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Exam SRM — Principal Components Analysis Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam SRM principal components analysis — proportion and cumulative variance explained from eigenvalues, choosing the number of components by threshold, Kaiser rule, and scree elbow, computing scores from loadings and standardized values, the effect of standardizing (covariance vs correlation matrix), total variance equal to $p$, and principal-components regression — each with a fully worked solution.

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Browse all 30 problems as a list
  1. Variance explained
    A correlation-matrix PCA on $4$ standardized variables returns eigenvalues $\lambda=(2.4,\,0.9,\,0.5,\,0.2)$. Calculate the proportion of variance explained by the first principal component. (A) $0.500$ (B) $0.600$ (C) $0.625$ (D) $0.727$ (E) $0.960$
    **Answer: (B).** The PVE of component $j$ is $\text{PVE}_j=\dfrac{\lambda_j}{\sum_k\lambda_k}$. Total variance $=2.4+0.9+0.5+0.2=4.0$ (as expected, the trace of a $4$-variable correlation matrix equals $p=4$). $\text{PVE}_1=\dfrac{2.4}{4.0}=0.600$. Dividing $\lambda_1$ by $\lambda_1$ alone or by a wrong total produces the distractors: $\dfrac{2.4}{2.4+0.9+0.5}=0.667$-type errors, while $0.625$ comes from using a total of $3.84$ (dropping $\lambda_4$ incorrectly as $0.16$).
  2. Variance explained
    A correlation-matrix PCA on $5$ standardized variables returns eigenvalues $\lambda=(3.0,\,1.0,\,0.5,\,0.3,\,0.2)$. Calculate the cumulative proportion of variance explained by the first two components. (A) $0.600$ (B) $0.750$ (C) $0.800$ (D) $0.875$ (E) $0.900$
    **Answer: (C).** Cumulative PVE of the first two components is $\dfrac{\lambda_1+\lambda_2}{\sum_k\lambda_k}$. Total variance $=3.0+1.0+0.5+0.3+0.2=5.0$ (matches $p=5$). $\dfrac{\lambda_1+\lambda_2}{5.0}=\dfrac{3.0+1.0}{5.0}=\dfrac{4.0}{5.0}=0.800$. Reporting only PC1 gives $3.0/5=0.600$ (A); using a total of $4$ instead of $5$ — i.e. wrongly assuming the eigenvalues sum to $p-1$ — gives $4/4=1.0$. Distractor (B) $0.750$ comes from the wrong total $\lambda_1+\lambda_2+\lambda_3+\lambda_4=4.8$ then $3.6/4.8$.
  3. Variance explained
    A covariance-matrix PCA (variables in the same units, **not** standardized) gives eigenvalues $\lambda=(60,\,24,\,12,\,4)$. Calculate the proportion of variance explained by the first component. (A) $0.250$ (B) $0.600$ (C) $0.625$ (D) $0.750$ (E) $0.800$
    **Answer: (B).** For a covariance-matrix PCA the eigenvalues need not sum to $p$; they sum to the total (unstandardized) variance, which is the trace of the covariance matrix. Total variance $=60+24+12+4=100$. $\text{PVE}_1=\dfrac{60}{100}=0.600$. A common error is to assume the eigenvalues sum to $p=4$ and compute $60/4$ (nonsensical) or to divide by only the retained eigenvalues. The first two together give $(60+24)/100=0.840$.
  4. Choosing components
    A covariance-matrix PCA on $5$ variables (same units, not standardized) gives eigenvalues $\lambda=(3.4,\,1.5,\,0.7,\,0.3,\,0.1)$. How many components are needed so that the cumulative proportion of variance explained is at least $90\%$? (A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$
    **Answer: (C).** For an unstandardized covariance-matrix PCA the eigenvalues sum to the total raw variance (the trace), not to $p$. Total variance $=3.4+1.5+0.7+0.3+0.1=6.0$. Cumulative PVE: PC1: $3.4/6.0\approx0.567$. PC1–2: $4.9/6.0\approx0.817$. PC1–3: $5.6/6.0\approx0.933$. The first time cumulative PVE reaches $90\%$ is at **three** components ($93.3\%\ge90\%$); two components give only $81.7\%$. Stopping at the first component past $80\%$ — a different threshold — gives the distractor $2$.
