# Determining confidence interval with one observation (for Poisson distribution)

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.

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 $$(X+2 - 1.96\sqrt{X+1},\; X+2 + 1.96\sqrt{X+1}).$$

Rationale: This adjusted 95% interval for smaller $$\lambda$$ is based on 'inverting' a standard test for $$H_0: \lambda = \lambda_0$$ vs. $$H_a: \lambda \ne \lambda_0,$$ with test statistic $$Z = \frac{X - \lambda_0}{\sqrt{\lambda_0}},$$ which rejects at the 5% level for $$|Z| \ge 1.96.$$ Specifically for given $$X,$$ the adjusted interval is found by solving a quadratic inequality for values $$\lambda_0$$ with $$|Z| < 1.96$$ and conflating $$1.96$$ with $$2$$ to obtain the terms with $$X + 2$$ and $$X+ 1.$$ In effect, the adjusted CI consists of non-rejectable hypothetical values of $$\lambda_0.$$ (One still assumes that $$Z$$ is approximately standard normal, but the additional assumption that $$1.96\sqrt{\lambda} \approx 1.96\sqrt{X}$$ is no longer required.)

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:

lam = seq(.5, 30, by=.0001); m = length(lam) # values of lambda
t = 0:200 # realistic values of T
LL = t+2 - 1.96*sqrt(t+1); UL = t+2 + 1.96*sqrt(t+1) # corresp. CIs
cov.pr = numeric(m)
for(i in 1:m) {
lam.i = lam[i] # pick a lambda
cov = (lam.i >= LL & lam.i <= UL) # TRUE if CI covers
cov.pr[i] = sum(dpois(t[cov], lam.i)) } # sum probs for covering T's
plot(lam, cov.pr, type="l", ylim=c(.8,1), lwd=2, xaxs="i")
abline(h=.95, col="green3")


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 $$\lambda.$$ 