A Worked Poisson Rate Interval on the Square-Root Scale

Why Use a Square-Root Scale?

For a Poisson count, the variance equals the mean. Consequently, when exposure is fixed, the variability of a rate estimate depends on the rate itself. A square-root transformation addresses that changing variability: to first order, the transformed estimate has a variance that does not depend on the unknown rate. This makes it possible to construct an interval on a more stable scale and then return its limits to the rate scale.

The result is an asymptotic interval, not an exact Poisson interval. Its justification concerns increasing exposure when the true rate is positive. The distinction matters especially when the observed count is zero or small.

Model and Transformation

Let $Y$ be the count of events observed over known exposure $t>0$. Assume $Y\sim\operatorname{Poisson}(t\lambda)$, where $\lambda\geq 0$ is the event rate per exposure unit. The exposure might be time at risk, person-time, or another known quantity appropriate to the study.

Definition: Under the Poisson rate model, the rate estimate is $\hat{\lambda}=Y/t$. The square-root transformation used here is $g(\lambda)=2\sqrt{\lambda}$, so the transformed estimate is $g(\hat{\lambda})=2\sqrt{Y/t}$. The factor of 2 gives the transformed estimate an asymptotic variance of $1/t$.

For $Y=y>0$, the likelihood as a function of $\lambda$ is proportional to $\lambda^y e^{-t\lambda}$, and its maximum occurs at $y/t$. For $y=0$, the likelihood is $e^{-t\lambda}$, which is maximized at the boundary $\lambda=0$. Thus $\hat{\lambda}=Y/t$ is the maximum likelihood estimator on the parameter space $\lambda\geq0$, including when the count is zero.

The Poisson mean and variance give $E[\hat{\lambda}]=\lambda$ and $\operatorname{Var}(\hat{\lambda})=\lambda/t$. The derivative of $g$ at a positive rate is $g'(\lambda)=1/\sqrt{\lambda}$. The first-order variance calculation is therefore

\bigl(g'(\lambda)\bigr)^2\operatorname{Var}(\hat{\lambda}) =\frac{1}{\lambda}\frac{\lambda}{t} =\frac{1}{t}.

The rate cancels. This is the variance-stabilizing property that motivates the interval. Because the transformed parameter is $2\sqrt{\lambda}$, its natural range is $[0,\infty)$, and its inverse is $g^{-1}(u)=u^2/4$ for $u\geq0$.

Asymptotic Distribution and Interval

Consider increasing exposure $t\to\infty$ while the true rate $\lambda>0$ remains fixed. The Poisson central limit theorem gives $(Y-t\lambda)/\sqrt{t\lambda}\xrightarrow{d}N(0,1)$. Equivalently, $\sqrt{t}(\hat{\lambda}-\lambda)\xrightarrow{d}N(0,\lambda)$. Applying the delta method to $g(\lambda)=2\sqrt{\lambda}$ yields the following result.

Theorem: If $Y\sim\operatorname{Poisson}(t\lambda)$, $t\to\infty$, and $0<\lambda<\infty$ is fixed, then $$ \sqrt{t}\left(2\sqrt{\hat{\lambda}}-2\sqrt{\lambda}\right) \xrightarrow{d}N(0,1). $$ Thus the transformed estimate $2\sqrt{\hat{\lambda}}$ has first-order asymptotic standard error $1/\sqrt{t}$.

Derivation. Since $\sqrt{t}(\hat{\lambda}-\lambda)\xrightarrow{d}N(0,\lambda)$, the delta method applies because $g$ is differentiable at the fixed positive $\lambda$. Its derivative is $1/\sqrt{\lambda}$, so

\sqrt{t}\bigl(g(\hat{\lambda})-g(\lambda)\bigr) \xrightarrow{d} N\left(0,\bigl(g'(\lambda)\bigr)^2\lambda\right) =N(0,1).

This establishes the stated limit distribution. In particular, it explains why the standard error on the transformed scale is $1/\sqrt{t}$, not a value that needs the unknown rate. $\square$

Let $z_{1-\alpha/2}$ denote the $1-\alpha/2$ quantile of the standard normal distribution. An asymptotic $100(1-\alpha)\%$ interval on the transformed scale is $2\sqrt{\hat{\lambda}}\pm z_{1-\alpha/2}/\sqrt{t}$. To return to the rate scale, first keep the transformed limits in the valid range $[0,\infty)$, then apply the increasing inverse $u^2/4$.

Rate interval: For a two-sided asymptotic interval with normal critical value $z=z_{1-\alpha/2}$, the back-transformed limits are $$ \left[ \frac{\{\max(0,\,2\sqrt{Y/t}-z/\sqrt{t})\}^{2}}{4}, \quad \frac{(2\sqrt{Y/t}+z/\sqrt{t})^{2}}{4} \right]. $$

The lower transformed endpoint is truncated at zero because a negative value is outside the range of $2\sqrt{\lambda}$. This truncation does not remove the true transformed rate, which is nonnegative. The upper endpoint is positive, so no corresponding adjustment is needed there.

Worked Calculations

Worked Example: 25 Events Over 10 Exposure Units

Suppose $Y=25$ events are observed over $t=10$ exposure units. Construct the approximate 95% interval, using $z_{0.975}=1.96$.

