Measurability and independence
Question 1
1(a). For f:U->V and A subset V, prove I_A(f(u))=I_{f^{-1}(A)}(u). (2 marks)
1(b). If G is a sigma-algebra on V, prove f^{-1}(G)={f^{-1}(A):A in G} is a sigma-algebra on U. (6
marks)
1(c). Construct three Bernoulli random variables that are pairwise independent but not mutually
independent. (4 marks)
1(d). Let Omega={1,2,3,4} and F={empty,Omega,{1,2},{3,4}}. Define X(1)=X(3)=0 and
X(2)=X(4)=1. Is X measurable? Give a non-constant measurable random variable on the same space. (4
marks)
1(e). A fair six-sided die is rolled seven times. Give a probability space and the cardinality of its power
set. (4 marks)
Total: 20 marks
MATH40005 Probability and Statistics Page 2
Worked solution and marking guidance
Question 1
1(a) (2 marks)
Recognition signal Indicator identities reduce to membership equivalence.
First key step Write the two “equals 1 iff” statements.
Historical basis 2023 Q2(a)(i)
Ability/method indicator/preimage identity; K02
Course-method solution. Both sides equal 1 exactly when f(u) belongs to A, equivalently u belongs to
f^{-1}(A), and equal 0 otherwise.
Marking points. 2 marks equivalence by cases
Common errors. Confusing image and preimage.
Final self-check. Test a point inside and outside the preimage.
Difficulty allocation: A:2; estimated normal time 3 min.
1(b) (6 marks)
Recognition signal Preimages preserve set operations exactly.
First key step Verify the three sigma-algebra axioms using preimage identities.
Historical basis 2023 Q2(a)(ii)
Ability/method preimage sigma-algebra proof; K02
Course-method solution. Preimages satisfy f^{-1}(empty)=empty, f^{-1}(A^c)=[f^{-1}(A)]^c, and
f^{-1}(union A_i)=union f^{-1}(A_i). Since G has the corresponding closure properties, the collection of
preimages is a sigma-algebra.
Marking points. 1 mark empty set; 2 marks complements; 2 marks countable unions; 1 mark conclusion
Common errors. Trying to prove f is one-to-one; it is unnecessary.
Final self-check. No probability measure is needed.
Difficulty allocation: A:2, B:2, D:2; estimated normal time 8 min.
1(c) (4 marks)
Recognition signal The standard parity construction separates pairwise from joint independence.
First key step Use two independent bits and their xor.
Historical basis 2026 Q2(a)
Ability/method pairwise-versus-mutual independence construction; K04
Course-method solution. Let U,V be independent Bernoulli(1/2), and W=U xor V. Each pair is uniform on
{0,1}^2, hence independent. But W is determined by U,V, and P(U=0,V=0,W=0)=1/4 differs from
(1/2)^3.
Marking points. 1 mark construction; 2 marks pairwise checks; 1 mark failure of mutual independence
Common errors. Giving three mutually independent variables.
Final self-check. The triple has only four possible outcomes, not eight.
Difficulty allocation: A:2, B:1, C:1; estimated normal time 5 min.
1(d) (4 marks)
Recognition signal Finite measurability is checked by singleton preimages.
First key step Test the preimage of 0.
Historical basis 2020 Q2(a); 2022 Q1(a)(ii)
Ability/method finite-space measurability check; K02
Course-method solution. No, because X^{-1}({0})={1,3} is not in F. A valid example is Y=0 on {1,2}
and Y=1 on {3,4}; its singleton preimages belong to F.
Marking points. 2 marks counterexample preimage; 2 marks valid example/justification
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Common errors. Checking only the image values.
Final self-check. A measurable variable must be constant on each atom {1,2} and {3,4}.
Difficulty allocation: A:2, B:1, C:1; estimated normal time 5 min.
1(e) (4 marks)
Recognition signal Repeated finite trials produce a Cartesian product; a finite power set has size
2^{|Omega|}.
First key step Count outcomes before subsets.
Historical basis 2021 Q1(b)(i)-(iii)
Ability/method product probability space and power set; K01,K05
Course-method solution. Omega={1,...,6}^7, |Omega|=6^7, F=P(Omega), P(A)=|A|/6^7. The power
set has cardinality 2^{6^7}.
Marking points. 1 mark sample space; 1 mark probability space; 2 marks power-set cardinality
Common errors. Answering 2^7 or 6^{2^7}.
