BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: A survival study gave \((1.63,2.55)\) as the \(95 \%\) linear confidence interval for the cumulative hazard function \(H\left(t_{0}\right)\).

Calculate the \(95 \%\) log-transformed confidence interval for \(H\left(t_{0}\right)\).

Options:
A
\(\quad(0.49,0.94)\)
B
\(\quad(0.84,3.34)\)
C
\((1.58,2.60)\)
D
\((1.68,2.50)\)
E
\(\quad(1.68,2.60)\)

Correct answer: E

Solution: The interval is centered at 2.09 and the plus/minus term is 0.46 which must equal \(1.96 \hat{\sigma}\) and so \(\hat{\sigma}=0.2347\). For the log-transformed interval we need \(\phi=e^{1.96(0.2347) / 2.09}=1.2462\). The lower limit is \(2.09 / 1.2462=1.68\) and the upper limit is \(2.09(1.2462)=2.60\).
Question: For a health insurance policy effective January 1 , the number of claims in a one-year period follows the Poisson distribution with mean \(\lambda=5\).

The following uniform \((0,1)\) random numbers are used in the given order to simulate the time of the first claim and the times between occurrences of subsequent claims.
\[
\begin{array}{lll}
0.605 & 0.529 & 0.782
\end{array}
\]

Calculate the simulated date of occurrence of the third claim.

Options:
A
Before May 1
B
On or after May 1, but before August 15
C
On or after August 15, but before September 1
D
On or after September 1, but before September 30
E
On or after September 30

Correct answer: C

Solution: The time intervals to the occurrence of the first claim, and between claims follow the exponential distribution with mean \(1 / 5\). Thus

The first claim occurs at \([-\ln (1-0.605)] / 5=0.186\)
The second claim occurs at \(0.186+[-\ln (1-0.529)] / 5=0.337\)
The third claim occurs at \(0.337+[-\ln (1-0.782)] / 5=0.642\)
\(0.642 \times 12=7.7\) months
Since month 1 is January, month \(7=\) July, month \(8=\) August, so this means the date will be in late August.
Question: The information associated with the maximum likelihood estimator of a parameter \(\theta\) is \(4 n\), where \(n\) is the number of observations.

Calculate the asymptotic variance of the maximum likelihood estimator of \(2 \theta\).

Options:
A
\(\quad 1 /(2 n)\)
B
\(\quad 1 / n\)
C
\(4 / n\)
D
\(8 n\)
E
\(16 n\)

Correct answer: B

Solution: From the information, the asymptotic variance of \(\hat{\theta}\) is \(1 / 4 n\). Then
\(\operatorname{Var}(2 \hat{\theta})=4 \operatorname{Var}(\hat{\theta})=4(1 / 4 n)=1 / n\)
Note that the delta method is not needed for this problem, although using it leads to the same answer.
Question: You are simulating the gain/loss from insurance where:
(i) Claim occurrences follow a Poisson process with \(\lambda=2 / 3\) per year.
(ii) Each claim amount is 1,2 or 3 with \(p(1)=0.25, p(2)=0.25\), and \(p(3)=0.50\).
(iii) Claim occurrences and amounts are independent.
(iv) The annual premium equals expected annual claims plus 1.8 times the standard deviation of annual claims.
(v) \(\quad i=0\)

You use the uniform \((0,1)\) values \(0.25,0.40,0.60\), and 0.80 and the inversion method to simulate time between claims.

You use the uniform ( 0,1 ) values \(0.30,0.60,0.20\), and 0.70 and the inversion method to simulate claim size.

Calculate the gain or loss from the insurer's viewpoint during the first 2 years from this simulation.

Options:
A
loss of 5
B
loss of 4
C
0
D
gain of 4
E
gain of 5

Correct answer: E

Solution: \(E(S)=\lambda E(X)=2 / 3(1 / 4+2 / 4+3 / 2)=2 / 3(9 / 4)=3 / 2\)
\(\operatorname{Var}(S)=\lambda E\left(X^{2}\right)=2 / 3(1 / 4+4 / 4+9 / 2)=23 / 6\)
So cumulative premium to time 2 is \(2(3 / 2+1.8 \sqrt{23 / 6})=10\), where the expression in parentheses is the annual premium.