The estimated rate is $\hat{\lambda}=25/10=2.5$ events per exposure unit. Its transformed estimate is $2\sqrt{2.5}=\sqrt{10}=3.16227766\ldots$, and the transformed-scale standard error is $1/\sqrt{10}=0.31622777\ldots$.

The margin is $1.96/\sqrt{10}=0.61967734\ldots$. Keeping the intermediate precision, the transformed limits are $3.16227766\ldots-0.61967734\ldots=2.54260032\ldots$ and $3.16227766\ldots+0.61967734\ldots=3.78195499\ldots$.

Both limits are positive. Squaring each and dividing by 4 gives $2.54260032^2/4\approx1.6162$ and $3.78195499^2/4\approx3.5758$. The approximate 95% rate interval is therefore $[1.6162,\,3.5758]$ events per exposure unit. The unrounded transformed margin is used in these calculations; rounding it early can change the displayed endpoints.

Worked Example: 9 Events Over 4 Exposure Units

Now let $Y=9$ and $t=4$. The estimated rate is $9/4=2.25$, so the transformed estimate is $2\sqrt{2.25}=3$. With $z=1.96$, the transformed standard error is $1/\sqrt{4}=0.5$, and the margin is $1.96(0.5)=0.98$.

The transformed limits are $3-0.98=2.02$ and $3+0.98=3.98$. Both are already in the transformation's range. Back-transforming gives $2.02^2/4=4.0804/4=1.0201$ and $3.98^2/4=15.8404/4=3.9601$.

Thus the approximate 95% interval is $[1.0201,\,3.9601]$. Its midpoint is not the rate estimate: back-transformation generally produces an interval that is not symmetric around $\hat{\lambda}=2.25$.

Worked Example: A Zero Count and a Boundary Limit

Suppose no events are observed over $t=20$ exposure units. The rate estimate is $\hat{\lambda}=0/20=0$, and its transformed estimate is $2\sqrt{0}=0$. The nominal transformed standard error formula gives $1/\sqrt{20}=0.2236068\ldots$. Using $z=1.96$, the untruncated transformed limits would be $-1.96/\sqrt{20}\approx-0.4383$ and $1.96/\sqrt{20}\approx0.4383$.

The lower limit must be set to zero, since transformed rates cannot be negative. The upper limit back-transforms to $0.4383^2/4\approx0.0480$, or, retaining the formula before rounding, $(1.96/\sqrt{20})^2/4=1.96^2/(80)=3.8416/80=0.04802$. The formula therefore produces $[0,\,0.04802]$.

This calculation illustrates the boundary adjustment; it does not establish that the interval has reliable 95% coverage in this case. The normal approximation was justified for a fixed positive rate as exposure increases, not at the boundary $\lambda=0$. A zero observed count can also occur at a positive rate, but the plug-in transformed estimate is then at the boundary. Interpret this output cautiously rather than treating it as an exact interval.

Why the Interval Has Its Stated Coverage

For fixed $\lambda>0$, write $\theta=2\sqrt{\lambda}$ and $\hat{\theta}=2\sqrt{\hat{\lambda}}$. The theorem gives $\sqrt{t}(\hat{\theta}-\theta)\xrightarrow{d}N(0,1)$. The probability that the untruncated transformed interval contains $\theta$ is

P\left(\left|\sqrt{t}(\hat{\theta}-\theta)\right|\leq z_{1-\alpha/2}\right).

The standard normal distribution is continuous at both endpoints, so convergence in distribution implies that this probability tends to $P(|Z|\leq z_{1-\alpha/2})=1-\alpha$, where $Z\sim N(0,1)$. Intersecting the interval with $[0,\infty)$ does not change whether it contains $\theta$, because $\theta\geq0$. Finally, $u\mapsto u^2/4$ is increasing on $[0,\infty)$, so back-transformation preserves containment of the true value. This proves the interval's asymptotic coverage for fixed $\lambda>0$. $\square$

Common Mistake and Practical Interpretation

A tempting but incorrect shortcut is to form a symmetric Wald interval directly for the rate using the estimated standard error $\sqrt{\hat{\lambda}/t}$, then assume it is equivalent to the square-root interval. It is not: the two procedures use different scales and generally give different endpoints. The purpose of this construction is to make the interval on the transformed scale first, then invert that scale's limits. It is also a mistake to square a negative transformed lower endpoint without adjustment: squaring would turn a value outside the transformation's range into a seemingly valid positive rate limit.

The interval is useful when the Poisson model is appropriate, exposure is known, and the expected count is large enough for the normal approximation to be credible. It is not an exact solution to sparse-count inference, and variance stabilization alone does not guarantee a good normal approximation for every observed sample. For a zero or very small count, the boundary and the approximation deserve explicit attention.

Check Your Understanding

Check that you can explain both the calculation and the conditions behind it.

For a Poisson count $Y$ over exposure $t$, what is the rate estimate, and what is the transformed estimate used here?
Why does the asymptotic standard error of $2\sqrt{\hat{\lambda}}$ depend on exposure but not on the unknown positive rate?
When a transformed lower endpoint is negative, what adjustment is made before back-transforming, and why?
For $Y=16$ and $t=8$, write the transformed estimate and the 95% transformed-scale margin using $z=1.96$.
Why should the zero-count calculation not automatically be interpreted as a valid 95% asymptotic interval?