# Solve for the covariance in the bivariate Poisson distribution

• GeorgeK
This means that for the bivariate Poisson distribution given above, the correlation between the two variables is bounded by the square root of the product of the two thetas, which are the parameters of the distribution. In summary, the correlation between the two variables in this distribution has a maximum and minimum bound based on the square root of the product of the parameters.

#### GeorgeK

Dear All,

The bivariate Poisson distribution is as follows,
$$$f(y_{s},y_{t})=e^{-(\theta_{s} + \theta_{t}+\theta_{st})}\frac{\theta_{s}^{y_{s}}}{y_{s}!}\frac{\theta_{t}^{y_{t}}}{y_{t}!} \sum_{k=0}^{min(y_{s},y_{t})} \binom{y_{s}}{k} \binom{y_{t}}{k} k!\left(\frac{\theta_{st}}{\theta_{s} \theta_{t}}\right)^k.$$$

Given that $$f(y_{s},y_{t}) >= 0$$, solve for $$\theta_{st}$$.

George

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Can someone please advise or give a comment or ask for more information if my question is not clear? I urgently/desperately need to know if this is solveable and how?

George

I am not familiar with the equation, but I wonder if there is a "solution". If the variables are correlated, the correlation (within bounds) is anything.

mathman said:
I am not familiar with the equation, but I wonder if there is a "solution". If the variables are correlated, the correlation (within bounds) is anything.

Right mathman (and of course to everyone else),

I am actually interested in finding what the bounds would be for the correlation but then [I thought] I first need to solve for the covariance. So, the reformed question is:

What are the bounds (maximum and minimum) for the correlation based on this bivariate Poisson Distribution?

George

In general, the absolute value of a covariance is bounded by the square root of the product of the variances.