Why Transform a Binomial Proportion?
If \(X\) successes are observed in \(n\) independent Bernoulli trials with success probability \(p\), the sample proportion \(\hat p=X/n\) estimates \(p\). Its variance is \(p(1-p)/n\), so the uncertainty in \(\hat p\) depends on the unknown probability and is smallest near zero or one. The arcsine square-root transformation puts the probability on a scale with approximately constant variance, making a simple normal interval possible.
This tutorial constructs that interval from an observed count, then converts its limits back to probabilities. The result is a first-order approximate confidence interval, not an exact binomial interval. Its normal approximation is most credible when the probability is away from the boundaries and the sample contains enough successes and failures.
Transformation and Approximate Standard Error
For \(0<p<1\), differentiation gives \(g'(p)=1/\sqrt{p(1-p)}\). Since \(\operatorname{Var}(\hat p)=p(1-p)/n\), the delta method gives the first-order variance of \(\hat\theta\) as \([g'(p)]^2\operatorname{Var}(\hat p)=1/n\). Thus the approximate standard error on the transformed scale is \(1/\sqrt n\), independent of \(p\).
Proof. For fixed \(0<p<1\), the binomial central limit theorem gives \(\sqrt n(\hat p-p)\xrightarrow{d}N(0,p(1-p))\). The function \(g\) is differentiable at \(p\), with derivative \(1/\sqrt{p(1-p)}\). By the delta method, \(\sqrt n\{g(\hat p)-g(p)\}\xrightarrow{d}N(0,[g'(p)]^2p(1-p))=N(0,1)\). The limiting variance is one, so the transformed estimator has first-order variance \(1/n\) and standard error \(1/\sqrt n\). \(\square\)
The condition \(0<p<1\) matters: the derivative used in the delta method is not finite at the boundaries. The constant standard error is a first-order result for interior probabilities; using it at an observed boundary count is a practical convention for forming the transformed interval, not a boundary version of this theorem.
Constructing and Back-Transforming the Interval
Let \(z_{1-\alpha/2}\) be the \(1-\alpha/2\) quantile of the standard normal distribution. The usual approximate \(1-\alpha\) interval on the transformed scale has untruncated limits \(\hat\theta-z_{1-\alpha/2}/\sqrt n\) and \(\hat\theta+z_{1-\alpha/2}/\sqrt n\). Because a transformed probability must lie in \([0,\pi]\), truncate these limits to that range:
The inverse transformation is continuous and increasing on \([0,\pi]\). Therefore, the corresponding probability-scale interval is the closed interval \([\sin^2(L_\theta/2),\sin^2(U_\theta/2)]\). Both endpoints are included. Truncation prevents a probability limit outside \([0,1]\); it does not make the normal approximation exact.
The coverage correspondence follows directly from monotonicity. For any fixed true \(p\in[0,1]\), write \(\theta=g(p)\in[0,\pi]\). The event that \(p\) belongs to the back-transformed interval is exactly the event that \(\theta\) belongs to the clipped transformed interval:
To check clipping, let \(a=\hat\theta-z_{1-\alpha/2}/\sqrt n\) and \(b=\hat\theta+z_{1-\alpha/2}/\sqrt n\). Since \(\theta\in[0,\pi]\), membership in \([\max(0,a),\min(\pi,b)]\) is equivalent to \(a\le\theta\le b\): clipping either endpoint to the parameter range cannot exclude a value of \(\theta\) already in that range. The inverse transformation preserves order and includes both endpoints, proving the equivalence. Thus the back-transformed interval has the same approximate coverage as the transformed-scale normal interval; its coverage is not thereby guaranteed to equal \(1-\alpha\) in a finite sample. \(\square\)
Worked Binomial Intervals
Worked Example: 36 Successes in 100 Trials
Suppose \(x=36\) successes are observed in \(n=100\) trials. Construct a 95% interval, using \(z_{0.975}=1.96\).
The observed proportion is \(\hat p=36/100=0.36\). Therefore \(\hat\theta=2\arcsin\sqrt{0.36}=2\arcsin(0.6)\approx1.2870\) radians. The transformed standard error is \(1/\sqrt{100}=0.1\), so the margin is \(1.96(0.1)=0.196\).
