Why Transform a Binomial Proportion?
For a binomial sample, the variability of the sample proportion depends on the unknown success probability. In particular, proportions near one-half have greater variance than proportions near zero or one. A transformation can put the estimate on a scale where its approximate variance is nearly constant, making a normal approximation easier to use across different values of the probability.
The arcsine-square-root transformation is obtained directly from the binomial variance function. It does not make the data exactly normal, nor does it remove the need to consider sample size and boundary values. Its role is to give a useful first-order approximation when the success probability is fixed away from zero and one and the sample size is sufficiently large.
Deriving the Transformation
Let \(X\sim\operatorname{Bin}(n,p)\), where \(n\) is the number of independent Bernoulli trials and \(0<p<1\) is the success probability. The sample proportion is \(\hat p=X/n\). Its expectation and variance are
Consider a differentiable, increasing transformation \(g\). The scalar delta method says that, when \(\hat p\) is concentrated near \(p\), the first-order approximation to the variance of \(g(\hat p)\) is \(g'(p)^2\operatorname{Var}(\hat p)\). To make this approximately independent of \(p\), choose \(g\) so that
for some positive constant \(C\). Taking the positive square root gives \(g'(p)=\sqrt{C}/\sqrt{p(1-p)}\). Integrating, and using \(u=\sqrt p\), yields
The additive constant does not affect variance. We use the doubled version \(g(p)=2\arcsin(\sqrt p)\), which makes the approximate variance \(1/n\). Another common convention is \(h(p)=\arcsin(\sqrt p)=g(p)/2\); its approximate variance is \(1/(4n)\). These are the same transformation up to a constant scale factor.
Approximate Normal Distribution
For fixed \(p\) with \(0<p<1\), the binomial central limit theorem gives \(\sqrt n(\hat p-p)\xrightarrow{d}N(0,p(1-p))\). The transformation is differentiable at \(p\), so the scalar delta method applies. The resulting limit explains why the transformed estimate has an approximately constant variance.
Equivalently, for large \(n\), \(g(\hat p)\) is approximately normal with mean \(g(p)\) and variance \(1/n\). This is a first-order approximation: it describes the transformed estimator near its target, not an exact finite-sample distribution.
To see the variance calculation explicitly, a first-order Taylor expansion around \(p\) gives
The binomial central limit theorem implies \(\sqrt n(\hat p-p)=O_p(1)\), so multiplying the Taylor remainder by \(\sqrt n\) gives \(o_p(1)\). Therefore,
The leading term converges in distribution to a normal variable with mean zero and variance
Adding a term that converges in probability to zero does not change this limiting distribution. This proves the stated result. \(\square\)
Worked Example: Checking the Variance Stabilization
Compare the approximate variance of the transformed proportion for \(n=100\) at \(p=0.25\) and \(p=0.50\).
At \(p=0.25\), \(\operatorname{Var}(\hat p)=0.25(0.75)/100=0.001875\), and \(g'(0.25)=1/\sqrt{0.25(0.75)}=1/\sqrt{0.1875}\). Hence the delta-method variance is
At \(p=0.50\), \(\operatorname{Var}(\hat p)=0.25/100=0.0025\) and \(g'(0.50)=1/\sqrt{0.25}=2\). Thus
The raw proportions have different variances, but the first-order transformed variances agree and equal \(1/n=0.01\). The equality is an approximation based on the delta method, not a claim that the two finite-sample transformed distributions are identical.
A Transformed Confidence Interval
The approximate normal distribution can be used to form a confidence interval. Let \(z_{1-\alpha/2}\) denote the \(1-\alpha/2\) quantile of the standard normal distribution. The transformed-scale interval for \(g(p)\) is centered at \(g(\hat p)\) with approximate standard error \(1/\sqrt n\). Its endpoints are then mapped back to the probability scale using the inverse transformation.
back to the \(p\)-scale. Since \(g^{-1}(y)=\sin^2(y/2)\) for \(0\leq y\leq\pi\), the interval is
The clipping at \(0\) and \(\pi\) keeps the transformed endpoints within the range of \(g\), so the back-transformed limits lie in \([0,1]\). For \(0<\hat p<1\), this interval follows by applying the increasing inverse transformation to the transformed-scale limits. Under the theorem’s fixed-interior-\(p\) assumptions, its coverage probability approaches \(1-\alpha\) as \(n\) grows. It is an approximate interval, not an exact binomial confidence interval.
Worked Example: A 95% Interval With 40 Successes
Suppose \(x=40\) successes are observed in \(n=100\) independent trials. Then \(\hat p=40/100=0.40\). Use \(z_{0.975}=1.96\).
First transform the estimate:
The transformed standard error is \(1/\sqrt{100}=0.10\), so the transformed limits are
Both limits fall in \([0,\pi]\). Apply \(g^{-1}(y)=\sin^2(y/2)\):
Thus the arcsine-based approximate 95% interval is about \((0.307,0.497)\). The endpoints are not obtained by simply adding and subtracting a standard error from \(0.40\); the normal approximation was made on the transformed scale and then inverted.
Worked Example: A Proportion Near Zero
Suppose \(x=9\) successes are observed in \(n=100\) trials, so \(\hat p=0.09\). Again use \(z_{0.975}=1.96\).
The transformed estimate and standard error are
The transformed limits are \(0.6094\pm0.196\), or approximately \((0.4134,0.8054)\). Inverting each endpoint gives
The interval remains on the probability scale and is asymmetric around \(0.09\), as a consequence of transforming back from a symmetric interval on the arcsine scale. The calculation illustrates the procedure; it does not guarantee accurate coverage for every small count or sample size.
Interpretation and a Common Pitfall
Variance stabilization is useful because the approximate uncertainty on the transformed scale is \(1/\sqrt n\), regardless of the fixed interior value of \(p\). It gives a principled way to construct an approximate interval by using the binomial variance function rather than assuming that the raw proportion has constant variability.
A common mistake is to treat \(g(\hat p)\) as if it were an estimate of \(p\) on the original scale, or to report \(g(\hat p)\pm z/\sqrt n\) as a confidence interval for \(p\). Those endpoints are on the transformed scale and are not probabilities. The correction is to apply \(g^{-1}(y)=\sin^2(y/2)\) to both transformed endpoints. Another error is to conclude that the transformed estimator is exactly normal with variance \(1/n\): both statements are only approximate at finite \(n\).
The theorem assumes \(p\) is fixed strictly between zero and one. When \(p\) is very close to a boundary, or when the observed count is small, the normal approximation may be poor even after transformation. If \(\hat p\) is exactly zero or one, the transformation itself is defined, but the interior delta-method argument does not establish good finite-sample coverage in that situation. The result should therefore be understood as a first-order approximation under its stated conditions.
Check Your Understanding
Use the definitions and derivations above to answer these questions.
- For \(X\sim\operatorname{Bin}(n,p)\), what feature of \(\operatorname{Var}(\hat p)\) motivates transforming \(\hat p\)?
- Differentiate \(g(p)=2\arcsin(\sqrt p)\) and show how \(g'(p)^2p(1-p)\) simplifies.
- What is the approximate variance of \(\arcsin(\sqrt{\hat p})\), rather than the doubled transform, for large \(n\)?
- Why must the endpoints of a transformed-scale interval be mapped through \(\sin^2(y/2)\) before they are interpreted as limits for \(p\)?
- State the restriction on \(p\) in the limiting normal result and explain why the result does not by itself guarantee accurate coverage when the observed count is small.