MATH40005 2026 Practice Paper 7

MATH40005 Probability and Statistics Resit Preparation Question 1: solution and marking guide 1(a). A map is measurable if the inverse image of every Borel set is in F; for this finite-valued map it is enough to check the inverse images of its values. Here X −1({0}) = {1, 3} /∈ F , so X is not measurable and hence is not a random variable on the stated measurable space. One valid non-constant example is Y (1) = Y (2) = 0 , Y (3) = Y (4) = 1 , because Y −1({0}) = {1, 2} and Y −1({1}) = {3, 4} are both in F. 4 marks; A:2, B:2 1(b). Non-negativity requires c ≥ 0. Normalisation gives 1 = c(1 + 2 + 3 + 4) = 10 c, so c = 1/10. Therefore FX (t) = 8 >>>>>>< >>>>>>: 0, t < 0, 1 10 , 0 ≤ t < 1, 3 10 , 1 ≤ t < 2, 6 10 , 2 ≤ t < 3, 1, t ≥ 3. 5 marks; A:3, C:2 1(c). The favourable selections are counted by 7 2 5 1 4 1 = 21 · 5 · 4 = 420 . There are 16 4 = 1820 equally likely four-token selections. Hence P(2R, 1B, 1G) = 420 1820 = 3 13 . 5 marks; A:3, B:2 1(d). Since X and Y are discrete, Im X and Im Y are countable. Thus Im Z ⊆ Im X ∪ Im Y is countable. For each z ∈ R, Z −1({z}) = X −1({z}) ∩ A ∪ Y −1({z}) ∩ Ac . Both sets in the union lie in F because X and Y are measurable, A, Ac ∈ F , and a sigma-algebra is closed under intersections and unions. Hence every singleton inverse image of Z is measurable, and Z is discrete. 6 marks; A:2, D:4 Question total: 20 marks 2

MATH40005 Probability and Statistics Resit Preparation Question 2: solution and marking guide 2(a). For m < 0, FM (m) = 0 ; for 0 ≤ m ≤ 2, FM (m) = nY i=1 P(Xi ≤ m) = m 2 n ; and for m ≥ 2, FM (m) = 1 . Differentiating on (0, 2) gives fM (m) = nmn−1 2n , 0 < m < 2, and zero otherwise. 5 marks; A:3, C:2 2(b). For t < 0, FT (t) = 0 . For t ≥ 0, FT (t) = P{− log(M /2) ≤ t} = P{M /2 ≥ e−t} = 1 − P{M < 2e−t} = 1 − e−nt. Thus T ∼ Exp(n). The use of < rather than ≤ is immaterial because M is continuous. 4 marks; B:1, C:1, D:2 2(c)(i). Normalisation requires 1 = c Z 1 0 Z x 0 (x + y) dy dx = c Z 1 0 3 2 x2 dx = c 2 , so c = 2. 3 marks; A:3 2(c)(ii). For 0 < x < 1, fX (x) = Z x 0 2(x + y) dy = 3x2. For 0 < y < 1, fY (y) = Z 1 y 2(x + y) dx = 1 + 2 y − 3y2. The variables are not independent: the support is triangular, so, for example, the joint density is zero at points with x < y even though both marginals are positive there. 4 marks; A:1, B:3 2(c)(iii). For 0 < x < 1 and 0 < y < x , fY |X=x(y | x) = 2(x + y) 3x2 . Therefore E(Y | X = x) = Z x 0 y 2(x + y) 3x2 dy = 5x 9 , 0 < x < 1. 4 marks; A:1, B:1, D:2 Question total: 20 marks 3

MATH40005 Probability and Statistics Resit Preparation Question 3: solution and marking guide 3(a). The standard normal-sample results are ¯X ∼ N µ, σ2 n , (n − 1)S2 σ2 ∼ χ2 n−1, and ¯X and S2 are independent. 4 marks; A:2, B:2 3(b). Using (n − 1)S2/σ2 ∼ χ2 9, the interval is 9s2 χ2 9,0.975 , 9s2 χ2 9,0.025 ! = 108 19.023 , 108 2.700 ≈ (5.68, 40.00). 4 marks; A:3, B:1 3(c). Since E( ¯X 2) = Var( ¯X) + [E( ¯X)]2 = σ2 n + µ2 and E(S2) = σ2, an unbiased estimator is cµ2 = ¯X 2 − S2 n . Indeed, E ¯X 2 − S2 n = µ2 + σ2 n − σ2 n = µ2. 5 marks; A:1, C:1, D:3 3(d). The median is 2.1. Including it in both halves gives q0.25 = 1.9, q 0.75 = 2.6, so IQR = 0.7. The Tukey fences are 1.9 − 1.5(0.7) = 0 .85, 2.6 + 1.5(0.7) = 3 .65. Thus 5.8 is the only outlier. 4 marks; A:2, C:1, D:1 3(e). After the data have been observed, the endpoints and the fixed parameter are no longer random in the frequentist model. The correct statement is that the procedure used to construct the interval produces intervals containing θ in 95% of repeated samples under the model. It is not correct to say that this particular realised interval contains θ with probability 0.95. 3 marks; B:3 Question total: 20 marks 4

MATH40005 Probability and Statistics Resit Preparation Question 4: solution and marking guide 4(a). Use bβ1 = P i(xi − ¯x)Yi Sxx , bβ0 = ¯Y − ¯xbβ1. First, Cov( ¯Y ,bβ1) = 1 nSxx X i,j (xj − ¯x) Cov(Yi, Yj) = σ2 nSxx X j (xj − ¯x) = 0 . Also Var(bβ1) = σ2/Sxx. Hence Cov(bβ0,bβ1) = Cov( ¯Y ,bβ1) − ¯x Var(bβ1) = − σ2 ¯x Sxx . 6 marks; A:1, B:1, C:2, D:2 4(b). The model does not fit well despite the high R2. The residuals show pronounced curvature rather than random scat- ter around zero, so the linear mean structure is inadequate. A reasonable improvement is to include a quadratic term, or equivalently fit a suitable transformation of x and then recheck the residuals. 3 marks; B:2, D:1 4(c). For observations in (0, 1), ℓ(θ) = n log θ + (θ − 1) nX i=1 log xi. Thus ℓ′(θ) = n θ + X i log xi, so bθ = − nP i log xi . Because ℓ′′(θ) = −n/θ2 < 0, this stationary point is the unique maximum whenever the observations lie in (0, 1) and are not all equal to 1 (an event of probability zero under the model). 6 marks; A:3, C:2, D:1 4(d). A bootstrap sample has size n and is obtained by sampling with replacement from the observed values x1, . . . , xn. Repetitions are therefore allowed, and some original observations may be absent. 2 marks; A:1, B:1 4(e). Correlation measures association, not causal direction. A high value can be generated by a common cause, time trend, selection bias, reverse causation, or coincidence. A causal conclusion requires an appropriate design and assumptions beyond the correlation coefficient. 3 marks; A:1, B:1, C:1 Question total: 20 marks 5