The transformed limits are \(1.2870-0.196=1.0910\) and \(1.2870+0.196=1.4830\). Both are already in \([0,\pi]\). Back-transforming gives \(\sin^2(1.0910/2)=\sin^2(0.5455)\approx0.2692\) and \(\sin^2(1.4830/2)=\sin^2(0.7415)\approx0.4562\).
The approximate 95% probability-scale interval is therefore \([0.2692,0.4562]\). Its endpoints are included; the interval is not the open interval between those values.
Worked Example: 4 Successes in 25 Trials
Now let \(x=4\) and \(n=25\), again using \(z_{0.975}=1.96\). Here \(\hat p=4/25=0.16\), and \(\hat\theta=2\arcsin\sqrt{0.16}=2\arcsin(0.4)\approx0.8230\).
The transformed standard error is \(1/\sqrt{25}=0.2\), giving a margin of \(1.96(0.2)=0.392\). The untruncated transformed limits are \(0.8230-0.392=0.4310\) and \(0.8230+0.392=1.2150\). Neither requires truncation.
The probability limits are \(\sin^2(0.4310/2)=\sin^2(0.2155)\approx0.0457\) and \(\sin^2(1.2150/2)=\sin^2(0.6075)\approx0.3258\). Thus the approximate interval is \([0.0457,0.3258]\).
The calculations follow the same recipe, but the count is small: there are only four successes. The interval’s numerical construction does not establish that its finite-sample coverage is close to 95%; the normal approximation should be treated cautiously here.
Worked Example: Zero Successes in 20 Trials
Suppose \(x=0\) out of \(n=20\) trials, and use \(z_{0.975}=1.96\). Then \(\hat p=0\) and \(\hat\theta=2\arcsin(0)=0\). The transformed standard error used in the interval recipe is \(1/\sqrt{20}\approx0.2236\), so the margin is \(1.96/\sqrt{20}\approx0.4383\).
Before truncation the transformed limits are \(-0.4383\) and \(0.4383\). Truncating to \([0,\pi]\) gives the closed interval \([0,0.4383]\). The probability limits are \(\sin^2(0/2)=0\) and \(\sin^2(0.4383/2)=\sin^2(0.21915)\approx0.0473\).
The back-transformed interval is \([0,0.0473]\), including zero. Its lower endpoint remains included after clipping and inversion. This boundary example illustrates the recipe, but the interior-probability normal limit theorem does not justify its nominal coverage at \(p=0\).
Common Mistakes and Interpretation
- Using the wrong transformed standard error. For \(g(p)=2\arcsin\sqrt p\), the first-order standard error is \(1/\sqrt n\). The alternative transformation \(\arcsin\sqrt p\) has half this scale and standard error \(1/(2\sqrt n)\); do not combine one transformation with the other’s standard error.
- Back-transforming only the point estimate. A confidence interval requires transforming both transformed-scale limits, not just \(\hat\theta\).
- Reporting open endpoints. A closed transformed interval maps under the continuous increasing inverse to a closed probability interval. Use square brackets, including when clipping sets a limit to zero or \(\pi\).
- Treating clipping as a coverage correction. Clipping enforces the parameter range and preserves the coverage event for probabilities in that range. It does not repair a poor normal approximation or guarantee nominal finite-sample coverage.
- Applying an interior asymptotic result uncritically at a boundary. The derivation assumes fixed \(0<p<1\). A zero or all-success count calls for particular caution, even though the computational recipe produces valid probability limits.
The value of the arcsine interval is its simple, approximately constant standard error on the transformed scale, together with limits that respect the probability range after back-transformation. Its limits are nonlinear in the observed proportion, so the interval need not be symmetric around \(\hat p\). Always report the count, sample size, confidence level, and the fact that the interval is approximate; those details help readers assess the strength of the normal approximation.
Check Your Understanding
Work through these questions using the interval construction above.
- For \(x=25\) out of \(n=100\), what is the transformed estimate and the transformed standard error?
- Why does the transformation \(2\arcsin\sqrt p\) have approximate standard error \(1/\sqrt n\), rather than \(1/(2\sqrt n)\)?
- For a 95% interval with \(x=0\), why must the transformed lower limit be clipped, and what happens to that endpoint after back-transformation?
- Explain why back-transforming the two endpoints preserves a closed interval.
- Does truncating the transformed limits guarantee 95% finite-sample coverage when the observed count is small? Explain.