Quiz: Lecture 17 — CLTs for Intrinsic Fréchet Means

This quiz tests your understanding of central limit theorems for Fréchet means on manifolds — sandwich covariance, cut locus challenges, non-i.i.d. CLTs, confidence region construction, and manifold-based inference. There are 10 multiple-choice questions, each worth 1 mark. After submitting, your score is shown, and any questions you got wrong are highlighted — but the correct answer is not revealed. You can then redo only the questions you missed.

Q1. The Bhattacharya–Patrangenaru (2005) CLT establishes that the sample Fréchet mean \(\hat{\mu}_n\) on a manifold satisfies:

A. \(\hat{\mu}_n\) converges to a Cauchy distribution B. \(\hat{\mu}_n\) is exactly normally distributed for all \(n\) C. In a smooth chart, \(\sqrt{n}\{\varphi(\hat{\mu}_n)-\varphi(\mu)\} \xrightarrow{d} \mathcal{N}(0, \Lambda^{-1} C (\Lambda^\top)^{-1})\), where \(\Lambda\) is the expected Hessian and \(C\) is the covariance of the score D. \(\sqrt{n}(\hat{\mu}_n - \mu) \xrightarrow{d} \mathcal{N}(0, \Sigma)\) with \(\Sigma = \operatorname{Cov}(X)\) — the same as in Euclidean space

Q2. The sandwich covariance \(\Lambda^{-1}C(\Lambda^\top)^{-1}\) in the manifold CLT consists of:

A. \(\Lambda\) = identity, \(C\) = sample covariance — no manifold adjustment needed B. With half-squared-distance loss, \(\Lambda = \mathbb{E}[\nabla^2 \frac{1}{2}d^2(\mu, X)]\) and \(C = \operatorname{Cov}\{\log_\mu(X)\}\) C. \(\Lambda\) = the Riemannian metric tensor, \(C\) = the curvature tensor D. \(\Lambda\) = Fisher information, \(C\) = identity

Q3. Why does the cut locus pose a fundamental challenge for CLTs on manifolds?

A. The CLT only holds for points on the cut locus B. The cut locus has no effect on asymptotic normality C. The function \(x\mapsto d^2(x,X)\) is nonsmooth when \(X\in\operatorname{Cut}(x)\), so observations near the moving cut locus can invalidate a Taylor expansion of the empirical score D. The CLT requires the cut locus to be empty

Q4. The Kendall–Le (2011) CLT extends the manifold CLT to:

A. Independent, non-identically distributed data, subject to a Lindeberg condition and additional Hessian regularity B. Only Gaussian data C. Only Euclidean spaces D. Only one-dimensional manifolds

Q5. How is an asymptotic confidence region for the Fréchet mean constructed on a manifold using the CLT?

A. Directly apply the Euclidean confidence interval formula in the ambient coordinate chart — no manifold adjustment is needed B. First construct an ellipsoid in the tangent space \(T_{\hat{\mu}_n}\mathcal{M}\) via the CLT pivot \(n \cdot v^\top \hat{\Sigma}^{-1} v \le \chi^2_{m,1-\alpha}\), then map it to the manifold via \(\exp_{\hat{\mu}_n}\) C. The confidence region is always a perfect geodesic ball of radius \(n^{-1/2}\) centered at \(\hat{\mu}_n\) D. Confidence regions cannot be constructed from the CLT — only permutation tests provide valid confidence sets on manifolds

Q6. On the SPD manifold with the log-Euclidean (LE) metric, the CLT simplifies because:

A. The LE metric requires a different CLT that has not been derived B. Matrix logarithm gives a global Euclidean isometry, so \(\operatorname{svec}(\log X)\) has an ordinary multivariate sample-mean CLT in \(\mathbb{R}^{m(m+1)/2}\) C. The CLT does not apply to the LE metric D. The manifold is flat, so no CLT is needed

Q7. Asymptotic confidence regions for Fréchet means on the SPD manifold are constructed via:

A. Using only permutation tests — the CLT does not provide confidence regions B. Bootstrapping the raw data and taking quantiles C. Assuming the manifold is Euclidean D. The CLT: \(\sqrt{n}\,\hat{\Lambda}_n \log_{\hat{\mu}_n}(\mu) \xrightarrow{d} \mathcal{N}(0, \hat{C}_n)\), which gives an ellipsoidal confidence region in the tangent space that can be mapped back to the manifold via \(\exp\)

Q8. The Wald test (based on the manifold CLT) and the permutation test (based on metric-space distances) are:

A. The Wald test is always superior B. Complementary — the Wald test exploits the manifold’s smooth structure for parametric efficiency, while the permutation test requires no manifold assumptions and works in any metric space C. Mutually exclusive — you can only use one D. Identical — they always give the same result

Q9. In the mean-centered chart (tangent space at the true mean \(\mu\)), the limiting covariance simplifies to:

A. \(4\Lambda^{-1}\operatorname{Cov}(\log_\mu(X))\Lambda^{-1}\) — the factor 4 comes from the Hessian of the squared distance \(d^2(x,\mu)\) relative to the log-map B. The identity matrix C. \(\operatorname{Cov}(X)\) — the raw Euclidean covariance D. Zero — the mean is known exactly

Q10. When does the manifold structure add inferential power beyond what metric-space methods provide?

A. Never — manifold methods are always equivalent to metric-space methods B. Only for one-dimensional manifolds C. When smooth tangent coordinates support directional confidence ellipsoids and useful covariance standardization; permutation methods remain valuable for exchangeability-based calibration D. Only when the data is exactly Gaussian