Introduction
A variance-stabilizing transformation is designed to make an estimator's leading variance less dependent on the unknown parameter. That is useful: a standard error that is approximately constant across parameter values can simplify comparisons and interval construction. But variance describes only one feature of a distribution. A transformed estimator may still be skewed, discrete, or far from normal in a finite sample.
The distinction is important in inference. A normal approximation requires a distributional argument, such as a central limit theorem and suitable regularity conditions. A transformation that stabilizes variance supplies no such argument on its own. We make that separation precise and then examine examples where skewness remains visible.
What First-Order Variance Stabilization Says
The definition concerns the leading term of the variance as \(n\) grows. For example, choosing \(g'(\theta)=1/\sqrt{V(\theta)}\) gives \(c=1\) when \(g\) is increasing. It does not assert that the exact finite-sample variance is constant, nor does it describe the estimator's skewness, tails, or support.
Here is the variance calculation behind the definition. Assume the delta-method approximation holds in mean square: \[ g(\hat{\theta}_n)-g(\theta) =g'(\theta)(\hat{\theta}_n-\theta)+R_n, \qquad E[R_n^2]=o(n^{-1}). \] Then \(n\operatorname{Var}(g(\hat{\theta}_n))\) has the same limit as \(g'(\theta)^2n\operatorname{Var}(\hat{\theta}_n)\).
Proof. Write \(A_n=g'(\theta)(\hat{\theta}_n-\theta)\). The variance difference satisfies \[ \left|\operatorname{Var}(A_n+R_n)-\operatorname{Var}(A_n)\right| \leq \operatorname{Var}(R_n) +2\sqrt{\operatorname{Var}(A_n)\operatorname{Var}(R_n)}. \] The first term is \(o(n^{-1})\). Since \(\operatorname{Var}(A_n)=O(n^{-1})\) and \(\operatorname{Var}(R_n)=o(n^{-1})\), the second term is also \(o(n^{-1})\). Thus \[ \operatorname{Var}(g(\hat{\theta}_n)) =g'(\theta)^2\operatorname{Var}(\hat{\theta}_n)+o(n^{-1}). \] Multiplying by \(n\) and using \(n\operatorname{Var}(\hat{\theta}_n)\to V(\theta)\) proves the result. \(\square\)
Variance Is Not a Distributional Result
The theorem controls a second moment. A normal approximation, by contrast, is a statement about the distribution of a centered and scaled estimator. For example, if the additional condition \[ \sqrt{n}(\hat{\theta}_n-\theta)\xrightarrow{d}N(0,V(\theta)) \] holds, then the delta method gives \[ \sqrt{n}\{g(\hat{\theta}_n)-g(\theta)\} \xrightarrow{d}N(0,g'(\theta)^2V(\theta)). \] When the transformation stabilizes variance at \(c\), this limit is \(N(0,c)\). The normality comes from the distributional limit assumed for \(\hat{\theta}_n\), not from the variance calculation. The delta method transfers an existing limit; it does not create one merely by making a variance constant.
A second-order expansion also shows why a smooth transformation need not remove skewness. For a sample mean, let \(X_1,\ldots,X_n\) be independent and identically distributed with mean \(\mu\), variance \(\sigma^2>0\), and third central moment \(\mu_3=E[(X_1-\mu)^3]\). Put \(\bar X=n^{-1}\sum_i X_i\). Assume the moments needed below are finite, \(g\) has three continuous derivatives near \(\mu\), \(g'(\mu)>0\), and the Taylor remainder is \(o(n^{-1})\) in \(L^3\). These conditions justify the moment expansion as well as the Taylor approximation.
Derivation. Write \(\Delta=\bar X-\mu\). The sample-mean moments are \[ E\Delta=0,\quad E\Delta^2=\frac{\sigma^2}{n},\quad E\Delta^3=\frac{\mu_3}{n^2},\quad E\Delta^4=\frac{3\sigma^4}{n^2}+o(n^{-2}). \] The centered Taylor expansion is \[ g(\bar X)-E[g(\bar X)] =g'(\mu)\Delta+ \frac{g''(\mu)}{2}\left(\Delta^2-\frac{\sigma^2}{n}\right)+R_n^*, \] where the remainder \(R_n^*\) contributes \(o(n^{-2})\) to the third central moment under the stated conditions. Cubing the two displayed leading terms, the contributions of order \(n^{-2}\) are \[ g'(\mu)^3 E\Delta^3+ \frac{3g'(\mu)^2g''(\mu)}{2} E\left[\Delta^2\left(\Delta^2-\frac{\sigma^2}{n}\right)\right]. \] The expectation in the second term is \(E\Delta^4-(E\Delta^2)^2=2\sigma^4/n^2+o(n^{-2})\). Therefore the third central moment is \[ \frac{g'(\mu)^3\mu_3+3g'(\mu)^2g''(\mu)\sigma^4}{n^2} +o(n^{-2}). \] The variance is \(g'(\mu)^2\sigma^2/n+o(n^{-1})\). Dividing the third central moment by this variance to the power \(3/2\) gives the stated skewness expansion. \(\square\)
The expression has two parts: the original sample-mean skewness and a term contributed by curvature. Curvature can reduce skewness, increase it, or—in a particular case—cancel its leading term. Stabilizing variance does not require that cancellation. Even when the leading skewness tends to zero, the finite-sample distribution need not be well approximated by a normal distribution at the sample size in hand.
