Exact tests, Poisson conditioning and discreteness
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). Assuming 365 equally likely independent birthdays, find the probability that among ten people
exactly one pair shares a birthday and all other birthdays are distinct. (5 marks)
1(b). Ten cups contain five tea and five coffee. A person selects five cups as tea. Under random guessing,
let X be the number of true tea cups selected. If X = 4, compute the one-sided exact p-value P (X ≥ 4)
and test at 0.05. (6 marks)
1(c). State and justify the conditional distribution of one of two independent Poisson counts given their
total. (5 marks)
1(d). Let X and Y be discrete random variables and A an event. Define Z = X on A and Z = Y on Ac.
Prove that Z is a discrete random variable. (4 marks)
T otal: 20 marks
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Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 1: solution and marking guide
1(a). Choose the pair, their birthday, and assign eight distinct birthdays from the remaining 364:10
2
365(364)8/36510.
5 marks; A:2, C:3
1(b). X ∼ HGeom(10, 5, 5). Thus P (X ≥ 4) = [
5
4
5
1
+
5
5
5
0
]/
10
5
= 26 /252 = 13 /126 ≈ 0.103. Fail
to reject random guessing at 0.05.
6 marks; B:2, D:4
1(c). For k = 0, . . . , n, independence and Poisson additivity give
Pr(X = k | X + Y = n) = e−λλk/k! e−µµn−k/(n − k)!
e−(λ+µ)(λ + µ)n/n! .
Cancelling common factors yields
n
k
λ
λ + µ
k µ
λ + µ
n−k
,
the pmf of Bin(n, λ/(λ + µ)).
5 marks; A:3, B:2
1(d). Its image is contained in the countable set Im X ∪ Im Y . For every z, {Z = z} = ( A ∩ {X =
z}) ∪ (Ac ∩ {Y = z}), which is measurable. Hence Z is discrete.
4 marks; A:3, C:1
Question total: 20 marks
26
Joint density, lower bounds and moment determinacy
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). A joint density is constant c on 0 < x < y < z < 1. Find c and E[XY Z ]. (7 marks)
2(b). Suppose E[X] = 1 , E[Y ] = 2 , E[XY ] = 1 , and 0 ≤ Y ≤ 4. Give a nontrivial lower bound on
sd(X). (5 marks)
2(c). Suppose all moments of U and V agree. Under what additional condition can you conclude they
have the same distribution? (4 marks)
2(d). If Y ∼ Beta(α, 1), prove that a median is 2−1/α. (4 marks)
T otal: 20 marks
23
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 2: solution and marking guide
2(a). The region has volume 1/6, so c = 6. Then E[XY Z ] = 6
R 1
0
R z
0
R y
0 xyz dx dy dz = 1/8.
7 marks; A:3, B:1, C:1, D:2
2(b). Cov(X, Y ) = 1 −2 = −1. Since Cov2 ≤ Var(X) Var(Y ) and Var(Y ) ≤ (4−0)2/4 = 4 , Var(X) ≥ 1/4,
so sd(X) ≥ 1/2.
5 marks; A:1, B:2, D:2
2(c). A sufficient condition is that both MGFs exist on an open interval containing zero. Their Taylor
series then have identical coefficients, so the MGFs agree near zero; MGF uniqueness gives identical
distributions. Without a moment-determinacy condition, equal moments need not suffice.
4 marks; A:2, B:1, C:1
2(d). The density is αyα−1 on (0, 1), so F (y) = yα. A median m satisfies mα = 1/2, hence m = 2 −1/α.
4 marks; A:2, B:1, C:1
Question total: 20 marks
27
Estimator construction, sample size and intervals
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). Independent Xi ∼ N (µi, σ2) have possibly different means. With S2 = (n − 1)−1 P(Xi − ¯X)2, find
E[S2]. (7 marks)
3(b). For i.i.d. observations with mean µ and variance σ2, construct an unbiased estimator of µ2 using
¯X and S2. (5 marks)
3(c). A population has unknown distribution but standard deviation at most 12. Find a sufficient sample
size so that the sample mean is within 3 of the true mean with confidence at least 0.99. (5 marks)
3(d). A normal sample has known standard deviation 4, sample size 16, and sample mean 10. Compute
a 95% confidence interval for the mean using z0.975 = 1.96, and state its frequentist interpretation. (3
marks)
T otal: 20 marks
24
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 3: solution and marking guide
3(a). Use P(Xi − ¯X)2 = P X 2
i − n ¯X 2. Now E[X 2
i ] = σ2 + µ2
i , E[ ¯X] = ¯µ, and Var( ¯X) = σ2/n. Hence
E
hX
(Xi − ¯X)2
i
= nσ2 +
X
µ2
i − {σ2 + n¯µ2} = (n − 1)σ2 +
X
(µi − ¯µ)2.
Therefore
E[S2] = σ2 + 1
n − 1
nX
i=1
(µi − ¯µ)2.
The second term vanishes only when all means are equal.
7 marks; A:2, B:2, C:1, D:2
3(b). Since E[ ¯X 2] = µ2 + σ2/n and E[S2] = σ2, ¯X 2 − S2/n is unbiased for µ2.
5 marks; A:2, B:1, C:1, D:1
3(c). Chebyshev gives P (| ¯X − µ| ≥ 3) ≤ 144/(9n) = 16 /n. Require 16/n ≤ 0.01, hence n ≥ 1600.
5 marks; A:2, B:1, C:1, D:1
3(d). The interval is 10±1.96(4/4) = 10 ±1.96, namely (8.04, 11.96). The parameter is fixed; the interval
procedure has 95% repeated-sampling coverage under the model.
3 marks; A:2, B:1
Question total: 20 marks
28
Bayesian updating, multiple testing and regression
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). Let Xi|θ be exponential with rate θ and let θ ∼ Γ(a, b), where b is the rate parameter. Derive and
identify the posterior. (6 marks)
4(b). One hundred tests use family-wise level 0.01. Under Bonferroni, which of 8 × 10−5, 2 × 10−4 and
10−6 are significant? (4 marks)
4(c). In simple linear regression with fixed xi and independent errors of variance σ2, show that Cov( ˆβ0, ˆβ1) =
−σ2 ¯x/Sxx. (6 marks)
4(d). Choose plots for (i) a time series, (ii) comparing medians across groups, (iii) checking normality,
and (iv) two quantitative variables. (4 marks)
T otal: 20 marks
25
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). Likelihood ∝ θne−θ ∑ xi. Multiplication by the prior yields a Gamma kernel, so θ|x ∼ Γ(a + n, b +P xi).
6 marks; A:2, B:2, D:2
4(b). The threshold is 10−4. Thus 8 × 10−5 and 10−6 are significant, while 2 × 10−4 is not.
4 marks; A:2, B:1, C:1
4(c). Write ˆβ1 = P(xi − ¯x)Yi/Sxx and ˆβ0 = ¯Y − ¯x ˆβ1. Since Cov( ¯Y , ˆβ1) = σ2[nSxx]−1 P(xi − ¯x) = 0 ,
the covariance equals −¯x Var(ˆβ1) = −¯xσ2/Sxx.
6 marks; A:2, B:2, D:2
4(d). (i) Time-series line plot; (ii) side-by-side boxplots; (iii) normal Q-Q plot; (iv) scatterplot.
4 marks; A:2, B:1, C:1
Question total: 20 marks
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