST, so-called Anscombe transformation, is given by [1,2]:



$$f(z) = 2\sqrt{z + 3/8}$$



Based on the first order Taylor expansion we can write (this is called Delta method in the literature) [3]:



$$\operatorname{Var}\{f(z)\} \approx \left( \left.\frac{df}{dz}\right|_{z=\mathbb{E}\{ Z \}} \right)^2 \operatorname{Var}\{Z\} = \frac{\operatorname{Var}\{Z\}}{\mathbb{E}\{ Z \} + 3/8}$$




Problem
I performed a Monte Carlo simulations to compare sample variance of the stabilized data $f(z)$ and the variance obtained by the above equation, i.e., $\operatorname{Var}\{f(z)\}$ both numerically as a sample variance and theoretically as follows:



$$\operatorname{Var}\{f(z)\} \approx \frac{\lambda}{\lambda + 3/8}$$



Moreover, I went further and derived second order approximation for $Var\{f(z)\}$.



There is a mismatch between variance of stabilized data $f(z)$ (green curve) and those obtained theoretically and numerically by means of the $\operatorname{Var}\{f(z)\}$. Can anyone explain me this inconsistency?




enter image description here




  1. Anscombe, F. J. (1948), "The transformation of Poisson, binomial and negative-binomial data", Biometrika 35 (3–4): 246–254, doi:10.1093/biomet/35.3-4.246, JSTOR 2332343

  2. http://en.wikipedia.org/wiki/Anscombe_transform

  3. Kendall's Advanced Theory of Statistics: Volume 1: Distribution Theory by Alan Stuart and Keith Ord (Apr 20, 2009), page. 351, eq. 10.14

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