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I would like to compute the power of two distributions which are poisson with $n_1$ being large and $n_2$ being medium sized. I have used the formula given by Krishnamoorthy and Thomson for the C-test (see below).

$$\sum^\infty_{k_1=0}\sum^\infty_{k_2=0}\frac{e^{-n_1\lambda_1}(n_1\lambda_1)^{k_1}}{k_1!}\frac{e^{-n_2\lambda_2}(n_2\lambda_2)^{k_2}}{k_2!}I[P(X_1\geq k_1|k_1+k_2, p(\lambda_1/ \lambda_2))\leq\alpha]$$

However, as $n_1$ or $n_2$ get large the exponential function goes to $0$ which makes taking the sum of the probability rather pointless.

Is there anyway I can compute the power with having an large uneven sample size?

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  • $\begingroup$ it seems no one can answer the question, it is in the end a computational problem I assume since I cannot calculate the exponential function for very large numbers [even with range reduction]. Also I am limited by computational estimations of $k!$. I will leave it as it is and hope someone has an idea. $\endgroup$ – DUWUDA Jul 24 '14 at 3:19
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    $\begingroup$ The authors of the paper give a straightforward method for doing the computation of the Poisson density that should apply equally well for computing their Equation 4.1. They also discuss where to stop calculating.The only thing left is to evaluate the indicator function, which just really needs the cumulative distribution function of the binomial distribution, also pretty straightforward---actually their discussion touches on how to do that a bit more efficiently, too. $\endgroup$ – jvbraun Nov 5 '14 at 22:30

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