Worked Example: The Arcsine Transformation Can Remain Discrete and Skewed
Let \(X_1,\ldots,X_n\) be Bernoulli observations with success probability \(p\), and let \(\hat p=\bar X\). Its variance is \(p(1-p)/n\). For \(g(p)=2\arcsin\sqrt p\), \[ g'(p)=\frac{1}{\sqrt{p(1-p)}}, \qquad g'(p)^2p(1-p)=1. \] Thus the transformed estimator has first-order variance \(1/n\) for \(0<p<1\).
Take the finite-sample case \(n=1\), \(p=0.1\). Then \(\hat p\) is 1 with probability \(0.1\) and 0 with probability \(0.9\). The transformed value is \(\pi\) with probability \(0.1\) and 0 with probability \(0.9\), so its distribution has only two possible values and cannot be normal. Its skewness is \[ \frac{1-2p}{\sqrt{p(1-p)}}= \frac{0.8}{\sqrt{0.09}}=\frac{8}{3}. \] This example does not say the asymptotic approximation is always poor; as \(n\) grows, a central limit theorem can support normality for fixed interior \(p\). It shows that stabilization itself does not make a discrete finite-sample distribution normal.
Worked Example: The Poisson Square-Root Transformation Changes Skewness but Does Not Guarantee Its Absence
Suppose \(X_1,\ldots,X_n\) are independent Poisson observations with mean \(\lambda>0\). The estimator \(\hat\lambda=\bar X\) has variance \(\lambda/n\). For \(g(\lambda)=2\sqrt{\lambda}\), \(g'(\lambda)=1/\sqrt{\lambda}\), and hence \(g'(\lambda)^2\lambda=1\).
For a Poisson variable, \(\sigma^2=\lambda\) and \(\mu_3=\lambda\). Also \(g''(\lambda)=-1/(2\lambda^{3/2})\), so \[ \frac{\mu_3}{\sigma^3}=\frac{1}{\sqrt{\lambda}}, \qquad 3\frac{g''(\lambda)}{g'(\lambda)}\sigma =3\left(-\frac{1}{2\lambda}\right)\sqrt{\lambda} =-\frac{3}{2\sqrt{\lambda}}. \] The expansion therefore gives \[ \operatorname{Skew}(2\sqrt{\bar X}) =-\frac{1}{2\sqrt{n\lambda}}+o(n^{-1/2}). \] The positive leading skewness of the original estimator is changed to a negative leading skewness; it is not eliminated. The transformed estimator also retains an atom at zero, since \(P(\bar X=0)=e^{-n\lambda}\). For fixed \(\lambda>0\), this probability vanishes as \(n\) grows, and the central limit theorem supports an asymptotic normal approximation. At finite \(n\), however, skewness and discreteness remain features to assess rather than assume away.
Worked Example: Constant Variance With Persistent Skewness
A simple estimator sequence shows directly that constant variance alone cannot imply normality. Let \(W\) have an exponential distribution with rate 1, and set \(Z=W-1\). Then \(E[Z]=0\), \(\operatorname{Var}(Z)=1\), and \(\operatorname{Skew}(Z)=2\). For a real parameter \(\theta\), define \[ \hat{\theta}_n=\theta+\frac{Z}{\sqrt n}. \] It is consistent because \(\hat{\theta}_n-\theta\to 0\) in probability, and its variance is exactly \(1/n\). The identity transformation therefore gives constant variance on the \(1/n\) scale.
But the centered and scaled estimator is \(\sqrt n(\hat{\theta}_n-\theta)=Z\) for every \(n\). Its distribution is not normal and its skewness remains exactly 2; it does not approach zero as \(n\) increases. This is a counterexample to the claim that a constant first-order variance forces normality. It is not a model for every common estimator; its purpose is to isolate what the variance condition does—and does not—say.
Why This Matters in Inference
A normal-based confidence interval can be unreliable when the transformed estimator remains strongly skewed, has substantial probability near a boundary, or is supported on a coarse discrete set. Variance stabilization may make standard errors more comparable across parameter values, but interval coverage also depends on the distributional approximation and on how the interval behaves after any back-transformation.
A common mistake is to reason: “The transformed variance is approximately constant, so the transformed estimator is approximately normal.” This confuses a second-moment statement with a distributional statement. The correction is to establish normality separately—for example, using a central limit theorem with its assumptions checked—and then judge whether the approximation is adequate at the actual sample size. A useful summary is:
- Variance stabilization controls the leading variance across parameter values.
- A central limit theorem or another distributional result supports normality.
- Finite-sample assessment checks whether skewness, discreteness, boundaries, or tail behavior still matter.
The two goals can work well together: a variance-stabilizing transformation can improve the scale on which a normal approximation is used. But they are distinct goals, and success at the first is not evidence by itself that the second has been achieved.
Check Your Understanding
Try these questions before moving on.
- State what first-order variance stabilization says about \(n\operatorname{Var}(g(\hat{\theta}_n))\), and name one distributional feature it does not determine.
- For a Bernoulli sample proportion, verify that \(g(p)=2\arcsin\sqrt p\) satisfies \(g'(p)^2p(1-p)=1\).
- In the sample-mean skewness expansion, identify the term contributed by the curvature \(g''(\mu)\). Can it have either sign?
- Why does a central limit theorem provide information that a variance calculation alone cannot provide?
- In the Poisson example, why does the transformed estimator's atom at zero not contradict asymptotic normality for fixed \(\lambda>0\)?