(a) (i) If X is a random variable with finite variance, show that lim (n→∞) n² P{|X| > n} = 0. (ii) In a certain recruitment…

(a)(i) Proof that n² P{|X| > n} → 0

Let X have finite variance, so E X² < ∞. Step 1: On the event {|X| > n} we have n² < X², so n² 1{|X| > n} ≤ X² 1{|X| > n}. Step 2: Taking expectations, 0 ≤ n² P{|X| > n} ≤ E[X² 1{|X| > n}]. Step 3: The events {|X| > n} decrease to the empty set (X is finite almost surely), and X² is integrable. By the dominated convergence theorem, E[X² 1{|X| > n}] → 0. Hence lim n² P{|X| > n} = 0 as n → ∞.

(a)(ii) Probability that the intelligent student was guessing

Let G be the event that he guessed and C the event that he ticks the correct option. Step 1: For an intelligent student, P(G) = 1 − 0·90 = 0·10 and P(not G) = 0·90. Step 2: If he guesses, P(C | G) = 1/4; if he knows, P(C | not G) = 1. Step 3: By Bayes' theorem, P(G | C) = (0·10 × 1/4) / (0·10 × 1/4 + 0·90 × 1) = 0·025/0·925 = 1/37 ≈ 0·027. The figures for the weaker student are not needed because the student is stated to be intelligent.

(b) Does the sequence obey the CLT?

The probabilities are valid for 0 ≤ λ < 1/2. Step 1: E Xₙ = 0 by symmetry, and E Xₙ² = n^(2λ) × 1/n^(2λ) = 1, so Var Xₙ = 1. Step 2: For Sₙ = X₁ + … + Xₙ, sₙ² = Var Sₙ = n. Step 3: E|Xₖ|⁴ = k^(4λ) × 1/k^(2λ) = k^(2λ). Step 4: Lyapunov's ratio with δ = 2 is (1/sₙ⁴) Σ E|Xₖ|⁴ = (1/n²) Σ k^(2λ) ≤ n^(1+2λ)/n² = n^(2λ−1), which tends to 0 because λ < 1/2. Step 5: Lyapunov's condition holds, so Sₙ/√n → N(0, 1) in distribution. The sequence obeys the CLT.

(c) SPRT, its OC function and ASN; SPRT for N(θ, 1)

Definition: for i.i.d. X₁, X₂, … with density f(x; θ), testing H₀: θ = θ₀ against H₁: θ = θ₁, let Λₙ = Π f(Xᵢ; θ₁)/f(Xᵢ; θ₀). With constants B < 1 < A, continue sampling while B < Λₙ < A, accept H₀ if Λₙ ≤ B, and accept H₁ if Λₙ ≥ A. By Wald's approximations A = (1 − β)/α and B = β/(1 − α). OC function: L(θ) = Pθ(accept H₀) ≈ (A^h − 1)/(A^h − B^h), where h = h(θ) ≠ 0 solves Eθ[(f₁/f₀)^h] = 1. ASN: Eθ(N) ≈ [L(θ) ln B + (1 − L(θ)) ln A]/Eθ(Z), where Z = ln(f₁/f₀). Application: θ₀ = 4, θ₁ = 5, α = 0·5, β = 0·2. Step 1: A = 0·8/0·5 = 1·6 and B = 0·2/0·5 = 0·4. So ln A = 0·470 and ln B = −0·916. Step 2: Λₙ = exp[Σ(Xᵢ − 4)²/2 − Σ(Xᵢ − 5)²/2] = exp(ΣXᵢ − 4·5n). Step 3: Continue sampling while 4·5n − 0·916 < ΣXᵢ < 4·5n + 0·470; accept H₀ if ΣXᵢ ≤ 4·5n − 0·916; accept H₁ if ΣXᵢ ≥ 4·5n + 0·470. Step 4: Here Z = X − 4·5, h(θ) = 9 − 2θ, L(4) = 0·5 = 1 − α and L(5) = 0·2 = β. The ASN is 0·446 at θ = 4 and 0·385 at θ = 5. These values are below 1 because α = 0·5 is very large.