I have a sample $X=(X_1,\dots,X_n)$ of $i.i.d$ Poisson variables such that $n=100,\overline{X}=8.8$.

My goal is to obtain a $80\%$ confidence interval for the parameter $\lambda=\theta$. That is, the confidence level $\gamma=0.8$.

I have looked through Confidence interval for a small number of iid Poisson, 95% Confidence interval of $\lambda$ for $X_1,...,X_n$ IID exponential with rate $\lambda$, Math. Confidence interval of Poisson-distributed r.v., Math. An exact and an approximate interval for a Poisson distribution which provide some formula derivation as well as formulae that give the numeric answers. My problem is that in all derivations there are missing steps, which I cannot restore. Below is what I have done so far:

Consider the statistic $T(X)=n\overline{X}=\sum_{i=1}^nX_i$. $T$ is a sum of Poisson r.v. and $T\in\text{Poisson}(n\theta)$, the c.d.f. of which is $$F_{\theta}(t)=\sum_{k=0}^t\frac{(n\theta)^k}{e^{n\theta}k!}$$

Taking a derivative with respect to $\theta$ gives: $$\frac{\partial F_{\theta}(t)}{\partial\theta}=\frac{kn^k\theta^{k-1}e^{n\theta}-n(n\theta)^ke^{n\theta}}{e^{2n\theta}k!}=\frac{n^k\theta^{k-1}(k-n\theta)}{e^{n\theta}k!}<0\quad\text{for }k<n\theta$$ which means that for such $k$ $F_{\theta}(T)>F_{\theta}(T+0)$.

The confidence interval is given by $$(\underline{\theta_n},\overline{\theta_n})=\{\theta: \gamma_1<F_{\theta}(T+0),F_{\theta}(T)<\gamma_2\}$$ where $\gamma_1,\gamma_2\in(0,1):\gamma_2-\gamma_1=\gamma$.

By definition of the c.d.f., \begin{align*}&F_{\theta}(T)=\sum_{k=0}^{n\overline{X}-1}\frac{(n\theta)^k}{e^{n\theta}k!}=\gamma_2\\&F_{\theta}(T+0)=\sum_{k=0}^{n\overline{X}}\frac{(n\theta)^k}{e^{n\theta}k!}=\gamma_1\end{align*} Using the relationship between $\text{Poisson}$ and $\chi^2$, \begin{align*}&F_{\theta}(T)=1-H_{2n\overline{X}}(2n\theta)\\&F_{\theta}(T+0)=1-H_{2(n\overline{X}-1)}(2n\theta)\end{align*} where $H_{2n\overline{X}}(x)$ is the c.d.f. of a $\chi^2(2n\overline{X})$ r.v. Assuming a symmetric CI?, \begin{align*}&1-F_{\theta}(T)=H_{2n\overline{X}}(2n\theta)=\frac{1-\gamma}{2}\quad\dagger\\&F_{\theta}(T+0)=1-H_{2(n\overline{X}-1)}(2n\theta)=\frac{1-\gamma}{2}\quad\ddagger\end{align*} which gives \begin{align*} &\overline{\theta_n}=\frac{1}{2n}\chi^2_{\frac{1+\gamma}{2}}(2n\overline{X})\\ &\underline{\theta_n}=\frac{1}{2n}\chi^2_{\frac{1-\gamma}{2}}(2n\overline{X}-2) \end{align*} I know how to use this formula to compute the CI in R, but I don't know how to argue to explain the transition to expressions $\dagger,\ddagger$. I also don't understand how to obtain the approximate interval $$\big(\overline{X}-z_{\gamma}\sqrt{\frac{\overline{X}}{n}}, \overline{X}+z_{\gamma}\sqrt{\frac{\overline{X}}{n}}\big)$$ where $z_{\gamma}$ is the $\gamma-$quantile of the standard normal distribution. I know I can view the $\chi^2$ distribution as a sum of squared normals that, by CLT, will tend to normal, but this argument doesn't seem to give the solution I need.

My questions are,

  1. How can I justify the transition to $\dagger,\ddagger$?
  2. Is there a way to derive the approximate CI from where I am now? What should I do to get this result?
  • 1
    $\begingroup$ See. The 'Wald' CI in your last display requires large $\bar X.$ There are many styles of CIs for Poisson $\lambda,$ each with its own advantages/disadvantages. $\endgroup$
    – BruceET
    Apr 23 at 14:49

1 Answer 1


In much the same way that the Agresti-Cooll CI for binomial $p$ approximately inverts the normal test for $H_0: p = p_0$ vs. $H_a: \ne,$ the following 95% CI for $\lambda$ approximately inverts the normal test for $H_0: \lambda = \lambda_0$ vs $H_a: \ne.$

Example: Suppose you have $X_1, \dots, X_n$ iid $\mathsf{Pois}(\lambda),$ so that $$T = \sum_{i=1}^n X_i \sim \mathsf{Pois}(n\lambda).$$ Then an approximate 95% CI for $\lambda$ is $$(T + 2)/n \pm (1.96\sqrt{T+1})/n.$$

set.seed(2022);  lam=10; n = 50
x = rpois(n, lam)
t = sum(x)
CI = (t + 2 + qnorm(c(.025,.975))*sqrt(t+1))/n
[1]  9.373022 11.146978  ## contains lam=10

Under CIs, the Wikipedia page on Poisson distributions suggests the CI with endpoints computed in R below:

 .5*qchisq(.025, 2*t)/n; .5*qchisq(.975, 2*t + 2)/n
 [1] 9.352981
 [1] 11.14577

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.