(a) Two events A and B are such that P(A) = 1/3, P(B) = 1/4 and P(A|B) + P(B|A) = 2/3. Evaluate the following: (i) P(A^c ∪ B^c)…

(a)(i) Evaluating P(Aᶜ ∪ Bᶜ)

Given P(A)=1/3 and P(B)=1/4. Let x = P(A∩B).

By definition of conditional probability:

P(A|B) = P(A∩B)/P(B) = x/(1/4) = 4x.

P(B|A) = P(A∩B)/P(A) = x/(1/3) = 3x.

Given P(A|B) + P(B|A) = 2/3, so 4x + 3x = 2/3 ⇒ 7x = 2/3 ⇒ x = 2/21.

Thus P(A∩B) = 2/21.

Now P(Aᶜ ∪ Bᶜ) = P((A∩B)ᶜ) = 1 − P(A∩B) = 1 − 2/21 = 19/21.

Final answer: 19/21.

(a)(ii) Evaluating P(A|Bᶜ) + P(B|Aᶜ)

P(Bᶜ) = 1 − 1/4 = 3/4.

P(A∩Bᶜ) = P(A) − P(A∩B) = 1/3 − 2/21 = 5/21.

So P(A|Bᶜ) = (5/21)/(3/4) = 20/63.

P(Aᶜ) = 1 − 1/3 = 2/3.

P(B∩Aᶜ) = P(B) − P(A∩B) = 1/4 − 2/21 = 13/84.

So P(B|Aᶜ) = (13/84)/(2/3) = 13/56.

Therefore, P(A|Bᶜ) + P(B|Aᶜ) = 20/63 + 13/56 = 160/504 + 117/504 = 277/504.

Final answer: 277/504.

(b)(i) Evaluating P(X ≥ 2, Y ≥ 1/2)

The joint probability function is f(x,y) = x y^(x−1)/3, x = 1,2,3, 0 < y < 1.

We need P(X ≥ 2 and Y ≥ 1/2). Since X is discrete and Y is continuous,

P(X ≥ 2, Y ≥ 1/2) = ∑ over x=2,3 of ∫ from y=1/2 to 1 of x y^(x−1)/3 dy.

For a fixed x, ∫ from 1/2 to 1 of x y^(x−1)/3 dy = (1/3)[yˣ] from 1/2 to 1 = (1/3)(1 − (1/2)ˣ).

For x = 2: (1/3)(1 − 1/4) = 1/4.

For x = 3: (1/3)(1 − 1/8) = 7/24.

Sum = 1/4 + 7/24 = 6/24 + 7/24 = 13/24.

Final answer: 13/24.

(b)(ii) Evaluating P(X ≥ 2)

Marginal probability of X:

P(X = x) = ∫ from y=0 to 1 of x y^(x−1)/3 dy = (1/3)[yˣ] from 0 to 1 = 1/3, for x = 1,2,3.

Thus X is uniform on {1,2,3}.

P(X ≥ 2) = P(X = 2) + P(X = 3) = 1/3 + 1/3 = 2/3.

Final answer: 2/3.

(c) WLLN Does Not Hold for ⟨Sₙ⟩

Let Zᵢ = Xᵢ − μ. For the sequence ⟨Sₙ⟩, the average of interest is

Cₙ = (1/n)∑ from k=1 to n of (Sₖ − E(Sₖ)) = (1/n)∑ from k=1 to n of (Sₖ − kμ).

Now Sₖ = ∑ from i=1 to k of Xᵢ, so

Cₙ = (1/n)∑ from k=1 to n of ∑ from i=1 to k of Zᵢ

= (1/n)∑ from i=1 to n of (n − i + 1)Zᵢ.

Let aₙ,ᵢ = (n − i + 1)/n. Then Cₙ = ∑ from i=1 to n of aₙ,ᵢ Zᵢ.

Since the Zᵢ are independent with mean 0 and variance σ²,

Var(Cₙ) = σ² ∑ from i=1 to n of aₙ,ᵢ²

= σ²/n² ∑ from j=1 to n of j²

= σ²/n² · n(n+1)(2n+1)/6

= σ²(n+1)(2n+1)/(6n)

= σ²(n/3 + 1/2 + 1/(6n)).

