Question 1
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). Construct three Bernoulli random variables that are pairwise independent but not mutually independent, and justify
both claims. (3 marks)
1(b). Calls arrive according to N ∼ Poi(λ). Independently, each call is labelled urgent with probability p and routine with
probability 1 − p. Let U and R be the urgent and routine counts. Show that U ∼ Poi(λp), R ∼ Poi(λ(1 − p)), and that U
and R are independent. (7 marks)
1(c). A fair coin is tossed repeatedly. Find the expected number of tosses required to observe two consecutive tails for the
first time. Derive your answer using a recursion. (5 marks)
1(d). A string is formed by arranging the multiset
{A, A, A, B, B, C, C, D, D, E}.
How many distinct strings are possible, and what is the probability that the three A’s are consecutive when a distinct string is
selected uniformly? (5 marks)
Total: 20 marks
2
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 1: solution and marking guide
1(a). Let U, V be independent fair Bernoulli variables and set W = U ⊕ V (addition modulo 2). Each pair is uniform on
{0, 1}2, hence pairwise independent. But W is determined by U and V , and for example P(U = 0 , V = 0 , W = 1) = 0 ̸=
1/8, so the three variables are not mutually independent.
3 marks; A:1, B:2
1(b). Conditionally on N = n, U ∼ Bin(n, p) and R = n − U . For u, r ≥ 0,
P(U = u, R = r) = P(N = u + r)
u + r
u
pu(1 − p)r
= e−λ λu+r
(u + r)!
(u + r)!
u!r! pu(1 − p)r
=
e−λp (λp)u
u!
e−λ(1−p) (λ(1 − p))r
r!
.
This factorises into the stated Poisson pmfs, proving both marginal laws and independence.
7 marks; A:3, C:2, D:2
1(c). Let E0 be the expected additional tosses when the current suffix contains no trailing tail, and E1 when the last toss was
a tail. Then
E0 = 1 + 1
2 E0 + 1
2 E1, E 1 = 1 + 1
2 E0 + 1
2 · 0.
The first equation gives E0 = 2 + E1. Substitution into the second yields E1 = 1 + E0/2, hence E0 = 6.
5 marks; A:2, B:2, D:1
1(d). The total number is
10!
3!2!2!2! .
For consecutive A’s, treat AAA as one block. The eight resulting objects have three repeated pairs, so the favourable count is
8!
2!2!2! .
Thus the probability is
8!/(2!3)
10!/(3!2!3) = 3! 8!
10! = 1
15 .
5 marks; A:4, D:1
Question total: 20 marks
2
Question 2
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). A joint density is constant on the triangular region 0 < x < y < 1 and zero elsewhere. Find the constant and compute
E(XY ). (6 marks)
2(b). Let X be continuous with CDF FX and density fX . Independently,
P(Y = 1) = 1
2 , P(Y = 2) = 1
3 , P(Y = 3) = 1
6 .
For Z = XY , find FZ(z) and fZ(z) in terms of FX and fX . (6 marks)
2(c). Let X1, . . . , Xn be independent Exp(λ) random variables. Find the distribution of M = min(X1, . . . , Xn). (4 marks)
2(d). Let X1, . . . , Xn be independent standard normal variables. Use moment generating functions to identify the distribution
of S = Pn
i=1 Xi. (4 marks)
Total: 20 marks
3
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 2: solution and marking guide
2(a). The triangle has area 1/2, so the constant is c = 2. Then
E(XY ) = 2
Z 1
0
Z y
0
xy dx dy = 2
Z 1
0
y y2
2 dy =
Z 1
0
y3 dy = 1
4 .
6 marks; A:3, B:1, C:2
2(b). Conditioning on Y and using independence,
FZ(z) = 1
2 FX (z) + 1
3 FX (z/2) + 1
6 FX (z/3).
Differentiating gives
fZ(z) = 1
2 fX (z) + 1
6 fX (z/2) + 1
18 fX (z/3).
The scale factors 1/y arise from the chain rule.
6 marks; A:1, B:2, C:1, D:2
2(c). For m ≥ 0,
P(M > m ) =
nY
i=1
P(Xi > m) = e−nλm.
Thus FM (m) = 1 − e−nλm for m ≥ 0, and M ∼ Exp(nλ).
4 marks; A:2, B:1, D:1
2(d). A standard normal has MGF MX (t) = et2/2. Independence gives
MS(t) =
nY
i=1
MXi (t) = ent2/2,
which is the MGF of N (0, n). Hence S ∼ N(0, n).
