Difference between $\mathrm{Poisson}(x_1)$, $\mathrm{Poisson}(x_2)$ and $\mathrm{BPoisson}(x_1, x_2)$

Question

I am trying to find out the difference between treating two random variables as poisson distributions, $\mathrm{Poisson}(x_1)$, $\mathrm{Poisson}(x_2)$, and using a bivariate poisson, $\mathrm{BPoisson}(x_1, x_2)$. I know the different forms these take but I am more interested in understanding the different assumptions behind their use.

Answer 1

Any multivariate distribution describes joint distribution of some variables. Karlis and Ntzoufras (2003) define bivariate Poisson distribution pmf as

$$ f(x,y) = \exp\{-(\lambda_1+\lambda_2+\lambda_3)\} \frac{\lambda_1^x}{x!} \frac{\lambda_2^y}{y!} \sum^{\min(x,y)}_{k=0} {x \choose k} {y \choose k} k!\left(\frac{\lambda_3}{\lambda_1\lambda_2}\right)^k $$

$E(X) = \lambda_1+\lambda_3$ and $E(Y) = \lambda_2+\lambda_3$, where $\mathrm{cov}(X,Y) = \lambda_3$, so you can treat $\lambda_3$ as a measure of dependence between the two marginal Poisson distributions. Example of such distribution is plotted below.

and two marginal distributions

by marginal distribution we mean how would joint distribution "look like" if we looked at it "from the $X$ side", or "from the $Y$ side". Marginal distributions of $X$ and $Y$ are Poisson distributions with parameters $\lambda_X = \lambda_1+\lambda_3$ and $\lambda_Y = \lambda_2+\lambda_3$.

What is different from looking only at two independent Poisson distributions is that we do not account anyhow for dependence between them (i.e. we assume $\lambda_3 = 0$). Using bivariate distribution lets you account for covariance between $X$ and $Y$.

Check also the following threads Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? and Deriving the bivariate Poisson distribution

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Karlis, D., & Ntzoufras, I. (2003). Analysis of sports data by using bivariate Poisson models. Journal of the Royal Statistical Society: Series D (The Statistician), 52(3), 381-393.