Propagate asymptotic uncertainty through a transformation

Try

Suppose an estimate Tₙ targets a positive value θ and √n(Tₙ−θ) approaches a centered normal with variance v. You need log θ instead. A tangent approximation gives log Tₙ−log θ approximately (Tₙ−θ)/θ, so the limiting variance should become v/θ². This is uncertainty propagation through a local slope. It requires the target to be inside the logarithm’s domain and the estimate to lie there with probability tending to one; otherwise the transformed statistic may not even be defined.

Predict

Predict what happens when the transformation has derivative zero. Squaring an estimate near θ=0 makes positive and negative errors point in the same direction. A first-order tangent is flat, so it cannot describe the leading nonzero variation. That does not mean the transformed estimate is exactly constant. It means the chosen expansion order and scaling have lost the interesting term. We will compare an ordinary differentiable point with this degeneracy using exact normal calculations.

Experiment

Let X̄ be a mean of n iid normal observations with variance one. Expand X̄²−μ²=2μ(X̄−μ)+(X̄−μ)². Track which term survives under each scaling.

True mean	Leading term	Nondegenerate scaling and limit
μ=1	2(X̄−1)	√n error tends to N(0,4)
μ=0	X̄²	nX̄² has χ²₁ law exactly

For μ=0, √n X̄ is exactly standard normal, so its square supplies the second row immediately. Explain why the first-order delta statement at zero still correctly gives a degenerate zero limit under √n scaling. It is incomplete for a nondegenerate approximation, not logically false.

Understand

The scalar delta method assumes √n(Tₙ−θ) converges in distribution to N(0,v) and g is differentiable at θ. Differentiability gives g(Tₙ)−g(θ)=g′(θ)(Tₙ−θ)+a remainder smaller than ∣Tₙ−θ∣ near θ. The assumed limit implies consistency and boundedness in probability of √n(Tₙ−θ). Multiplying the remainder by √n therefore gives a term tending to zero in probability. Slutsky yields a centered normal limit with variance [g′(θ)]²v, permitting a degenerate normal if the derivative vanishes.

For a vector estimate with limiting covariance Σ and a differentiable scalar g, replace the slope by its gradient: the variance is ∇gᵀΣ∇g. The covariance terms can reinforce or cancel variation. Dropping them because two estimates have different names is incorrect. Their joint sampling law, not the notation, determines dependence. A vector-valued transformation uses its Jacobian on both sides of Σ.

Apply

For a Bernoulli proportion with interior p, transform to log odds g(p)=log(p/(1−p)). Its derivative is 1/[p(1−p)]. The limiting variance of √n(g(p̂)−g(p)) is therefore 1/[p(1−p)]. For a finite sample, p̂ can equal zero or one, so the log odds may be infinite even though this event’s probability tends to zero at a fixed interior p. A practical finite-sample correction changes the estimator and should be disclosed rather than silently called the same exact procedure.

For a reciprocal g(x)=1/x at θ=0, there is no finite transformed target and no derivative at that point. A second-order expansion does not fix an undefined target. Distinguish a zero derivative, which may invite higher-order analysis, from a transformation that is not defined at the population value.

Continue

Your solution should identify the original limit, transformation domain, derivative or gradient, and resulting covariance. Then check whether an exact distribution is available as a stronger reference. Continue to Exact normal pivots, where orthogonal normal geometry gives finite-sample laws without a large-n approximation. The difference between exact and asymptotic reasoning remains central even when their numerical answers happen to be close.