Times between claims are determined by \(-(1 / \lambda) \log (1-u)\) and are \(0.43,0.77,1.37,2.41\). So 2 claims before time 2 (second claim is at 1.20 ; third is at 2.57 )

Sizes are \(2,3,1,3\), where only the first two matter.
The gain to the insurer is \(10-(2+3)=5\).
Question: You are given:
(i) A random sample of five observations from a population is:
\[
\begin{array}{lllll}
0.2 & 0.7 & 0.9 & 1.1 & 1.3
\end{array}
\]
(ii) You use the Kolmogorov-Smirnov test for testing the null hypothesis, \(H_{0}\), that the probability density function for the population is:
\[
f(x)=\frac{4}{(1+x)^{5}}, \quad x>0
\]
(iii) Critical values for the Kolmogorov-Smirnov test are:

\begin{tabular}{|l|l|l|l|l|}
\hline Level of Significance & 0.10 & 0.05 & 0.025 & 0.01 \\
\hline Critical Value & \(\frac{1.22}{\sqrt{n}}\) & \(\frac{1.36}{\sqrt{n}}\) & \(\frac{1.48}{\sqrt{n}}\) & \(\frac{1.63}{\sqrt{n}}\) \\
\hline
\end{tabular}

Determine the result of the test.

Options:
A
Do not reject \(H_{0}\) at the 0.10 significance level.
B
Reject \(H_{0}\) at the 0.10 significance level, but not at the 0.05 significance level.
C
Reject \(H_{0}\) at the 0.05 significance level, but not at the 0.025 significance level.
D
Reject \(H_{0}\) at the 0.025 significance level, but not at the 0.01 significance level.
E
Reject \(H_{0}\) at the 0.01 significance level.

Correct answer: D

Solution: The CDF is \(F(x)=1-\frac{1}{(1+x)^{4}}\)

\begin{tabular}{|c|c|c|c|}
\hline Observation \((x)\) & \(F(x)\) & compare to: & Maximum difference \\
\hline 0.2 & 0.518 & \(0,0.2\) & 0.518 \\
\hline 0.7 & 0.880 & \(0.2,0.4\) & 0.680 \\
\hline 0.9 & 0.923 & \(0.4,0.6\) & 0.523 \\
\hline 1.1 & 0.949 & \(0.6,0.8\) & 0.349 \\
\hline 1.3 & 0.964 & \(0.8,1.0\) & 0.164 \\
\hline
\end{tabular}

The K-S statistic is the maximum from the last column, 0.680.
Critical values are: \(0.546,0.608,0.662\), and 0.729 for the given levels of significance. The test statistic is between 0.662 ( \(2.5 \%\) ) and \(0.729(1.0 \%)\) and therefore the test is rejected at 0.025 and not at 0.01 .
Question: The random variable \(N\) has a mixed distribution:
(i) With probability \(p, N\) has a binomial distribution with \(q=0.5\) and \(m=2\).
(ii) With probability \(1-p, N\) has a binomial distribution with \(q=0.5\) and \(m=4\).

Which of the following is a correct expression for \(\operatorname{Pr}(N=2)\) ?

Options:
A
\(\quad 0.125 p^{2}\)
B
\(0.375+0.125 p\)
C
\(\quad 0.375+0.125 p^{2}\)
D
\(\quad 0.375-0.125 p^{2}\)
E
\(0.375-0.125 p\)

Correct answer: E

Solution: With probability \(p, \operatorname{Pr}(N=2)=0.5^{2}=0.25\). With probability \((1-p)\), \(\operatorname{Pr}(\mathrm{N}=2)=\binom{4}{2} 0.5^{4}=0.375 . \quad \operatorname{Pr}(N=2)=p(0.25)+(1-p)(0.375)=0.375-0.125 p\)
Question: You are given the following four pairs of observations:
\[
x_{1}=(-1,0), \quad x_{2}=(1,1), \quad x_{3}=(2,-1), \text { and } x_{4}=(5,10) .
\]

A hierarchical clustering algorithm is used with complete linkage and Euclidean distance.

