(a) Let X and Y be two independent random variables following exponential distribution with mean 1/(λ) and 1/(μ) respectively, λ…

(a) Wald's SPRT based on Bernoulli indicators

Let p = P(X ≥ Y). Since X and Y are independent exponential with rates λ and μ,

p = ∫₀∞ P(X ≥ y) μ e^(−μy) dy = ∫₀∞ e^(−λy) μ e^(−μy) dy = μ/(λ+μ).

Ties have probability zero. Under H : λ = μ, p₀ = μ/(μ+μ) = 1/2. Under K : λ = 2μ, p₁ = μ/(2μ+μ) = 1/3. Hence U₁, U₂, ..., Uₙ are iid Bernoulli(p), with p₀ = 1/2 under H and p₁ = 1/3 under K. Let Sₙ = Σ Uᵢ.

For H : p = p₀ and K : p = p₁, the likelihood ratio is

Λₙ = ∏ [p₁^Uᵢ (1−p₁)^(1−Uᵢ)] / [p₀^Uᵢ (1−p₀)^(1−Uᵢ)] = (p₁/p₀)^Sₙ ((1−p₁)/(1−p₀))^(n−Sₙ) = (2/3)^Sₙ (4/3)^(n−Sₙ).

Therefore

log Λₙ = n log(4/3) − Sₙ log 2.

Let A = log(β/(1−α)) and B = log((1−β)/α), assuming 0 < α, β < 1 and α + β < 1. Wald’s SPRT is:

• if log Λₙ ≤ A, stop and accept H;
• if log Λₙ ≥ B, stop and reject H, i.e. accept K;
• if A < log Λₙ < B, continue sampling.

Equivalently, in terms of Sₙ:

• Accept H if Sₙ ≥ [n log(4/3) − log(β/(1−α))]/log 2.
• Reject H if Sₙ ≤ [n log(4/3) − log((1−β)/α)]/log 2.
• Continue if [n log(4/3) − log((1−β)/α)]/log 2 < Sₙ < [n log(4/3) − log(β/(1−α))]/log 2.

This is the required Wald SPRT based on the Uᵢ’s.

(b) Convergence in probability

Let Yᵢ ~ U(−1, 1) independently.

(i) For ε > 0,

P(|Yᵢ/i| > ε) = P(|Yᵢ| > iε).

Since |Yᵢ| ≤ 1 almost surely, this probability is 0 whenever iε ≥ 1. Hence for i > 1/ε, it is exactly 0. Therefore

P(|Yᵢ/i| > ε) → 0 as i → ∞.

Thus {Yᵢ/i} converges in probability to 0.

(ii) For ε > 0,

P(|Yᵢⁱ| > ε) = P(|Yᵢ| > ε^(1/i)).

If ε ≥ 1, this probability is 0. If 0 < ε < 1,

P(|Yᵢ| > ε^(1/i)) = 1 − ε^(1/i) = 1 − exp((log ε)/i).

As i → ∞, exp((log ε)/i) → 1, so the probability tends to 0. Hence {Yᵢⁱ} converges in probability to 0.

(c) Complete sufficient statistic and best unbiased estimator for U(−θ, θ)

The joint density of X₁, X₂, ..., Xₙ is

f(x₁, ..., xₙ; θ) = (2θ)^(−n), if −θ ≤ xᵢ ≤ θ for all i, and 0 otherwise.

This is equivalent to θ ≥ maxᵢ |xᵢ|. Therefore, by the factorization theorem, T = max₁≤i≤n |Xᵢ| is sufficient for θ.

The distribution of T is

P(T ≤ t) = P(|Xᵢ| ≤ t for all i) = (t/θ)ⁿ, 0 ≤ t ≤ θ.

Hence the density of T is

f_T(t) = n tⁿ⁻¹ / θⁿ, 0 < t < θ.

To prove completeness, suppose Eθ[g(T)] = 0 for all θ > 0. Then

∫₀θ g(t) n tⁿ⁻¹ / θⁿ dt = 0,

so

∫₀θ g(t) tⁿ⁻¹ dt = 0 for all θ > 0.

Differentiating with respect to θ gives g(θ)θⁿ⁻¹ = 0 almost everywhere, hence g = 0 almost everywhere. Thus T is complete sufficient.

Now

Eθ[T] = ∫₀θ t n tⁿ⁻¹ / θⁿ dt = nθ/(n+1).

Therefore

θ̂ = (n+1)/n T = (n+1)/n max₁≤i≤n |Xᵢ|

satisfies Eθ[θ̂] = θ. Since θ̂ is unbiased and is a function of the complete sufficient statistic T, by the Lehmann–Scheffé theorem it is the best unbiased estimator of θ.