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I want to calculate the confidence interval for $\lambda$ from a small ($n=10$) set of repeated observations from a Poisson distribution. That is, I have $X_1, \dots, X_{10}$ which I believe are i.i.d. and I want a 95% confidence interval given these observations for $\lambda$.

I know I can use $\bar x \pm 1.96\frac{s}{\sqrt{n}}$ when $n$ is large but that doesn't seem to be a good assumption here. I also know for a single instance $y$ you can use 

qchisq(0.025,2*y)/2, qchisq(0.975,2*(y+1))/2

so my problem appears to be somewhere between the two.

Hugh
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  • Not sure whether the $n$ large c.i. is mean to be $\bar x\pm1.96 \frac{s}{\sqrt{n}}$ or $\bar x\pm 1.96\sqrt{\lambda/n}$. Not confident enough to accept or reject edit. – Hugh Apr 17 '13 at 03:27

1 Answers1

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Since it is small sample you can calculate exact distribution to arrive at confidence interval.

$Z= X_1+ \cdots + X_{10} \sim $ Poisson($10 \lambda$). The mle of $\lambda = \hat{\lambda} = \bar{X}= Z/10$. From here you can find the confidence interval.

EDIT: Confidence interval

$Z= 10* \bar{X} \sim$ Poisson($10 \lambda$). Let $\lambda_L$ and $\lambda_U$ be the confidence limits of $\lambda$. And $\lambda_{L^{*}}$ and $\lambda_{U^{*}}$ be the confidence limits of $10\lambda$.

$ \lambda_{U^{*}}=10\lambda_U, \quad $ and $ \lambda_{L^{*}}=10\lambda_L \quad $

Then

$\exp(-\lambda_{L^{*}}) \sum\limits_{j=z}^{\infty}\frac{\lambda_{L^{*}}^j}{j!}= \frac{\alpha}{2}, \quad $ and $ \quad \exp(-\lambda_{U^{*}}) \sum\limits_{j=0}^{z}\frac{\lambda_{U^{*}}^j}{j!}= \frac{\alpha}{2}$

From the relationship between poisson and gamma (chisqure) it can write as

$10*\lambda_L =0.5 \chi^2_{2z,\alpha/2}, \quad$ $10*\lambda_L =0.5 \chi^2_{2(z+1),1-\alpha/2}$ which leads the CI of $\lambda$ as

(qchisq(0.025,2*z)/(2*10), qchisq(0.975,2*(z+1))/(2*10))

vinux
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  • So the confidence interval is $\frac{`qchisq(0.025,2*z)/2`}{10} \leq \lambda \leq \frac{`qchisq(0.975,2*(z+1))/2`}{10}$ – Hugh Apr 17 '13 at 03:27
  • How are you connecting poisson with chi-square? Use `qpois` – vinux Apr 17 '13 at 04:26
  • http://en.wikipedia.org/wiki/Poisson_distribution#Confidence_interval The wikipedia page says that the $\chi^2$ method is the way to calculate confidence intervals. – Hugh Apr 17 '13 at 22:48
  • The expression in wiki is for one observation. I will search the expression for sample of size 10 and come back to you. I will edit my answer and add the solution. – vinux Apr 18 '13 at 03:08
  • Your solution is correct. Sorry if I confuse you. – vinux Apr 18 '13 at 03:24