Final self-check. The event-space cardinality is vastly larger than the sample-space cardinality.
Difficulty allocation: A:2, B:2; estimated normal time 5 min.
Total: 20 marks
MATH40005 Probability and Statistics Page 3
MGFs and random-variable transformations
Question 2
2(a). Let X~N(0,1), Y~Gamma(alpha,beta) in rate form, independently, and Z=sqrt(Y)X. Find the MGF of
Z and its domain. (7 marks)
2(b). Let X be continuous with density f_X and let independent Y take values 1,...,10. For Z=XY, derive
f_Z and verify it is a density. (6 marks)
2(c). A joint density is constant c on 0<x<y<z<1. Find c and E[XYZ]. (7 marks)
Total: 20 marks
MATH40005 Probability and Statistics Page 3
Worked solution and marking guidance
Question 2
2(a) (7 marks)
Recognition signal A random scale mixture signals conditional expectation.
First key step Condition on the scale variable Y.
Historical basis 2021 Q2(d)
Ability/method MGF of a scale mixture; K12,K23,K24
Course-method solution. Condition on Y=y: E[e^{uZ}|Y=y]=M_X(u sqrt(y))=exp(u^2 y/2). Thus
M_Z(u)=M_Y(u^2/2)=[beta/(beta-u^2/2)]^alpha, requiring u^2/2<beta, i.e. |u|<sqrt(2beta).
Marking points. 2 marks conditioning; 1 mark normal MGF; 2 marks gamma MGF; 1 mark domain; 1
mark result
Common errors. Replacing E[exp(u^2Y/2)] by exp(u^2E[Y]/2).
Final self-check. At u=0 the MGF is 1.
Difficulty allocation: A:1, B:1, C:2, D:3; estimated normal time 10 min.
2(b) (6 marks)
Recognition signal A product with a positive discrete multiplier is a finite mixture of scaled densities.
First key step Condition on Y and use the CDF method.
Historical basis 2022 Q2(c)(i)
Ability/method continuous-discrete mixture transform; K15,K20,K23
Course-method solution. Conditioning on Y, F_Z(z)=sum_{y=1}^{10} F_X(z/y)P(Y=y). Differentiating
gives f_Z(z)=sum_{y=1}^{10} [1/y]f_X(z/y)P(Y=y). It is non-negative and integrates to sum_y
P(Y=y)=1 after the change v=z/y.
Marking points. 1 mark condition/CDF; 2 marks differentiation; 1 mark non-negativity; 2 marks integral
check
Common errors. Omitting the Jacobian 1/y.
Final self-check. Each component density integrates to its mixture weight.
Difficulty allocation: A:2, B:2, D:2; estimated normal time 9 min.
2(c) (7 marks)
Recognition signal An ordered simplex has nested bounds.
First key step Normalize first, then reuse the same bounds for the moment.
Historical basis 2020 Q3(c)
Ability/method three-dimensional normalization and LOTUS; K17,K22,K20
Course-method solution. The ordered region has volume 1/6, so c=6. Direct LOTUS gives E[XYZ]=6
int_0^1 int_0^z int_0^y xyz dx dy dz=1/8.
Marking points. 2 marks normalization/constant; 3 marks triple integral; 2 marks result
Common errors. Assuming X,Y,Z independent.
Final self-check. The order-statistic symmetry gives c=3!=6.
Difficulty allocation: A:2, B:1, C:1, D:3; estimated normal time 10 min.
Total: 20 marks
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Exact tests, quartiles and confidence bounds
Question 3
3(a). Ten cups contain five tea and five coffee. A person selects five cups as tea and identifies four
correctly. Under random guessing, perform a one-sided exact test at level 0.05. (5 marks)
3(b). For the data {14.3,8.3,15.9,12.7,9.4,24.0,7.2,6.1,11.6,10.5}, compute the lower and upper
quartiles using the median-of-halves convention, then apply Tukey’s rule for outliers. (4 marks)
3(c). Combined data favor a placebo, but within each of two risk groups a drug has a lower disease rate.
Name the phenomenon and explain how it can occur. (3 marks)
3(d). Potato weights have unknown distribution and only the bound sigma < 10 g is known. Using
the bound 10 g to construct a distribution-free 99% Chebyshev interval, find the minimum integer n
that guarantees its total width is strictly less than 4 g. (5 marks)
3(e). Define Type I and Type II errors and explain qualitatively how increasing alpha affects them. (3
marks)
Total: 20 marks
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Worked solution and marking guidance
Question 3
3(a) (5 marks)
Recognition signal Fixed-size classification without replacement gives a hypergeometric null.