4 marks; A:2, B:1, D:1
Question total: 20 marks
3
Question 3
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). Suppose EX = 2, EY = 1, E(XY ) = 0 , and 0 ≤ Y ≤ 3. Obtain a nontrivial lower bound on sd (X). (5 marks)
3(b). Suppose all nonnegative integer moments of U and V are equal. State a sufficient additional condition under which U
and V must have the same distribution, and justify the conclusion. (4 marks)
3(c). Let Xi be i.i.d. with mean θ and variance σ2. For
T = 1
n + 1
nX
i=1
Xi,
find the bias, variance and MSE of T as an estimator of θ. (5 marks)
3(d). A normal population has known standard deviation 3. For n = 25 observations, ¯x = 18.4. Compute a 98% confidence
interval for µ using z0.99 = 2.326, and interpret it correctly. (3 marks)
3(e). A population has unknown distribution but standard deviation at most10. Using Chebyshev’s inequality, find a sufficient
sample size for the sample mean to be within 2 of the population mean with confidence at least 0.99. (3 marks)
Total: 20 marks
4
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 3: solution and marking guide
3(a). Here
Cov(X, Y ) = 0 − (2)(1) = −2.
By Cauchy–Schwarz, Cov(X, Y )2 ≤ Var(X) Var(Y ). Since Y ∈ [0, 3],
Var(Y ) ≤ (3 − 0)2
4 = 9
4 .
Therefore
Var(X) ≥ 4
9/4 = 16
9 , sd(X) ≥ 4
3 .
5 marks; A:1, C:1, D:3
3(b). A sufficient condition is that both MGFs exist on an open interval containing 0. Equality of all moments then gives
identical Taylor series for the MGFs about 0, so the MGFs agree on a neighbourhood of 0. Since an MGF, when it exists
on such a neighbourhood, uniquely determines the distribution, U and V have the same distribution. Without a determinacy
condition, equality of all moments alone need not suffice.
4 marks; A:1, B:1, C:1, D:1
3(c). We have
ET = n
n + 1 θ, Bias(T ) = − θ
n + 1 ,
and, by independence,
Var(T ) = nσ2
(n + 1)2 .
Hence
MSE(T ) = θ2 + nσ2
(n + 1)2 .
5 marks; A:2, B:3
3(d). The interval is
18.4 ± 2.326 3√
25 = 18.4 ± 1.3956,
namely approximately (17.00, 19.80). The construction has 98% repeated-sampling coverage under the normal model with
the stated known variance; it is not a posterior probability statement about this realised interval.
3 marks; A:2, B:1
3(e). Chebyshev gives
P(| ¯X − µ| ≥ 2) ≤ σ2/n
4 ≤ 100
4n = 25
n .
To make this at most 0.01, take n ≥ 2500.
3 marks; A:2, B:1
Question total: 20 marks
4
Question 4
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). Let X1, . . . , Xn | θ be independent geometric variables with P(Xi = x | θ) = θ(1 − θ)x−1 for x = 1 , 2, . . .. If
θ ∼ Beta(a, b), derive the posterior distribution. (5 marks)
4(b). Derive the least-squares slope and intercept for the model yi = β0 + β1xi + ei. (6 marks)
4(c). The residual plot below accompanies a fitted straight line. Does the model fit well? (3 marks)
4(d). In each of two risk groups, a treatment has a lower failure rate than control, but the combined data show a higher failure
rate under treatment. Name the phenomenon and explain one mechanism that can cause it. (3 marks)
4(e). The data are {4, 7, 11, 13, 19}. Is {19, 7, 7, 4, 13} a bootstrap sample? Is {4, 7, 11, 13} a bootstrap sample? Justify.
(3 marks)
0 5 10 15 20
−0.5
0
0.5
xi
Residual bεi
Total: 20 marks
5
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). The likelihood is
L(θ) ∝ θn(1 − θ)
∑
i xi−n.
Multiplying by the prior kernel θa−1(1 − θ)b−1 gives
π(θ | x) ∝ θa+n−1(1 − θ)b+∑
i xi−n−1.
Hence
θ | x ∼ Beta
a + n, b +
X
i
xi − n
!
.
5 marks; A:1, B:1, C:2, D:1
4(b). Minimise Q(b0, b1) =P
i(yi − b0 − b1xi)2. The normal equations are
X
i
(yi − b0 − b1xi) = 0 ,
X
i
xi(yi − b0 − b1xi) = 0 .
The first gives b0 = ¯y − b1 ¯x. Substitution into the second gives
bβ1 = Sxy
Sxx
, bβ0 = ¯y −bβ1 ¯x.
The residual sum of squares is a convex quadratic, so this solution is the global minimum when Sxx > 0.
6 marks; A:1, B:1, C:2, D:2
4(c). Y es, the residuals show no obvious trend, curvature or changing spread; they are scattered around zero with roughly
constant variability. On this diagnostic alone, the linear mean and constant-variance assumptions look reasonable. This does
not prove all assumptions, but there is no visible contradiction.
3 marks; B:2, D:1
4(d). This is Simpson’s paradox. It can occur when treatment allocation is imbalanced across groups with very different
baseline risks. For example, many more high-risk patients may receive treatment, so aggregation changes the weights and
reverses the within-group comparison.
3 marks; A:1, B:1, C:1
4(e). The first is a bootstrap sample: it has size five and every value is drawn from the observed data, with replacement. The
second is not a standard bootstrap sample because it has size four rather than the original sample size five.
3 marks; A:3
Question total: 20 marks
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