When you make only one measurement, how do you determine 95% confidence interval for the Poisson distribution?
Let's say that I make one measurement of a quantity $x$ and obtain $x=10$. Also it is known that the population $x$ is Poisson-distributed. If I want to calculate 95% confidence interval for the population mean, then how can I find the corresponding z-score?
One says the margin of error for 95% is $1.96\lambda$. I wonder how to obtain the $1.96$. Please note that in the example above the sample size is one.
Best Answer
If $X \sim \mathsf{Pois}(\lambda).$ then $E(X) = \lambda$ and $SD(X) = \sqrt{\lambda}.$ For sufficiently large $\lambda,$ the random variable $X$ is approximately normally distributed. Then one says that $Z = \frac{X -\lambda}{\sqrt{\lambda}}$ is approximately standard normal, so that $$P\left(-1.96 < \frac{X -\lambda}{\sqrt{\lambda}} < 1.96\right) \approx 0.95.$$ This gives rise to $P(X - 1.96\sqrt{\lambda} < \lambda < X + 1.96\sqrt{\lambda})\approx0.95.$ Again, for sufficiently large $\lambda,$ one says that $1.96\sqrt{\lambda} \approx 1.96\sqrt{X}.$ So finally, an approximate 95% confidence interval for $\lambda$ is of the form $$(X - 1.96\sqrt{X},\;X + 1.96\sqrt{X}).$$
This type of interval was proposed by Wald as asymptotically accurate for $\lambda \rightarrow \infty.$ It works reasonably well for $\lambda > 50.$ For smaller $\lambda,$ a confidence interval with somewhat closer to 95% coverage is $$(X+2 - 1.96\sqrt{X+1},\; X+2 + 1.96\sqrt{X+1}).$$
For both styles of CIs, an approximate 90% confidence interval is shorter, using $\pm 1.645$ instead of $\pm 1.96.$
Because a Poisson distribution is discrete, actual coverage probabilities can vary by a surprising amount with a small change in the value of $\lambda.$ Here is a graph that shows actual coverage probabilities of the second (small $\lambda)$ type of "95%" confidence interval given above, for many values of $\lambda$ between $0.5$ and $30.$ For $\lambda > 5,$ coverage probabilities are not far from 95%.
The figure was made using the following R code:
Addendum: By contrast, making obvious minor changes in the program above, we have the following graph showing the true coverage probabilities of a "95%" Wald confidence interval for $\lambda.$