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Solving an exercise I found myself with this problem: the solution $c$ needs to verify both $$\sum_{k=1}^c n\lambda^k\frac{e^{-n\lambda}}{k!}\leq \alpha$$ and $$1-\sum_{k=1}^{c} n\lambda^k\frac{e^{-n\lambda}}{k!}\geq \alpha$$

The values $n,\lambda,\alpha$ are known. Can an equation like this be solved for $c$?

Cure
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  • Yes of course.... – Brethlosze Oct 20 '17 at 02:14
  • @hyprfrcb: I'm not confident these inequalities can be "solved for $c$". It is not stated what values $n,\lambda,\alpha$ are, or even that their values are known to someone who is asked to solve for $c$. – hardmath Oct 20 '17 at 02:34
  • @hardmath I edited that. The values $n,\lambda,\alpha$ are known. Also, there was a typo, the denominator is $k!$ instead of $k$. – Cure Oct 20 '17 at 05:41
  • So with $C := n e^{n \lambda}$ you try to solve $C \sum_{k=1}^c \frac{\lambda^k}{k!} \le \min (\alpha, 1-\alpha)$ ? – Martin R Oct 20 '17 at 10:58
  • @MartinR Yes, that's what I want to solve. – Cure Oct 20 '17 at 11:10
  • Is $\lambda$ positive? Then the series is increasing, and a solution would exist exactly if $C \lambda \le \min(\alpha, 1 - \alpha)$ – Martin R Oct 20 '17 at 11:14
  • @MartinR I don't know if $\lambda$ is positive in this case. This is part of a series of equations I need to solve, for general $\lambda$, but the inequation I posted were found under the assumption $\lambda>0$. The inequalities invert if $\lambda < 0$. Is there a closed form tor the solution? – Cure Oct 20 '17 at 11:46
  • Closed forms for the partial sums of the exponential series are not "nice", see for example https://math.stackexchange.com/questions/50746/general-term-formula-of-s-n-sum-limits-i-1n-fracaii. – Martin R Oct 20 '17 at 11:48
  • The second summation originally ran to upper limit $c+1$, but now to $c$ (just like the first summation). Is this perhaps an unintentional edit? – hardmath Oct 20 '17 at 12:02
  • @hardmath It's not. I changed it because I was supposed to have $\text{Pr}(\sum_{i=1}^{n}X_i\geq c+1 |\lambda=\lambda_0)>\alpha$ which I took to be $1-\text{Pr}(\sum_{i=1}^{n}X_i\leq c |\lambda=\lambda_0)=1-\sum_{k=1}^{c} n\lambda^k\frac{e^{-n\lambda}}{k!}\geq \alpha$ because the $X_i$ are independent Poisson distributed variables. – Cure Oct 20 '17 at 15:26

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