  5. Loadings & scores
    The first loading vector for two standardized variables is $\phi_1=(0.6,\,0.8)^{\top}$. An observation has standardized values $x_1=1.2$, $x_2=-0.5$. Calculate its first-component score $z=\sum_j\phi_j x_j$. (A) $-0.28$ (B) $0.32$ (C) $0.40$ (D) $1.12$ (E) $1.36$
    **Answer: (B).** First confirm the loadings are normalized: $0.6^{2}+0.8^{2}=0.36+0.64=1$. Good. The score is the projection $z=\phi_{1}x_1+\phi_{2}x_2=0.6(1.2)+0.8(-0.5)=0.72-0.40=0.32$. Dropping the sign on $x_2$ gives $0.72+0.40=1.12$ (D); swapping the loadings ($0.8,0.6$) gives $0.96-0.30=0.66$; and adding the loadings times the values without signs differently produces the other distractors.
  6. Scaling
    Two standardized variables have correlation $\rho=0.6$. The correlation matrix is $\begin{pmatrix}1 & 0.6\\ 0.6 & 1\end{pmatrix}$. Calculate the proportion of variance explained by the first principal component. (A) $0.500$ (B) $0.600$ (C) $0.700$ (D) $0.800$ (E) $0.840$
    **Answer: (D).** For a $2\times2$ correlation matrix with off-diagonal $\rho$, the eigenvalues are $1+\rho$ and $1-\rho$. $\lambda_1=1+0.6=1.6$ and $\lambda_2=1-0.6=0.4$. They sum to $2=p$, as required for two standardized variables. $\text{PVE}_1=\dfrac{1.6}{2}=0.800$. Using $\rho$ itself as a proportion gives $0.600$ (B); using $\rho^2=0.36$ in some form, or $(1+\rho)/(2+\text{something})$, produces the other distractors. The correct answer is $(1+\rho)/2$.
  7. Loadings & scores
    A PCA on $4$ standardized variables yields the first loading vector $\phi_1=(0.5,\,0.5,\,0.5,\,0.5)^{\top}$. An observation has standardized values $(x_1,x_2,x_3,x_4)=(2,\,-1,\,3,\,0)$. Calculate its first-component score. (A) $1.0$ (B) $2.0$ (C) $3.0$ (D) $4.0$ (E) $6.0$
    **Answer: (B).** Normalization check: $4\times0.5^{2}=4(0.25)=1$, so $\phi_1$ is a valid unit loading vector. Because all loadings equal $0.5$, the score is $0.5$ times the sum of the standardized values: $z=0.5(2-1+3+0)=0.5(4)=2.0$. Forgetting the $0.5$ weight and just summing gives $4.0$ (D); summing the magnitudes $|2|+|1|+|3|+|0|=6$ then halving gives $3.0$ (C). The first component here is an equal-weight "overall size" index.
  8. Variance explained
    A PCA reports that PC1 explains $40\%$ and PC2 explains $24\%$ of the total variance, and the total variance is $25$. Calculate the eigenvalue $\lambda_1$. (A) $0.40$ (B) $6.0$ (C) $9.0$ (D) $10.0$ (E) $16.0$
    **Answer: (D).** Since $\text{PVE}_j=\dfrac{\lambda_j}{\sum_k\lambda_k}$ and the total $\sum_k\lambda_k=25$, the eigenvalue is $\lambda_j=\text{PVE}_j\times\sum_k\lambda_k$. $\lambda_1=0.40\times25=10.0$. (For reference $\lambda_2=0.24\times25=6.0$, distractor B.) Reporting the PVE itself ($0.40$) is distractor (A); multiplying the two PVEs or adding eigenvalues gives the others.