Calculate the intercluster dissimilarity between \(\left\{x_{1}, x_{2}\right\}\) and \(\left\{x_{4}\right\}\).

Options:
A
2.2
B
3.2
C
9.9
D
10.8
E
11.7

Correct answer: E

Solution: First, calculate the distance between pairs of elements in each set. There are two pairs here:
\(x_{1}, x_{4}: \sqrt{(-1-5)^{2}+(0-10)^{2}}=\sqrt{136}=11.66\)
\(x_{2}, x_{4}: \sqrt{(1-5)^{2}+(1-10)^{2}}=\sqrt{97}=9.85\).

For complete linkage, the dissimilarity measure used is the maximum, which is 11.66 .
Question: Determine which of the following statements is/are true.
I. The number of clusters must be pre-specified for both \(K\)-means and hierarchical clustering.
II. The \(K\)-means clustering algorithm is less sensitive to the presence of outliers than the hierarchical clustering algorithm.
III. The \(K\)-means clustering algorithm requires random assignments while the hierarchical clustering algorithm does not.

Options:
A
I only
B
II only
C
III only
D
I, II and II
E
The correct answer is not given by (A), (B), (C), or (D)

Correct answer: C

Solution: I is false because the number of clusters is pre-specified in the \(K\)-means algorithm but not for the hierarchical algorithm.

II is also false because both algorithms force each observation to a cluster so that both may be heavily distorted by the presence of outliers.

III is true.
Question: You are given:
i) The random walk model
\[
y_{t}=y_{0}+c_{1}+c_{2}+\cdots+c_{t}
\]
where \(c_{t}, t=0,1,2, \ldots, T\) denote observations from a white noise process.
ii) The following nine observed values of \(c_{t}\) :

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline\(t\) & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\
\hline\(c_{t}\) & 2 & 3 & 5 & 3 & 4 & 2 & 4 & 1 & 2 \\
\hline
\end{tabular}
iii) The average value of \(c_{1}, c_{2}, \ldots, c_{10}\) is 2 .
iv) The 9 step ahead forecast of \(y_{19}, \hat{y}_{19}\), is estimated based on the observed value of \(y_{10}\).

Calculate the forecast error, \(y_{19}-\hat{y}_{19}\).

Options:
A
1
B
2
C
3
D
8
E
18

Correct answer: D

Solution: \(y_{10}=y_{0}+c_{1}+\cdots+c_{10}=y_{0}+20\)
\(y_{19}=y_{10}+c_{11}+\cdots+c_{19}=y_{0}+20+c_{11}+\cdots+c_{19}=y_{0}+20+26=y_{0}+46\)
\(\hat{y}_{19}=y_{10}+\hat{c}_{11}+\cdots+\hat{c}_{19}=y_{0}+20+9(2)=y_{0}+38\)
\(y_{19}-\hat{y}_{19}=\left(y_{0}+46\right)-\left(y_{0}+38\right)=8\).
Question: You are given:
i) The random walk model
\[
y_{t}=y_{0}+c_{1}+c_{2}+\cdots+c_{t}
\]
where \(c_{t}, t=0,1,2, \ldots T\) denote observations from a white noise process.
ii) The following ten observed values of \(y_{t}\).

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline\(t\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline\(y_{t}\) & 2 & 5 & 10 & 13 & 18 & 20 & 24 & 25 & 27 & 30 \\
\hline
\end{tabular}
iii) \(\quad y_{0}=0\)

Calculate the standard error of the 9 step-ahead forecast, \(\hat{y}_{19}\).

Options:
A
\(4 / 3\)
B
4
C
9
D
12
E
16

Correct answer: B

Solution: \(c_{t}=y_{t}-y_{t-1}\) and hence \(c_{1}, c_{2}, \ldots c_{10}=2,3,5,3,5,2,4,1,2,3\).
The mean of the \(c\) values is 3 , the variance is \((1+0+4+0+4+1+1+4+1+0) / 9=16 / 9\). The standard deviation is \(4 / 3\). The standard error of the forecast is \((4 / 3) \sqrt{9}=4\).
Question: Consider the following statements:
I. Principal Component Analysis (PCA) provide low-dimensional linear surfaces that are closest to the observations.
II. The first principal component is the line in p-dimensional space that is closest to the observations.
III. PCA finds a low dimension representation of a dataset that contains as much variation as possible.
IV. PCA serves as a tool for data visualization.