First key step Use the upper tail including the observed result.
Historical basis 2024 Q5(b) variant
Ability/method exact hypergeometric test; K10,K44
Course-method solution. Under H0, X~Hypergeometric(10,5,5). The p-value
P(X>=4)=[C(5,4)C(5,1)+C(5,5)C(5,0)]/C(10,5)=26/252=13/126≈0.103. Fail to reject H0.
Marking points. 1 mark model/hypotheses; 2 marks tail; 1 mark value; 1 mark conclusion
Common errors. Using only P(X=4).
Final self-check. The p-value exceeds 0.05.
Difficulty allocation: A:2, B:1, C:1, D:1; estimated normal time 7 min.
3(b) (4 marks)
Recognition signal Tukey outliers are based on quartiles and 1.5 IQR fences.
First key step Sort before splitting the sample.
Historical basis 2023 Q4(d)
Ability/method quartiles and Tukey outliers; K46
Course-method solution. Sorted: 6.1,7.2,8.3,9.4,10.5,11.6,12.7,14.3,15.9,24.0. Lower half median
q1=8.3; upper half median q3=14.3. IQR=6.0, fences [-0.7,23.3], so 24.0 is the only outlier.
Marking points. 1 mark sorting/quartiles; 1 mark IQR; 1 mark fences; 1 mark outlier
Common errors. Including both middle observations in halves for even n.
Final self-check. The identified value must lie outside a computed fence.
Difficulty allocation: A:2, B:1, C:1; estimated normal time 6 min.
3(c) (3 marks)
Recognition signal A reversal after aggregation is the signature of Simpson’s paradox.
First key step Compare subgroup weights and baseline risks.
Historical basis 2021 Q6(d); 2023 Q5(d)
Ability/method Simpson reversal/confounding; K51
Course-method solution. This is Simpson’s paradox. It occurs when treatment allocation is unbalanced
across groups with different baseline risks; aggregation changes the weights and can reverse the
within-group comparison. The risk group is a confounder.
Marking points. 1 mark name; 1 mark unequal weighting; 1 mark contextual explanation
Common errors. Calling it random error without explaining the weighting mechanism.
Final self-check. Stratified comparisons preserve the within-group direction.
Difficulty allocation: A:1, B:1, C:1; estimated normal time 4 min.
3(d) (5 marks)
Recognition signal No distribution is specified, so use the variance bound rather than normal theory.
First key step Translate confidence and width separately.
Historical basis 2021 Q4(b)
Ability/method Chebyshev sample-size design; K45,K28
Course-method solution. Since only the bound sigma < 10 is known, report the interval
Xbar +/- 10/sqrt(0.01n) = Xbar +/- 100/sqrt(n). Chebyshev gives coverage at least 0.99.
Its total width is 200/sqrt(n). The strict requirement 200/sqrt(n) < 4 is equivalent to
n > 2500. Hence the minimum integer is n = 2501.
Marking points. 1 mark use known bound 10; 1 mark half-width 100/sqrt(n); 1 mark total
width; 1 mark strict inequality n > 2500; 1 mark minimum integer n = 2501.
Common errors. Using the unknown actual sigma in a reported interval; accepting n = 2500, for which
the bound-based width is exactly 4; using z critical values.
Final self-check. 200/sqrt(2501) < 4, whereas 200/sqrt(2500) = 4.
Difficulty allocation: A:1, B:1, C:1, D:2; estimated normal time 8 min.
3(e) (3 marks)
Recognition signal The two errors condition on different truths.
First key step State the decision and the true state.
Historical basis 2022 Q5(a); 2025 Q5(c)
Ability/method testing error trade-off; K42
Course-method solution. Type I error is rejecting a true H0, with probability alpha for a level-alpha test.
Type II error is failing to reject a false H0. With other design features fixed, increasing alpha enlarges
the rejection region, increasing Type I error and generally decreasing Type II error/increasing power.
Marking points. 1 mark each definition; 1 mark trade-off
Common errors. Saying both decrease when alpha increases.
Final self-check. A larger rejection region makes rejection more frequent under both null and
alternative.
Difficulty allocation: A:1, B:1, C:1; estimated normal time 4 min.