For σ² > 0, this tends to ∞ as n → ∞. In Chebyshev's (Markov's) form, WLLN holds if Bₙ/n² → 0, where Bₙ = Var(S₁ + … + Sₙ) = σ²n(n+1)(2n+1)/6; here Bₙ/n² → ∞, so this sufficient condition is violated. Because that condition is only sufficient, the failure is confirmed as follows. If the weak law held for ⟨Sₙ⟩, we would need Cₙ → 0 in probability.

To see this fails, set sₙ² = Var(Cₙ). Since aₙ,ᵢ ≤ 1 and sₙ → ∞, Lindeberg’s condition holds: for any ε > 0,

(1/sₙ²)∑ from i=1 to n of E[aₙ,ᵢ² Zᵢ² 1{|aₙ,ᵢZᵢ| > εsₙ}]

≤ (1/σ²) E[Z₁² 1{|Z₁| > εsₙ}] → 0.

Hence by the Lindeberg–Feller central limit theorem, Cₙ/sₙ → N(0,1) in distribution.

But if Cₙ → 0 in probability, then Cₙ/sₙ → 0 in probability, contradicting a non-degenerate normal limit.

Therefore Cₙ does not converge to 0 in probability, so the weak law of large numbers does not hold for the sequence ⟨Sₙ⟩. (The statement is for the non-degenerate case σ² > 0; if σ² = 0, the sequence is degenerate.)

Final answer: WLLN fails for ⟨Sₙ⟩ because the centred average Cₙ does not converge to 0 in probability.

(d) Criteria of a Good Estimator; Sufficiency under Poisson

Criteria of a good estimator:

• Unbiasedness: E(T) = θ for all θ.

• Consistency: Tₙ → θ in probability as n → ∞.

• Efficiency: among unbiased estimators, it should have minimum variance; in regular cases it should attain the Cramér–Rao lower bound.

• Sufficiency: T should contain all information in the sample about θ; equivalently, the conditional distribution of the sample given T should not depend on θ.

• Completeness: together with sufficiency, it helps obtain a unique minimum-variance unbiased estimator.

• Minimum mean squared error, robustness and invariance are also desirable in practical problems.

Let X₁, X₂ be iid Poisson(λ). Their joint probability mass function is

P(X₁=x₁, X₂=x₂) = e^(−λ)λˣ¹/x₁! · e^(−λ)λˣ²/x₂!

= e^(−2λ)λ^(x₁+x₂)/(x₁! x₂!).

For T = X₁ + X₂, let t = x₁ + x₂. Then

P(X₁=x₁, X₂=x₂) = e^(−2λ)λᵗ/(x₁! x₂!).

By the factorization theorem, this is g(t,λ)h(x₁,x₂) with g(t,λ) = e^(−2λ)λᵗ and h(x₁,x₂) = 1/(x₁! x₂!), which does not depend on λ. Hence T = X₁ + X₂ is sufficient for λ.

Now take U = X₁ + 2X₂. Consider two sample points (2,0) and (0,1). For both, U = 2.

But

P(X₁=2, X₂=0)/P(X₁=0, X₂=1)

= [e^(−2λ)λ²/(2! 0!)]/[e^(−2λ)λ¹/(0! 1!)]

= λ/2.

This ratio depends on λ. If U were sufficient, then for the same value of U, the ratio of joint probabilities would be independent of λ by the factorization theorem. Since it is not, U = X₁ + 2X₂ is not sufficient for λ.

Final answer: T = X₁ + X₂ is sufficient; T = X₁ + 2X₂ is not sufficient.

(e) Critical Region for the Two-Sided Z-Test

We test H₀: μ = 100 against H₁: μ ≠ 100.

Given n = 50, σ² = 200, so σ = √200 = 10√2.

The standard error of X̄ is σ/√n = √200/√50 = √(200/50) = √4 = 2.

The test statistic is Z = (X̄ − 100)/2.

Under H₀, Z ~ N(0,1). For a two-sided test at α = 0.05, the critical values are ±1.96.

Thus the critical region is Z < −1.96 or Z > 1.96.

Equivalently, (X̄ − 100)/2 < −1.96 or (X̄ − 100)/2 > 1.96

⇒ X̄ < 100 − 3.92 or X̄ > 100 + 3.92

⇒ X̄ < 96.08 or X̄ > 103.92.

So reject H₀ if |X̄ − 100| > 3.92.

Final answer: Critical region is X̄ < 96.08 or X̄ > 103.92.