Determine which of the statements are correct.

Options:
A
Statements I, II, and III only
B
Statements I, II, and IV only
C
Statements I, III, and IV only
D
Statements II, III, and IV only
E
Statements I, II, III, and IV are all correct

Correct answer: E

Solution: Statement I is correct - Principal components provide low-dimensional linear surfaces that are closest to the observations.

Statement II is correct - The first principal component is the line in p-dimensional space that is closest to the observations.

Statement III is correct - PCA finds a low dimension representation of a dataset that contains as much variation as possible.

Statement IV is correct - PCA serves as a tool for data visualization.
Question: Consider the following statements:
I. The proportion of variance explained by an additional principal component never decreases as more principal components are added.
II. The cumulative proportion of variance explained never decreases as more principal components are added.
III. Using all possible principal components provides the best understanding of the data.
IV. A scree plot provides a method for determining the number of principal components to use.

Determine which of the statements are correct.

Options:
A
Statements I and II only
B
Statements I and III only
C
Statements I and IV only
D
Statements II and III only
E
Statements II and IV only

Correct answer: E

Solution: Statement I is incorrect - The proportion of variance explained by an additional principal component decreases or stays the same as more principal components are added. Statement II is correct - The cumulative proportion of variance explained increases or stays the same as more principal components are added.

Statement III is incorrect - We want to use the least number of principal components required to get the best understanding of the data.

Statement IV is correct - Typically, the number of principal components is chosen based on a scree plot.
Question: Determine which of the following statements describe the advantages of using a subset selection process, such as backward selection.
I. Doing so will likely result in a simpler model
II. Doing so will likely improve prediction accuracy
III. The results are likely to be easier to interpret

Options:
A
I only
B
II only
C
III only
D
I, II, and III
E
The correct answer is not given by (A), (B), (C), or (D)

Correct answer: D

Solution: Subset selection procedures will tend to remove the irrelevant variables from the predictors, thus resulting in a simpler and easier to interpret model. Accuracy will likely be improved due to reduction in variance.
Question: A classification tree is being constructed to predict if an insurance policy will lapse. A random sample of 100 policies contains 30 that lapsed. You are considering two splits:

Split 1: One node has 20 observations with 12 lapses and one node has 80 observations with 18 lapses.
Split 2: One node has 10 observations with 8 lapses and one node has 90 observations with 22 lapses.
The total Gini index after a split is the weighted average of the Gini index at each node, with the weights proportional to the number of observations in each node.
The total entropy after a split is the weighted average of the entropy at each node, with the weights proportional to the number of observations in each node.

Determine which of the following statements is/are true?
I. Split 1 is preferred based on the total Gini index.
II. Split 1 is preferred based on the total entropy.
III. Split 1 is preferred based on having fewer classification errors.

Options:
A
I only
B
II only
C
III only
D
I, II, and III
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: E

Solution: The total Gini index for Split 1 is \(2[20(12 / 20)(8 / 20)+80(18 / 80)(62 / 80)] / 100=0.375\) and for Split 2 is \(2[10(8 / 10)(2 / 10)+90(22 / 90)(68 / 90)] / 100=0.3644\). Smaller is better, so Split 2 is preferred. The factor of 2 is due to summing two identical terms (which occurs when there are only two classes).
The total entropy for Split 1 is \(-[20(12 / 20) \ln (12 / 20)+20(8 / 20) \ln (8 / 20)+ 80(18 / 80) \ln (18 / 80)+80(62 / 80) \ln (62 / 80)] / 100=0.5611\) and for Split 2 is \(-[10(8 / 10) \ln (8 / 10)+10(2 / 10) \ln (2 / 10)+90(22 / 90) \ln (22 / 90)+90(68 / 90) \ln (68 / 90)] / 100=\) 0.5506. Smaller is better, so Split 2 is preferred.

