# Deriving the bivariate Poisson distribution

I've recently encountered the bivariate Poisson distribution, but I'm a little confused as to how it can be derived.

The distribution is given by:

$P(X = x, Y = y) = e^{-(\theta_{1}+\theta_{2}+\theta_{0})} \displaystyle\frac{\theta_{1}^{x}}{x!}\frac{\theta_{2}^{y}}{y!} \sum_{i=0}^{min(x,y)}\binom{x}{i}\binom{y}{i}i!\left(\frac{\theta_{0}}{\theta_{1}\theta_{2}}\right)^{i}$

From what I can gather, the $\theta_{0}$ term is a measure of correlation between $X$ and $Y$; hence, when $X$ and $Y$ are independent, $\theta_{0} = 0$ and the distribution simply becomes the product of two univariate Poisson distributions.

Bearing this in mind, my confusion is predicated on the summation term - I'm assuming this term explains the the correlation between $X$ and $Y$.

It seems to me that the summand constitutes some sort of product of binomial cumulative distribution functions where the probability of "success" is given by $\left(\frac{\theta_{0}}{\theta_{1}\theta_{2}}\right)$ and the probability of "failure" is given by $i!^{\frac{1}{min(x,y)-i}}$, because $\left(i!^{\frac{1}{min(x,y)-i!}}\right)^{(min(x,y)-i)} = i!$, but I could be way off with this.

Could somebody provide some assistance on how this distribution can be derived? Also, if it could be included in any answer how this model might be extended to a multivariate scenario (say three or more random variables), that would be great!

(Finally, I have noted that there was a similar question posted before (Understanding the bivariate Poisson distribution), but the derivation wasn't actually explored.)

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Shouldn't the first term with exponent be $e^{-\left( \theta_1 + \theta_2 + \theta_0 \right)}$ instead of $e^{\theta_1 + \theta_2 + \theta_0}$ ? – Gilles Jul 21 '14 at 14:26
@Giles Sorry, I misread your comment initially - yes, you're correct; the term should read $e^{-(\theta_{1}+\theta_{2}+\theta_{0})}$. Thanks for catching that! – user9171 Jul 21 '14 at 15:03
In general, it's not "the" for multivariate versions of univariate distributions, with a few conventional exceptions ("the" multivariate normal for example). There are many ways of getting multivariate extensions, depending on which features are most important to have. Different authors may have different multivariate versions of common univariate distributions. So generally, one might say something like "a multivariate Poisson", or 'So-and-so's bivariate Poisson". This one is a pretty natural one, but not the only one. – Glen_b Jul 21 '14 at 23:38
(ctd) ... e.g. some authors look for a multivariate distribution capable of negative dependence, a capability this one doesn't possess. – Glen_b Jul 21 '14 at 23:45
@Glen_b Thanks for that! – user9171 Jul 22 '14 at 10:41

## 2 Answers

In a slide presentation, Karlis and Ntzoufras define a bivariate Poisson as the distribution of $(X,Y)=(X_1+X_0,X_2+X_0)$ where the $X_i$ independently have Poisson $\theta_i$ distributions. Recall that having such a distribution means

$$\Pr(X_i=k) = e^{-\theta_i}\frac{\theta_i^k}{k!}$$

for $k=0, 1, 2, \ldots.$

The event $(X,Y)=(x,y)$ is the disjoint union of the events

$$(X_0,X_1,X_2) = (i,x-i,y-i)$$

for all $i$ that make all three components non-negative integers, from which we may deduce that $0 \le i \le \min(x,y)$. Because the $X_i$ are independent their probabilities multiply, whence

$$F_{(\theta_0,\theta_1,\theta_2)}(x,y)=\Pr((X,Y)=(x,y)) \\= \sum_{i=0}^{\min(x,y)} \Pr(X_0=i)\Pr(X_1=x-i)\Pr(X_2=y-i).$$

This is a formula; we are done. But to see that it is equivalent to the formula in the question, use the definition of the Poisson distribution to write these probabilities in terms of the parameters $\theta_i$ and (assuming neither of $\theta_1,\theta_2$ is zero) re-work it algebraically to look as much as possible like the product $\Pr(X_1=x)\Pr(X_2=y)$:

\eqalign{ F_{(\theta_0,\theta_1,\theta_2)}(x,y)&= \sum_{i=0}^{\min(x,y)} \left( e^{-\theta_0} \frac{\theta_0^i}{i!}\right) \left( e^{-\theta_1} \frac{\theta_1^{x-i}}{(x-i)!}\right) \left( e^{-\theta_2} \frac{\theta_2^{y-i}}{(y-i)!}\right) \\ &=e^{-(\theta_1+\theta_2)}\frac{\theta_1^x}{x!}\frac{\theta_2^y}{y!}\left(e^{-\theta_0}\sum_{i=0}^{\min(x,y)} \frac{\theta_0^i}{i!}\frac{x!\theta_1^{-i}}{(x-i)!}\frac{y!\theta_2^{-i}}{(y-i)!}\right). }

