Poisson confidence interval using the pivotal method I am trying to build a confidence interval for the Poisson distribution using the pivotal method. I have the theory down but I am struggling to come up with $h(Y, \lambda)$, the probability distribution which does not depend on the parameter. 
Can anyone help me out?
 A: As far as I know, there isn't really a pivotal quantity for $\lambda$*, though it's possible to construct approximately pivotal quantities if $n\lambda$ isn't small. (I include $n$ there just in case you have multiple observations from the same Poisson. In many cases you'll just have the one count. From here on I'll just refer to $X$ and $\lambda$ as if we were using a single $X$.)
* However, that's not to say nothing can be done. You can derive an interval for $\lambda$ from the relationship between the Poisson and the chi-square (see the end of this section). I can't say that it counts as pivotal, though.
For example, $\sqrt{X+\frac{3}{8}}$ is approximately normal with nearly constant variance (this is related to the Anscombe transform for the Poisson), so could be used to construct an approximate pivotal quantity (by subtracting its mean for example); $\sqrt{X}+\sqrt{X+1}$ - Freeman-Tukey - is another, similar choice from which you could obtain an approximately pivotal quantity. (Indeed, the confidence interval here relies on just such an approach)
To my recollection, some papers have given other quantities - if I remember where I've seen these, I'll add references.
With large values of $\lambda$, the simpler $\frac{X-\lambda}{\sqrt \lambda}$ might be used as an approximately pivotal quantity.
But with small $\lambda$ (more generally small $n\lambda$), there's really not much that can be done - I don't think you get a function of the parameter and the data whose distribution doesn't depend on the parameter. If your $\lambda$ is down around 0.5 or 1 or 2, say, there's not much you can really do about it... the large spikes at 0, 1 and 2 change substantially in relative probability with $\lambda$ and no transformation is going to alter that.
