Let's say we have $k$ vectors each containing $n$ non-negative integers (counts), and we know that each of those vectors are distributed by a Poisson, each with a very different mean. I am wondering whether there is a way to normalize each of those $k$ vectors such that each of resulting $k$ vectors is approximately distributed by a Poisson with mean 1. That is, I am looking for a Poisson counterpart of subtracting the mean value from each of Gaussian vectors which result in each vector being a 0-mean Gaussian.
2 Answers
I don't think you can use a linear transform like you can with normally distributed RV's as the expectation and variance will not be equal which is forced under Poisson distributions (variance will be the constant multiple of your expectation).
The easiest way would just be to use the inverse CDF of your Poisson with mean = $\lambda$ then put this [0,1] through the CDF for a Poisson $\lambda$ = 1.
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$\begingroup$ This is correct. The question, it seems to me, is whether this will yield a situation that actually helps anything. $\endgroup$ Commented Sep 6, 2017 at 0:54
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$\begingroup$ Then is there a way to standardize negative binomial distribution using linear transform, given that there is an additional parameter for dispersion? $\endgroup$– user5054Commented Sep 6, 2017 at 6:14
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$\begingroup$ @user5054, it would be the same: you would fit your data to a NB by maximum likelihood, then you would pass each realization through that NB's inverse CDF to get a right-tailed probability, & pass that p through the Poisson(1)'s CDF to get a converted quantile. $\endgroup$ Commented Sep 6, 2017 at 11:47
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$\begingroup$ The problem with these count data distributions when trying to standardise using linear transforms is the supports for all these distributions only lie on $[0,1]$. Therefore from first glance immediately you can only multiply-scale by some function of the parameters to change the distribution rather than add-subtract some function of the parameters because your support for the resultant distribution will automatically be wrong. And since multiply-scaling affects both first and second moments of the distribution, and we cant add/subtract to readjust for the other we generally cant rely on LT. $\endgroup$– Dale CCommented Sep 7, 2017 at 3:30
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$\begingroup$ @TilefishPoele Why "put this [0,1]"? Where does that interval come from? A Poisson random variable with mean 1 can take a value bigger than 1, right? $\endgroup$– user5054Commented Sep 13, 2017 at 22:12
The variance stabilizing transformation of the Poisson distribution is to take the square root. Once you have done that, the variance is approximately 1/4. So to change to a variance of 1 you would just need to $2\cdot\sqrt{\lambda_k}$ for each of your $k$ vectors.
This still does not make the means the same though for each of your vectors. To do that you would still need to subtract the mean of the transformed data.
Also see the Wikipedia page on the Anscombe transform for additional options with Poisson data. Note all of these transforms frequently recommend the mean of the series be about 5, under that and they just have too few of values and will never look symmetric. That is a limitation, even with the CDF transform recommended by Tilefish.
I have not seen any simple transforms recommended for negative binomial distributions, so the CDF approach may be the best option. In this article though I do some simulations and show that simply adding 1 sigma (for control charting) after the $2\cdot\sqrt{\lambda_k}$ transform produces pretty close to nominal coverage.
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$\begingroup$ Thanks for the information, but the resulting vectors do not contain integers any more after those transformations, right? $\endgroup$– user5054Commented Sep 6, 2017 at 19:18
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[normalization]
tag, you'll see it refers to making "values lie within a specified range". I think you mean standardization. At any rate, why would you do this? What would the result mean? $\endgroup$