Solve for the covariance in the bivariate Poisson distribution

[GeorgeK]

Dear All,

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

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

Many thanks in advance,

George

 

[GeorgeK]

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?

Many thanks in advance,

George

 

[mathman]

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.

 

[GeorgeK]

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

 

[mathman]

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