(a) A production unit manufacturing surgical masks is concerned about the quality of their masks. A random sample of n masks are…

(a) Sample size for estimating a proportion

Let p be the true proportion of defective masks and p̂ the sample proportion. For large n, by the normal approximation, p̂ ≈ N(p, p(1-p)/n). We need P(p̂ lies in p ± 0.1) = 0.95. Thus P(|p̂ − p| ≤ 0.1) = 0.95, so 0.1 = 1.96 sqrt(p(1-p)/n). Since p is unknown, use the maximum possible value p(1-p) ≤ 1/4. Hence n ≥ (1.96)² (1/4) / (0.1)² = 3.8416 × 0.25 / 0.01 = 96.04. Therefore the required sample size is n = 97 masks. If a prior guess p₀ is available, replace 1/4 by p₀(1-p₀). The normal approximation is valid when np and n(1-p) are at least about 5.

(b) Conditional probability from the policy counts

Let A = auto, H = home, M = medical. Define: a = only auto, x = auto and home only, y = auto and medical only, z = home and medical only, t = all three. Given: h = only home = 30, m = only medical = 10. From (iii): a + x + y = 98. From (iv): m + y + z = 27 ⇒ 10 + y + z = 27 ⇒ y + z = 17. From (v): y + t = 13.

The total number of policy-holders is 150, so a + x + y + z + t + h + m = 150. Using a + x + y = 98 and h + m = 40, 98 + z + t + 40 = 150 ⇒ z + t = 12. Now y + z = 17 and z + t = 12 give y − t = 5, so y = t + 5. But y + t = 13, hence (t + 5) + t = 13 ⇒ 2t = 8 ⇒ t = 4. Then y = 9 and z = 8.

Thus the number having medical insurance is m + y + z + t = 10 + 9 + 8 + 4 = 31. The number having both home and medical insurance is z + t = 8 + 4 = 12. Therefore, P(H | M) = 12/31. (Here (v) is read as all holders of both auto and medical insurance, including those with all three.) Answer: 12/31.

(c) Conditional means given the order of two exponentials

Since X and Y are iid exponential with mean λ, f_X(x) = (1/λ) e^−x/λ, x > 0. Also P(Z = 1) = P(X < Y) = 1/2.

Now E[X | Z = 1] = E[X | X < Y] = [∫₀^∞ ∫_x^∞ x f_X(x) f_Y(y) dy dx] / P(X < Y). The inner integral is f_X(x) P(Y > x) = (1/λ) e^−x/λ e^−x/λ. So the numerator is (1/λ) ∫₀^∞ x e^−2x/λ dx = (1/λ)(λ²/4) = λ/4. Dividing by 1/2 gives E[X | Z = 1] = λ/2.

Similarly, E[X | Z = 0] = [∫₀^∞ ∫₀^x x f_X(x) f_Y(y) dy dx] / P(X ≥ Y). The inner integral is f_X(x) F_X(x) = (1/λ) e^−x/λ(1 − e^−x/λ). Thus the numerator is (1/λ)[∫₀^∞ x e^−x/λ dx − ∫₀^∞ x e^−2x/λ dx] = (1/λ)(λ² − λ²/4) = 3λ/4. Dividing by 1/2 gives E[X | Z = 0] = 3λ/2.

Therefore, E[X | Z = 1] + E[X | Z = 0] = λ/2 + 3λ/2 = 2λ. Answer: 2λ.

(d) Cramér–Rao bound for an exponential family

The density is f(x, θ) = log(θ)/(θ − 1) · θ^x, 0 < x < 1, θ > 1. Write η = log θ. Then f(x, θ) = η/(e^η − 1) e^ηx = exp(ηx − A(η)), where A(η) = log((e^η − 1)/η). Thus this is a one-parameter exponential family with sufficient statistic x. For a sample of size n, the likelihood depends on the data through S = Σ Xᵢ, and the family is regular.

For one observation, E[X] = A′(η) = e^η/(e^η − 1) − 1/η = θ/(θ − 1) − 1/log θ. Also Var(X) = A″(η) = 1/(log θ)² − θ/(θ − 1)².

Take g(θ) = E_θ[X] = θ/(θ − 1) − 1/log θ. Then X̄ = (1/n) Σ Xᵢ is unbiased for g(θ). Its variance is Var(X̄) = Var(X)/n. The derivative is g′(θ) = A″(η)/θ. The Fisher information for n observations is I_n(θ) = n A″(η)/θ². Therefore the Cramér–Rao lower bound for g(θ) is (g′(θ))² / I_n(θ) = (A″(η)²/θ²) / (n A″(η)/θ²) = A″(η)/n = Var(X)/n = Var(X̄). Hence the bound is attained.

Yes. The function is g(θ) = θ/(θ − 1) − 1/log θ, and X̄ is the unbiased estimator attaining the C-R lower bound.

(e) Cauchy family: MLR and sufficiency

(i) For θ₂ > θ₁, the likelihood ratio is R(x) = f(x, θ₂)/f(x, θ₁) = [θ₂/(π(θ₂² + x²))] / [θ₁/(π(θ₁² + x²))] = (θ₂/θ₁) · (θ₁² + x²)/(θ₂² + x²). Then d/dx log R(x) = 2x/(θ₁² + x²) − 2x/(θ₂² + x²) = 2x(θ₂² − θ₁²) / [(θ₁² + x²)(θ₂² + x²)]. For x > 0 this derivative is positive, while for x < 0 it is negative. Hence R(x) decreases on (−∞, 0) and increases on (0, ∞). It is not monotone over the whole real line. Therefore the Cauchy family does not have a Monotone Likelihood Ratio.

(ii) Since f(x, θ) = θ/(π(θ² + x²)) = θ/(π(θ² + |x|²)), the density depends on x only through |x|. By the factorization theorem, f(x, θ) = g(|x|, θ) h(x), with h(x) = 1 and g(t, θ) = θ/(π(θ² + t²)). Hence |X| is sufficient for θ.

Let Y = |X|. For y > 0, f_Y(y, θ) = f_X(y, θ) + f_X(−y, θ) = 2θ/(π(θ² + y²)). For θ₂ > θ₁, the likelihood ratio is r(y) = f_Y(y, θ₂)/f_Y(y, θ₁) = (θ₂/θ₁) · (θ₁² + y²)/(θ₂² + y²). Then d/dy log r(y) = 2y(θ₂² − θ₁²) / [(θ₁² + y²)(θ₂² + y²)] > 0 for y > 0. Thus r(y) is nondecreasing in y. Therefore the distribution of |X| does have a Monotone Likelihood Ratio.