Quiz: Lecture 5 — Fréchet Means: Computation – Lectures on Non-Euclidean Data Analysis

Q1. In a Hadamard space, the proximal map (resolvent) \(J_{\lambda}^{f}(x)\) is defined as:

A. \(\arg\min_{y \in \mathcal{M}}\{f(y) + \lambda\, d^2(y, x)\}\) B. \(\arg\min_{y \in \mathcal{M}}\{f(y) + \frac{1}{2\lambda}d^2(y, x)\}\) C. \(\arg\min_{y \in \mathcal{M}}\{\lambda f(y) + d^2(y, x)\}\) D. \(\arg\min_{y \in \mathcal{M}}\{f(y) - \frac{1}{2\lambda}d^2(y, x)\}\)

Q2. Why is the minimizer of the proximal subproblem unique in a Hadamard space?

A. Because \(f\) is strictly convex and the squared-distance term is strongly convex along geodesics B. Because the space is finite-dimensional C. Because \(f\) is always differentiable D. Because the proximal map uses only one data point at a time

Q3. For the Fréchet mean objective \(f_i(x) = w_i\,d^2(x, X_i)\), the closed-form component resolvent update along the geodesic from current iterate \(y\) to data point \(a\) has step size:

A. \(t = \frac{\lambda w_i}{1 + \lambda w_i}\) B. \(t = \frac{2\lambda w_i}{1 + 2\lambda w_i}\) C. \(t = \frac{\lambda w_i}{2 + \lambda w_i}\) D. \(t = \lambda w_i\) (linear approximation)

Q4. In the split proximal-point algorithm, what is the advantage of applying the proximal step to one summand at a time rather than to the full objective?

A. Each component step has a closed-form geodesic interpolation — avoiding a full minimization B. The full objective is non-convex, so splitting is required for convergence C. Splitting guarantees convergence in one cycle through the data D. The full proximal map cannot be evaluated because \(f\) is not differentiable

Q5. The step-size sequence \((\lambda_k)\) for the PPA must satisfy:

A. \(\sum \lambda_k < \infty\) and \(\sum \lambda_k^2 = \infty\) B. \(\sum \lambda_k = \infty\) and \(\sum \lambda_k^2 < \infty\) C. \(\lambda_k \to 0\) at any rate D. \(\lambda_k\) must be constant: \(\lambda_k = c\) for all \(k\)

Q6. Which of the following step-size sequences satisfies the PPA convergence conditions?

A. \(\lambda_k = c/k\) (harmonic decay) B. \(\lambda_k = c/k^2\) (quadratic decay) C. \(\lambda_k = c/k^{1/2}\) (square-root decay) D. \(\lambda_k = c\) (constant)

Q7. In the randomized split PPA, at each iteration the algorithm:

A. Moves along the geodesic toward all data points simultaneously B. Picks a random data point and moves the iterate toward it along the geodesic C. Cycles through data points in a fixed, deterministic order D. Projects the iterate onto the tangent space at each data point

Q8. As \(\lambda \to \infty\), the component resolvent step size \(t = \frac{2\lambda w}{1 + 2\lambda w}\) approaches:

A. 0 (the iterate does not move) B. 1 (the iterate moves all the way to the data point) C. \(\frac{1}{2}\) (the iterate moves to the midpoint) D. \(w\) (the iterate moves proportionally to the weight)

Q9. Why does the interactive sphere example restrict the data points to the octant \(\{x,y,z \ge 0\}\)?

A. Euclidean and geodesic distances are identical within the octant B. The octant is geodesically convex, so geodesics between its points—and hence the PPA iterates—remain inside it C. The component resolvent has a closed form only in this octant D. Restricting to the octant makes a diminishing step-size sequence unnecessary

Q10. Why does the randomized split PPA converge almost surely in a locally compact Hadamard space?

A. Because the objective is convex and the step-size conditions ensure both sufficient exploration (\(\sum \lambda_k = \infty\)) and damping (\(\sum \lambda_k^2 < \infty\)) B. Because the algorithm evaluates the gradient at every iteration C. Because each data point is visited exactly once per cycle D. Because the Fréchet mean has a closed-form solution in Hadamard spaces