If you really want to--it is somewhat suggestive--you can re-express the terms in the sum using the binomial coefficients $\binom{x}{i}=x!/((x-i)!i!)$ and $\binom{y}{i}$, yielding

$$F_{(\theta_0,\theta_1,\theta_2)}(x,y) = e^{-(\theta_0+\theta_1+\theta_2)}\frac{\theta_1^x}{x!}\frac{\theta_2^y}{y!}\sum_{i=0}^{\min(x,y)}i!\binom{x}{i}\binom{y}{i}\left(\frac{\theta_0}{\theta_1\theta_2}\right)^i,$$

exactly as in the question.

Generalization to multivariate scenarios could proceed in several ways, depending on the flexibility needed. The simplest would contemplate the distribution of

$$(X_1+X_0, X_2+X_0, \ldots, X_d+X_0)$$

for independent Poisson distributed variates $X_0, X_1, \ldots,X_d$. For more flexibility additional variables could be introduced. For instance, use independent Poisson $\eta_i$ variables $Y_1, \ldots, Y_d$ and consider the multivariate distribution of the $X_i + (Y_i + Y_{i+1} + \cdots + Y_d)$, $i=1, 2, \ldots, d.$

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kudos ! Btw, shouldn't the second $e^{-\theta_0}$ in the big parenthesis prior to the last step be $e^{-\theta_2}$ ? – Gilles Jul 21 '14 at 14:10
@Gilles Thank you for catching the typo--I fixed it. The initial exponent of $\theta_0+\theta_1$ needed to be $\theta_1+\theta_2$; the $e^{-\theta_0}$ within the parentheses is correct. – whuber Jul 21 '14 at 14:17
@whuber Thanks a million! That's a perfect answer! – user9171 Jul 21 '14 at 16:07
@whuber Great answer! I still don't see why the event $(X,Y) = (x,y)$ should be the disjoint union of the events $(X_0,X_1,X_2)=(i,x-i,y-i)$. I guess this is only true for $i=0$. Perhaps you meant $(X,Y) \leq (x,y)$ (component-wise)? But is that sufficient to characterize the distribution function? – vanguard2k Sep 4 '15 at 9:32
@vanguard2k I don't understand your comment. Are you asserting those events are not disjoint? (Yet they must be, for they have distinct values of $X_0$.) Or are you asserting they are not exhaustive? (If so, what value(s) of $(X,Y)$ do you think have not been included?) – whuber Sep 4 '15 at 12:25

Here is a way to derive the bivariate poisson distribution.

Let $X_0, X_1, X_2$ be independent poisson random variables with parameters $\theta_0, \theta_1, \theta_2$. Then we define $Y_1=X_0+X_1, Y_2 = X_0+X_2$. The variable $X_0$, common to both $Y_1$ an $Y_2$, causes the pair $(Y_1, Y_2)$ to be correlated. Then we must calculate the probability mass funtion:

\begin{align} P(Y_1=y_1, Y_2=y_2) &= P(X_0+X_1=y_1, X_0+X_2=y_2) \\ &= \sum_{x_0=0}^{\min(y_1, y_2)} P(X_0=x_0) P(X_1=y_1-x_0) P(X_2=y_2-y_0) \\ &= \sum_{x_0=0}^{\min(y_1, y_2)} e^{-\theta_0} \frac{{\theta_0}^{x_0}}{x_0!} e^{-\theta_1} \frac{{\theta_1}^{y_1-x_0}}{(y_1-x_0)!} e^{-\theta_2} \frac{{\theta_2}^{y_2-x_0}}{(y_2-x_0)!} \\ &= e^{-\theta_0-\theta_1-\theta_2} {\theta_1}^{y_1} {\theta_2}^{y_2} \sum_{x_0=0}^{\min(y_1,y_2)} \left(\frac{\theta_0}{\theta_1 \theta_2}\right)^{x_0} x_0! \binom{y_1}{x_0} \binom{y_2}{x_0} \end{align}
Hope this helps!

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Hi Kjetil--I fixed the problems with the $\TeX$ formatting (but, wishing to change as little as possible, left several typos intact). I do not understand why you are posting a replica of the derivation in my earlier answer, especially when you lost some crucial factors along the way which cause the final result to be incorrect. Is there a particular point you are trying to make? – whuber Jul 21 '14 at 15:23
whuber: I started to write my answer before your answer where posted! else, I would not have written it. – kjetil b halvorsen Jul 21 '14 at 15:26