CFA Level II Quantitative Methods Practice Questions
Question 1
An analyst estimates a regression of a country's equity returns on local credit growth only. The true model also includes real exports growth, which has a positive effect on equity returns and is negatively correlated with credit growth in the sample. What is the effect of omitting export growth on the estimated credit growth coefficient?
- A. Biased upward and inconsistent, because exports raise returns
- B. Biased downward and inconsistent
- C. Unbiased, with larger standard errors because of the missing variable
- D. Unbiased and consistent, since the omitted variable is not in the fitted equation
Show answer and explanation
Correct answer: B. Biased downward and inconsistent
Omitted variable bias is approximately (effect of omitted variable) x (its correlation with the included variable). Positive effect times negative correlation gives a negative bias, so the credit coefficient is biased downward, and the bias does not vanish as the sample grows.
- A. The sign of the bias is the product of the omitted variable's effect and its correlation with the regressor, which is negative here.
- B. Correct: positive effect x negative correlation = downward bias, inconsistent.
- C. Omitting a correlated relevant variable biases the slope; it is not merely a standard-error problem.
- D. A variable left out of the fitted equation is exactly what creates the bias when it is correlated with a regressor and affects the dependent variable.
Question 2
An analyst has 40 corporate bonds described by duration, spread and rating score and does not know how many groups are appropriate. She wants an output that shows how bonds merge into progressively larger clusters. Which statement correctly describes the method that suits her need?
- A. Agglomerative hierarchical clustering starts with one cluster of all bonds and repeatedly splits it, requiring the cluster count in advance
- B. K-means clustering builds a dendrogram and does not need the number of clusters specified in advance
- C. K-means clustering is preferred because it repeatedly merges the closest pairs of bonds until a single cluster remains
- D. Agglomerative hierarchical clustering starts with each bond as its own cluster and produces a dendrogram, so the number of clusters is chosen afterward
Show answer and explanation
Correct answer: D. Agglomerative hierarchical clustering starts with each bond as its own cluster and produces a dendrogram, so the number of clusters is chosen afterward
Agglomerative hierarchical clustering is bottom-up: every observation begins as a cluster and the closest pairs merge, displayed in a dendrogram, so k can be chosen afterward. K-means needs k set up front and reassigns observations to nearest centroids.
- A. Starting with one cluster and splitting describes divisive clustering, and neither hierarchical form requires the count in advance.
- B. K-means requires the number of clusters in advance and produces centroids, not a dendrogram.
- C. Repeated merging of nearest pairs describes agglomerative clustering, not k-means, which iterates centroid assignment.
- D. Correct: bottom-up merging, dendrogram, cluster count chosen after the fact.
Question 3
An analyst finds ARCH(1) effects in daily returns of an equity index: the squared residual follows e2(t+1) = 0.0002 + 0.5 e2(t). Today's residual is 0.03. What is the forecast of next period's residual variance?
- A. 0.00065
- B. 0.00045
- C. 0.00110
- D. 0.00020
Show answer and explanation
Correct answer: A. 0.00065
ARCH(1): variance(t+1) = a0 + a1 e(t)^2 = 0.0002 + 0.5(0.03^2) = 0.0002 + 0.5(0.0009) = 0.0002 + 0.00045 = 0.00065. A significant a1 means error variance depends on the prior squared error, invalidating ordinary standard errors.
- A. Correct: 0.0002 + 0.00045 = 0.00065.
- B. Omits the intercept a0: 0.5(0.0009) = 0.00045.
- C. Adds the squared residual without weighting by a1: 0.0002 + 0.0009 = 0.00110.
- D. Uses only the intercept, ignoring the conditional dependence on today's shock.
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An analyst finds that two commodity price series each contain a unit root. She regresses one on the other and tests the residuals for a unit root with an Engle-Granger-type test. The test rejects the null of a unit root in the residuals. What should she conclude?
- A. The regression is spurious, so both series must be first-differenced before modeling any relationship
- B. The series are cointegrated, so a long-run levels relationship can be modeled
- C. The series are cointegrated, which implies that each individual series is itself stationary
- D. The series are not cointegrated, because two series that each have a unit root cannot share a relationship
Show answer and explanation
Correct answer: B. The series are cointegrated, so a long-run levels relationship can be modeled
If the residuals of the levels regression are stationary, the two unit-root series share a common stochastic trend: they are cointegrated. The levels regression is then valid, though each series individually remains nonstationary.
- A. A spurious regression is indicated when the residuals retain a unit root; here they do not.
- B. Correct: stationary residuals mean cointegration and a valid long-run relationship.
- C. Cointegration does not make the individual series stationary; only a linear combination is stationary.
- D. Cointegration specifically concerns nonstationary series that move together; having unit roots is a precondition, not a barrier.