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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.

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  • $\begingroup$ If you mouseover the [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? $\endgroup$ – gung - Reinstate Monica Sep 6 '17 at 0:37
  • $\begingroup$ I think normalization is a much more suitable term than standardization here. Standardization is defined as "shifting and rescaling data to assure they have zero mean and unit variance" which is not at all what I am trying to do. Given the resulting values for different vectors will be in similar ranges once they are all Poisson with mean 1, normalization is suitable enough term here if not the best. $\endgroup$ – user5054 Sep 6 '17 at 0:45
  • $\begingroup$ My purpose is to make each of the vectors (or the random variables) comparable and to be able to apply statistical methods on those data without a scale bias. $\endgroup$ – user5054 Sep 6 '17 at 0:46
  • $\begingroup$ The analogy is closer to standardizing, but it doesn't matter that much, you can keep the tag if you prefer. I don't understand the purpose or how it will be served by transforming the variables. I'm not sure what you mean by avoiding a "scale bias". A concrete example might help to clarify your situation. $\endgroup$ – gung - Reinstate Monica Sep 6 '17 at 0:53
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    $\begingroup$ I agree that the analogy is closer to standardization. The definition at the tag on this site is not a general definition but a is defining standardization for normal distribution. I still added the tag "standardization" to the question. The reply at math.stackexchange.com/questions/431310/… says that Poisson with mean 1 is the standard Poisson. What I want is to standardize Poisson. $\endgroup$ – user5054 Sep 6 '17 at 6:07
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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. $\endgroup$ – gung - Reinstate Monica Sep 6 '17 at 0:54
  • $\begingroup$ Then is there a way to standardize negative binomial distribution using linear transform, given that there is an additional parameter for dispersion? $\endgroup$ – user5054 Sep 6 '17 at 6:14
  • $\begingroup$ @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. $\endgroup$ – gung - Reinstate Monica Sep 6 '17 at 11:47
  • $\begingroup$ 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 C Sep 7 '17 at 3:30
  • $\begingroup$ @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? $\endgroup$ – user5054 Sep 13 '17 at 22:12
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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? $\endgroup$ – user5054 Sep 6 '17 at 19:18
  • $\begingroup$ No, they won't. $\endgroup$ – Andy W Sep 6 '17 at 19:45

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