For Split 1, there are \(8+18=26\) errors and for Split 2 there are \(2+22=24\) errors. With fewer errors, Split 2 is preferred.
Question: Determine which of the following statements about random forests is/are true?
I. If the number of predictors used at each split is equal to the total number of available predictors, the result is the same as using bagging.
II. When building a specific tree, the same subset of predictor variables is used at each split.
III. Random forests are an improvement over bagging because the trees are decorrelated.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: C

Solution: II is false because with random forest a new subset of predictors is selected for each split.
Question: You are given the following results from a regression model.

\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{c} 
Observation \\
number \((i)\)
\end{tabular} & \(y_{i}\) & \(\hat{f}\left(x_{i}\right)\) \\
\hline 1 & 2 & 4 \\
\hline 2 & 5 & 3 \\
\hline 3 & 6 & 9 \\
\hline 4 & 8 & 3 \\
\hline 5 & 4 & 6 \\
\hline
\end{tabular}

Calculate the sum of squared errors (SSE).

Options:
A
-35
B
-5
C
5
D
35
E
46

Correct answer: E

Solution: Solution: SSE is sum of the squared differences between the observed and predicted values. That is, \(\left[(2-4)^{2}+(5-3)^{2}+(6-9)^{2}+(8-3)^{2}+(4-6)^{2}\right]=46\).
Question: Determine which of the following statements is/are true for a simple linear relationship, \(y=\beta_{0}+\beta_{1} x+\varepsilon\).
I. If \(\varepsilon=0\), the \(95 \%\) confidence interval is equal to the \(95 \%\) prediction interval.
II. The prediction interval is always at least as wide as the confidence interval.
III. The prediction interval quantifies the possible range for \(E(y \mid x)\).

Options:
A
I only
B
II only
C
III only
D
I, II, and III
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: E

Solution: I is true. The prediction interval includes the irreducible error, but in this case it is zero.
II is true. Because it includes the irreducible error, the prediction interval is at least as wide as the confidence interval.
III. is false. It is the confidence interval that quantifies this range.
Question: From an investigation of the residuals of fitting a linear regression by ordinary least squares it is clear that the spread of the residuals increases as the predicted values increase. Observed values of the dependent variable range from 0 to 100.

Determine which of the following statements is/are true with regard to transforming the dependent variable to make the variance of the residuals more constant.
I. Taking the logarithm of one plus the value of the dependent variable may make the variance of the residuals more constant.
II. A square root transformation may make the variance of the residuals more constant.
III. A logit transformation may make the variance of the residuals more constant.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: B

Solution: Adding a constant to the dependent variable avoids the problem of the logarithm of zero being negative infinity. In general, a log transformation may make the variance constant. Hence I is true. Power transformations with the power less than one, such as the square root transformation, may make the variance constant. Hence II is true. A logit transformation requires that the variable take on values between 0 and 1 and hence cannot be used here.
Question: You are performing a \(K\)-means clustering algorithm on a set of data. The data has been initialized randomly with 3 clusters as follows:

\begin{tabular}{|l|l|}
\hline Cluster & Data Point \\
\hline A & (2, -1) \\
\hline A & \((-1,2)\) \\
\hline A & (-2, 1) \\
\hline A & \((1,2)\) \\
\hline B & \((4,0)\) \\
\hline B & \((4,-1)\) \\
\hline B & \((0,-2)\) \\
\hline B & \((0,-5)\) \\
\hline C & (-1, 0) \\
\hline C & \((3,8)\) \\
\hline C & (-2, 0) \\
\hline C & \((0,0)\) \\
\hline
\end{tabular}

A single iteration of the algorithm is performed using the Euclidian distance between points and the cluster containing the fewest number of data points is identified.

Calculate the number of data points in this cluster.

