Free SOA Exam SRM (Statistics for Risk Modeling) Unsupervised Learning Techniques Practice Questions

D is correct.

Each eigenvalue of the covariance (or correlation) matrix equals the variance of the data when projected onto the corresponding principal component. Larger eigenvalues indicate directions that capture more variation in the data.

D is correct.

A scree plot has the component number on the x-axis and the corresponding eigenvalue on the y-axis. It is used to visually identify the point at which additional components provide diminishing returns in variance explained.

A is correct.

Cluster 1: $\{1, 3, 5\}$, centroid $\bar{x}_1 = (1+3+5)/3 = 3.0$

Cluster 1 WCSS:
- $(1-3)^2 = 4$
- $(3-3)^2 = 0$
- $(5-3)^2 = 4$
- Sum = $4 + 0 + 4 = 8.0$

Cluster 2: $\{9, 11, 13\}$, centroid $\bar{x}_2 = (9+11+13)/3 = 11.0$

Cluster 2 WCSS:
- $(9-11)^2 = 4$
- $(11-11)^2 = 0$
- $(13-11)^2 = 4$
- Sum = $4 + 0 + 4 = 8.0$

Total WCSS $= 8.0 + 8.0 = 16.0$.

Watch the naming convention. This total is what ISLR's clustering lab calls the total within-cluster sum of squares and what R's kmeans reports as tot.withinss. ISLR separately defines the within-cluster variation of a cluster as $\frac{1}{|C_k|} \sum_{i, i' \in C_k} (x_i - x_{i'})^2$, whose sum runs over ordered pairs and so equals $2 \sum_{i \in C_k} (x_i - \bar{x}_k)^2$. For these two clusters that doubled quantity is $2(8.0 + 8.0) = 32.0$. Both versions have the same minimizer, so K-means behaves identically either way, but when a prompt asks for total within-cluster variation rather than the sum of squares, double the figure computed here.