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I have datasets from two bivariate poisson distributions, $BVP_x(\lambda_1, \lambda_2, \lambda_{12})$, and $BVP_y(\lambda_3, \lambda_4, \lambda_{34})$ respectively.

Now we know the correlation coefficient for these two distributions can be defined as $\rho_x = \frac{\lambda_{12}}{\sqrt{(\lambda_1 + \lambda_{12})(\lambda_2 + \lambda_{12})}}$ and$\rho_y = \frac{\lambda_{34}}{\sqrt{(\lambda_3 + \lambda_{34})(\lambda_4 + \lambda_{34})}}$,

Is there a standard test for testing $H_0 : \rho_x = \rho_y$ vs $H_1 : \rho_x \neq \rho_y$, I understand this doable by likelihood ratio tests, but not sure if there is already existing distribution for testing the parameters.

I know for bivariate normal distribution Fisher's transformation and $z$ test would work. Would that also work for bivariate poisson.

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1 Answer 1

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Yes, the Fisher z-transform would be a reasonable test (even if the underlying distributions are non-normal). The reason for this is that the distribution of the correlation coefficient is nearly normal with the transform. Furthermore, this is empirical in nature (as opposed to something derived from a distribution...though I'm certain there has been a proof using the central limit theorem by this point in time).

The reference you might wish to review is: https://www.jstor.org/stable/20156574.

Hope this helps.

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  • $\begingroup$ Thanks for the reply, if I go by the definition of Fisher's transformation wiki, "If (X, Y) has a bivariate normal distribution with correlation ρ and the pairs (Xi, Yi) are independent and identically distributed, then z is approximately normally distributed with..". Don't you think this is a hard constraint? $\endgroup$
    – CuriousS
    Commented May 29, 2023 at 15:01
  • $\begingroup$ I would have to investigate this a bit further, but I believe the assumption of normality is for when you are looking at single sample estimates. If you are comparing two correlations, I am not sure this assumption is required, as the standard error of the transform for each will be approximately $\frac{1}{\sqrt{N_i-3}}$. $\endgroup$
    – Gregg H
    Commented May 29, 2023 at 15:11

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