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If $Y$ is Poisson$(\lambda)$ distributed, how do we obtain this confidence interval for $\lambda$:

$$ \left (\frac{(\sqrt Y+\sqrt{Y + 1}-z_{(α⁄2)} )^2 - 1}{4},\ \frac{(\sqrt Y+\sqrt{Y +1}+z_{(α⁄2)} )^2 - 1}{4} \right ) $$

Glen_b
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Raziye
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  • The only reference I can find with this formula (Patil&Kulkarni, 2012, *Comparison of confidence intervals for the Poisson mean*, REVSTAT) simply mentions the original paper by Freeman and Tukey, which itself doesn't contain this formula, nor, at first sight, anything that would obviously give it. It may be that it's possible to derive it as a limiting case of the binomial interval there, but Molenaar (1973, Biometrics, "*A simple approximation to the Poisson, Binomial, ...*") offers a different interval based on the limit of the binomial case in Freeman and Tukey ...ctd – Glen_b Jun 30 '16 at 00:48
  • ctd... so I am unconvinced that it comes directly from there (but maybe can be had via a similar argument). – Glen_b Jun 30 '16 at 00:51
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    Not an answer, but some additional information in regards to the OP's comments: 1. A definition of the FT CI appears here, along with a discussion of/comparison with other Poisson CIs: [Byrne and Kabaila](http://www.tandfonline.com/doi/pdf/10.1081/STA-200052109) 2. The original paper by Freeman and Tukey can be found here: [Freeman and Tukey](https://projecteuclid.org/download/pdf_1/euclid.aoms/1177729756). The term $\sqrt{Y} + \sqrt{Y+1}$ derives from the variance stabilising transformation. – Maurits Evers Jun 30 '16 at 00:51
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    The Byrne and Kabaila reference in Maurits' comment gives the above interval, and also refers to Freeman and Tukey (1950) for it, so presumably it does follow directly from what is there. However, as mentioned above I didn't see the direct connection at first glance. I might get time to have a closer look later. – Glen_b Jun 30 '16 at 01:05
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    I think I may have it: there's an approximation $E(\sqrt{Y}+\sqrt{Y+1})\approx \sqrt{4\lambda+1}$. This is not explicitly given in Freeman and Tukey, 1950, but is mentioned in other works where Tukey is an author or co-author, for example in connection with the (closely related) double root residual. However, perhaps it can be derived from the binomial case in F&T 1950. That's sufficient to derive the interval. If nobody beats me to it I'll try to write it up as an answer. – Glen_b Jun 30 '16 at 01:20

1 Answers1

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Taking these as given:

$E(\sqrt{Y}+\sqrt{Y+1})\approx \sqrt{4\lambda+1}$

$\text{Var}(\sqrt{Y}+\sqrt{Y+1})\approx 1$

and assuming $\lambda$ is large enough that $(\sqrt{Y}+\sqrt{Y+1})$ is approximately normal, then

$P(-z_{1-\alpha/2}<\sqrt{4\lambda+1}-(\sqrt{Y}+\sqrt{Y+1})<z_{1-\alpha/2})\approx 1-\alpha$

$P((\sqrt{Y}+\sqrt{Y+1})-z_{1-\alpha/2}<\sqrt{4\lambda+1}<(\sqrt{Y}+\sqrt{Y+1})+z_{1-\alpha/2})\approx 1-\alpha$

Further assume that $\lambda$ and $\alpha$ are large enough that the lower limit ($(\sqrt{Y}+\sqrt{Y+1})-z_{1-\alpha/2}$) is non-negative. For $\alpha=0.05$ this means $Y$ should be at least $1$ (though for a reasonably good approximation you'll want it larger than that anyway) -- and the case $Y=0$ would require a different approximation, as it does for the double root residual. If your observed $Y$ is $0$ you simply shouldn't be using this sort of approximation there (and down that far any hope of a two-tailed $1-\alpha$ interval with $\alpha$ at typical values is gone as well).

Then we can square all the terms without complication (we won't be flipping any inequalities, having to consider separate cases for $\pm$, or introducing any extraneous solutions):

$P((\sqrt{Y}+\sqrt{Y+1}-z_{1-\alpha/2})^2<{4\lambda+1}<(\sqrt{Y}+\sqrt{Y+1}+z_{1-\alpha/2})^2)\approx 1-\alpha$

$P(\frac14 [(\sqrt{Y}+\sqrt{Y+1}-z_{1-\alpha/2})^2-1]<{\lambda}<\frac14[(\sqrt{Y}+\sqrt{Y+1}+z_{1-\alpha/2})^2-1]\approx 1-\alpha$

which is the required interval.

Note that since $\lambda$ must be positive, we now require $Y$ to be at least 2 to get even a feasible answer for the lower limit at the 5% level (though it won't give a very good approximation for the coverage down that far) -- don't try to use this at all when $Y\leq 1$.

Indeed I'd suggest using it when $Y\leq 4$ or so isn't a great idea; once you get much above $4$ it tends to work at least reasonably well at typical levels and nearby $\lambda$ values. (Your own utility function or circumstances may vary, don't take that value as in any way prescriptive or even worth quoting).

Glen_b
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  • Thaaaaaaaaaaaaaaaaaaank you!!!!! I don't know how to thank you!!! Really really really thank you!!! @Glen_b – Raziye Jul 13 '16 at 16:07