Total: 20 marks
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MLE, regression and estimators
Question 4
4(a). Independent positive observations have lognormal density proportional to exp[-(log
x-theta)^2/(2sigma^2)]/x, where sigma^2 is known. Find the MLE of theta. (6 marks)
4(b). Show that the fitted least-squares line in simple linear regression passes through (xbar,ybar). (4
marks)
4(c). Suppose all non-negative integer moments of U and V agree. State a sufficient condition for
equality in distribution and justify. (4 marks)
4(d). Describe a percentile bootstrap 95% confidence interval for a median. (3 marks)
4(e). For i.i.d. observations with mean mu and variance sigma^2, construct an unbiased estimator of
mu^2 using Xbar and the unbiased sample variance S^2. (3 marks)
Total: 20 marks
MATH40005 Probability and Statistics Page 5
Worked solution and marking guidance
Question 4
4(a) (6 marks)
Recognition signal Known-variance lognormal location estimation becomes least squares on log data.
First key step Discard theta-free terms and differentiate.
Historical basis 2022 Q6(a)
Ability/method regular log-likelihood MLE; K53
Course-method solution. The log-likelihood in theta is constant -[1/(2sigma^2)]sum(log x_i-theta)^2.
Differentiating gives sum(log x_i-theta)=0, so theta_hat=n^{-1}sum log x_i. The second derivative is
-n/sigma^2<0.
Marking points. 2 marks log-likelihood; 2 marks score equation; 1 mark maximum check; 1 mark
estimator
Common errors. Using the arithmetic mean of x_i.
Final self-check. The estimate is the sample mean of transformed observations.
Difficulty allocation: A:2, B:1, C:1, D:2; estimated normal time 8 min.
4(b) (4 marks)
Recognition signal The intercept normal equation forces mean residual zero.
First key step Differentiate RSS with respect to beta0.
Historical basis 2022 Q6(d)
Ability/method OLS mean-point property; K48
Course-method solution. The normal equation for the intercept is sum(y_i-beta0-beta1x_i)=0, so
beta0_hat=ybar-beta1_hat xbar. Hence at x=xbar, yhat=beta0_hat+beta1_hat xbar=ybar.
Marking points. 2 marks normal equation; 1 mark intercept relation; 1 mark point conclusion
Common errors. Using the result without justification.
Final self-check. The fitted residuals sum to zero.
Difficulty allocation: A:3, B:1; estimated normal time 5 min.
4(c) (4 marks)
Recognition signal Moment equality requires a determinacy condition.
First key step Use MGF existence near zero.
Historical basis 2025 Q3(c)(i)
Ability/method moment determinacy; K34,K12
Course-method solution. If both MGFs exist on an open interval around 0, their Taylor expansions have
the same moment coefficients, so the MGFs agree there. MGF uniqueness yields equal distributions.
Marking points. 1 mark condition; 2 marks MGF equality; 1 mark uniqueness
Common errors. Claiming the conclusion without a condition.
Final self-check. MGFs that agree near zero uniquely determine distributions.
Difficulty allocation: A:1, B:1, C:1, D:1; estimated normal time 5 min.
4(d) (3 marks)
Recognition signal The bootstrap distribution is the distribution of replicate medians.
First key step Resample with replacement at the original size.
Historical basis 2025 Q6(c)
Ability/method bootstrap percentile interval; K52
Course-method solution. Generate many size-n resamples with replacement, compute each median,
and take the empirical 2.5th and 97.5th percentiles.
Marking points. 1 mark resampling; 1 mark statistic; 1 mark percentiles
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Common errors. Taking percentiles of raw data.
Final self-check. The endpoints come from bootstrap medians.
Difficulty allocation: A:2, B:1; estimated normal time 4 min.
4(e) (3 marks)
Recognition signal Correct the upward bias of Xbar^2 using an unbiased variance estimate.
First key step Compute E[Xbar^2].
Historical basis 2021 Q4(a)(iii)
Ability/method constructing an unbiased estimator; K32
Course-method solution. Since E[Xbar^2]=mu^2+sigma^2/n and E[S^2]=sigma^2, the estimator
Xbar^2-S^2/n is unbiased for mu^2.
Marking points. 1 mark moment identity; 1 mark estimator; 1 mark unbiasedness
Common errors. Using Xbar^2 alone.
Final self-check. Taking expectations returns exactly mu^2.
Difficulty allocation: A:2, B:1; estimated normal time 5 min.
Total: 20 marks
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