Options:
A
0
B
1
C
2
D
3
E
4

Correct answer: D

Solution: The cluster centers are \(\mathrm{A}:(0,1), \mathrm{B}:(2,-2)\), and \(\mathrm{C}:(0,2)\). The new assignments are:

\begin{tabular}{|l|l|l|}
\hline Cluster & Data Point & New Cluster \\
\hline A & ( \(2,-1\) ) & B \\
\hline A & (-1, 2) & C \\
\hline A & (-2, 1) & A \\
\hline A & \((1,2)\) & C \\
\hline B & \((4,0)\) & B \\
\hline B & \((4,-1)\) & B \\
\hline B & \((0,-2)\) & B \\
\hline B & \((0,-5)\) & B \\
\hline C & (-1, 0) & A \\
\hline C & \((3,8)\) & C \\
\hline C & (-2, 0) & A \\
\hline C & \((0,0)\) & A \\
\hline
\end{tabular}

Cluster C has the fewest points with three.
Question: For a simple linear regression model the sum of squares of the residuals is \(\sum_{i=1}^{25} e_{i}^{2}=230\) and the \(R^{2}\) statistic is 0.64.

Calculate the total sum of squares (TSS) for this model.

Options:
A
\(\quad 605.94\)
B
638.89
C
690.77
D
701.59
E
750.87

Correct answer: B

Solution: TSS \(=(\) Residual Sum of Squares \() /\left(1-R^{2}\right)=230 /(1-0.64)=638.89\)
Question: A random walk is expressed as
\[
y_{t}=y_{t-1}+c_{t} \text { for } t=1,2, \ldots
\]
where
\[
E\left(c_{t}\right)=\mu_{c} \text { and } \operatorname{Var}\left(c_{t}\right)=\sigma_{c}^{2}, \quad t=1,2, \ldots
\]

Determine which statements is/are true with respect to a random walk model.
I. If \(\mu_{C} \neq 0\), then the random walk is nonstationary in the mean.
II. If \(\sigma_{c}^{2}=0\), then the random walk is nonstationary in the variance.
III. If \(\sigma_{c}^{2}>0\), then the random walk is nonstationary in the variance.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: C

Solution: I is true because the mean \(E\left(y_{t}\right)=y_{0}+t \mu_{c}\) depends on \(t\).
II is false because the variance \(\operatorname{Var}\left(y_{t}\right)=t \sigma_{c}^{2}=0\) does not depend in \(t\).
III is true because the variance depends on \(t\).
Question: Determine which of the following statements concerning decision tree pruning is/are true.
I. The recursive binary splitting method can lead to overfitting the data.
II. A tree with more splits tends to have lower variance.
III. When using the cost complexity pruning method, \(\alpha=0\) results in a very large tree.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: C

Solution: I is true because the method optimizes with respect to the training set, but may perform poorly on the test set.
II is false because additional splits tends to increase variance due to adding to the complexity of the model.

III is true because in this case only the training error is measured.
Question: Determine which of the following considerations may make decision trees preferable to other statistical learning methods.
I. Decision trees are easily interpretable.
II. Decision trees can be displayed graphically.
III. Decision trees are easier to explain than linear regression methods.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: E

Solution: All three statements are true. See Section 8.1 of An Introduction to Statistical Learning. The statement that trees are easier to explain than linear regression methods may not be obvious. For those familiar with regression but just learning about trees, the reverse may be the case. However, for those not familiar with regression, relating the dependent variable to the independent variables, especially if the dependent variable has been transformed, can be difficult to explain.
Question: Principal component analysis is applied to a large data set with four variables. Loadings for the first four principal components are estimated.
Determine which of the following statements is/are true with respect the loadings.
I. The loadings are unique.
II. For a given principal component, the sum of the squares of the loadings across the four variables is one.
III. Together, the four principal components explain \(100 \%\) of the variance.

Options:
A
None
B
I and II only
C
I and III only
D
II and III only
E
The correct answer is not given by (A), (B), (C), or (D).

Correct answer: D

Solution: I is false because the loadings are unique only up to a sign flip.
II is true. Principal components are designed to maximize variance. If there are no constraints on the magnitude of the loadings, the variance can be made arbitrarily large. The PCA algorithm's constraint is that the sum of the squares of the loadings equals 1 . III is true because four components can capture all the variation in four variables, provided there are at least four data points (note that the problem states that the data set is large).