(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10…

(a)

Let p = P(E), q = P(F), r = P(G). Pairwise independence gives P(E∩F) = pq = 1/10 and P(F∩G) = qr = 3/10. Hence p = 1/(10q) and r = 3/(10q).

Now Eᶜ ∪ G is the complement of E ∩ Gᶜ. Therefore P(Eᶜ ∪ G) = 1 − P(E ∩ Gᶜ).

Because E and G are independent, P(E∩G) = pr. So P(E ∩ Gᶜ) = P(E) − P(E∩G) = p − pr = p(1−r). Thus P(Eᶜ ∪ G) = 1 − p(1−r).

We need p(1−r) ≤ 1/12. Substitute p and r: p(1−r) = [1/(10q)] [1 − 3/(10q)] = 1/(10q) − 3/(100q²). Then p(1−r) ≤ 1/12 ⇔ 1/(10q) − 3/(100q²) ≤ 1/12 ⇔ multiply by 300q²: 30q − 9 ≤ 25q² ⇔ 25q² − 30q + 9 ≥ 0 ⇔ (5q − 3)² ≥ 0, which is always true. Hence p(1−r) ≤ 1/12. Therefore P(Eᶜ ∪ G) = 1 − p(1−r) ≥ 1 − 1/12 = 11/12.

Therefore P(Eᶜ ∪ G) ≥ 11/12.

(b)

As printed, M(t₁,t₂) = e^(t₁+2e^(t₂)−3) gives M(0,0) = e^(−1) ≠ 1, so it cannot be a joint moment generating function. The intended valid form that matches the conclusion is M(t₁,t₂) = exp(e^(t₁) + 2e^(t₂) − 3). Assume this standard correction.

Since X and Y are independent, M(t₁,t₂) = M_X(t₁) M_Y(t₂). Now exp(e^(t₁) + 2e^(t₂) − 3) = exp(e^(t₁) − 1) · exp(2e^(t₂) − 2).

The mgf of Poisson(λ) is exp(λ(e^t − 1)). Hence M_X(t₁) = exp(1·(e^(t₁) − 1)) ⇒ X ∼ Poisson(1), M_Y(t₂) = exp(2·(e^(t₂) − 1)) ⇒ Y ∼ Poisson(2).

Therefore X + Y ∼ Poisson(1+2) = Poisson(3). So for S = X + Y, P(S = k) = e^(−3) 3^k / k! , k = 0,1,2,...

Thus P(X+Y = 0) = P(S=0) = e^(−3), P(X+Y = 2) = P(S=2) = e^(−3) 3² / 2! = 9e^(−3)/2.

Hence 2P(X+Y = 2) = 2 · 9e^(−3)/2 = 9e^(−3) = 9P(X+Y = 0).

Therefore 2P(X+Y = 2) = 9P(X+Y = 0).

(c)

Let S_n = ∑ (i=1 to n) X_i, where X_i ∼ U(0,1). Then E(X_i) = 1/2, Var(X_i) = 1/12. So E(S_n) = n/2, Var(S_n) = n/12, standard deviation √(n/12).

By the Lindeberg–Lévy Central Limit Theorem, (S_n − n/2) / √(n/12) → N(0,1) in distribution.

Now P(S_n ≤ n/2 + √(n/144)) = P( (S_n − n/2)/√(n/12) ≤ √(n/144)/√(n/12) ).

The standardized upper limit is √(n/144)/√(n/12) = √(12/144) = √(1/12) = 1/√12 = √3/6.

Given √3 = 1·74, √3/6 = 1·74/6 = 0·29.

Therefore the limiting probability is Φ(0·29) = 0·6141.

Hence Lim_n→∞ P(∑ (i=1 to n) X_i ≤ n/2 + √(n/144)) = 0·6141.

(d)

(i) Both μ and σ² unknown. The joint density of X₁, X₂, ..., Xₙ is f(x₁,...,xₙ; μ,σ²) = (2πσ²)^(−n/2) exp[ −(1/(2σ²)) ∑ (i=1 to n) (x_i − μ)² ].

Expand the exponent: ∑ (x_i − μ)² = ∑ x_i² − 2μ ∑ x_i + nμ². Thus f = (2πσ²)^(−n/2) exp[ −∑ x_i²/(2σ²) + μ∑ x_i/σ² − nμ²/(2σ²) ].

By the Neyman–Fisher factorization theorem, the density depends on the sample only through ∑ X_i and ∑ X_i². Therefore a sufficient statistic for (μ,σ²) when both are unknown is T = (∑ X_i, ∑ X_i²). Equivalently, T = (X̄, S²), where X̄ is the sample mean and S² is the sample variance.

(ii) If σ² is known. Then the joint density becomes f = (2πσ²)^(−n/2) exp[ −∑ (x_i − μ)²/(2σ²) ] = (2πσ²)^(−n/2) exp[ −∑ x_i²/(2σ²) ] exp[ μ∑ x_i/σ² − nμ²/(2σ²) ].

The factor depending on both μ and the data is exp[ μ∑ x_i/σ² ]. Hence by factorization, the sufficient statistic for μ is ∑ X_i, or equivalently X̄.

Therefore, for known σ², ∑ X_i is sufficient for μ.

(e)

By the Neyman–Pearson Lemma, the best test rejects H₀ for large values of the likelihood ratio Λ(x) = f₁(x)/f₀(x).

Compute Λ(x): x=1: 0·05/0·01 = 5 x=2: 0·04/0·01 = 4 x=3: 0·03/0·01 = 3 x=4: 0·02/0·01 = 2 x=5: 0·01/0·01 = 1 x=6: 0·85/0·95 = 17/19 < 1.

The largest likelihood ratios occur at x = 1, 2, 3, 4, 5. We need size 0·03 under H₀. Under H₀, P(X=1)=0·01, P(X=2)=0·01, P(X=3)=0·01. So rejecting H₀ when X ∈ {1,2,3} gives size α = 0·01 + 0·01 + 0·01 = 0·03.

Thus the best critical region is C = {1,2,3}. Reject H₀ if X = 1, 2, or 3.

Type-II error is the probability of accepting H₀ when H₁ is true: β = P_H₁(X ∈ {4,5,6}) = 0·02 + 0·01 + 0·85 = 0·88.

The power of the test is 1 − β = 0·12.

Since power = 0·12 > size = 0·03, the test is unbiased.

Therefore the best test of size 0·03 rejects H₀ for X = 1,2,3; its type-II error is 0·88; and it is unbiased because its power exceeds its size.