(a) Out of 1000 persons born, only 900 reach the age of 15 years, and out of every 1000 who reach the age of 15 years, 950 reach…

(a) Probability of attaining age 51

Let A₁₅, A₅₀, A₅₁ be the events of reaching ages 15, 50 and 51. Step 1: P(A₁₅) = 900/1000 = 9/10. Step 2: P(A₅₀ | A₁₅) = 950/1000 = 19/20. Step 3: 40 of every 1000 aged 50 die within a year, so P(A₅₁ | A₅₀) = 960/1000 = 24/25. Step 4: Since A₅₁ ⊂ A₅₀ ⊂ A₁₅, the multiplication rule gives P(A₅₁) = (9/10)(19/20)(24/25) = 4104/5000 = 513/625 = 0·8208. The probability that a newborn attains age 51 is 0·8208 (about 821 of every 1000 born).

(b) Cumulative distribution function and P(X > 3/2)

F(x) = ∫ f(t) dt from −∞ to x. Step 1: For x < 0, F(x) = 0. Step 2: For 0 ≤ x < 1, F(x) = ∫₀ˣ t/2 dt = x²/4, so F(1) = 1/4. Step 3: For 1 ≤ x < 2, F(x) = 1/4 + (x − 1)/2 = x/2 − 1/4, so F(2) = 3/4. Step 4: For 2 ≤ x < 3, F(x) = 3/4 + ∫₂ˣ (3 − t)/2 dt = 3/4 + (3x/2 − x²/4 − 2) = (6x − x² − 5)/4, so F(3) = 1. Step 5: For x ≥ 3, F(x) = 1. Hence F(x) = 0 for x < 0; x²/4 for 0 ≤ x < 1; x/2 − 1/4 for 1 ≤ x < 2; (6x − x² − 5)/4 for 2 ≤ x < 3; 1 for x ≥ 3. Since 3/2 lies in [1, 2), F(3/2) = 3/4 − 1/4 = 1/2. P(X > 3/2) = 1 − F(3/2) = 1/2.

(c) Condition on α for the weak law of large numbers

Let Sₙ = X₁ + … + Xₙ. Step 1: E(Xₙ) = 0 and Var(Xₙ) = n^(2α). Step 2: By independence, Var(Sₙ) = Σ i^(2α) for i = 1 to n, which is of order n^(2α+1)/(2α+1). Step 3: By Chebyshev, P(|Sₙ/n| > ε) ≤ Var(Sₙ)/(n²ε²), which is of order n^(2α−1). This tends to 0 when 2α − 1 < 0, so the law holds for α < 1/2. Step 4: Necessity. For α = 1/2, Var(Sₙ/n) tends to 1/2 and, by Lindeberg's CLT, Sₙ/n tends in distribution to N(0, 1/2), not to 0. For α > 1/2, the standard deviation of Sₙ/n grows like n^(α−1/2) and Sₙ/n does not converge in probability to 0. (This agrees with Feller's necessary and sufficient condition for independent variables.) The sequence obeys the WLLN if and only if 0 < α < 1/2.

(d) Run test of randomness

H₀: the sequence is random; H₁: it is not random. Step 1: Counting the 48 letters gives n₁ = 36 H's and n₂ = 12 T's. The number of runs is R = 19. Step 2: Mean μ = 1 + 2n₁n₂/n = 1 + 2(36)(12)/48 = 19. Step 3: Variance σ² = 2n₁n₂(2n₁n₂ − n)/[n²(n − 1)] = 864 × 816/(2304 × 47) = 6·511, so σ = 2·552. Step 4: Z = (R − μ)/σ = (19 − 19)/2·552 = 0. Step 5: For the two-sided test at 5%, the critical value is Z₀.₀₂₅ = 1·96. Since |Z| = 0 < 1·96, H₀ is not rejected. The sequence is consistent with randomness at the 5% level.

(e) Prior and posterior distributions; Bayes estimator under squared error loss

The prior π(θ) expresses belief about the parameter before the data are seen. The posterior π(θ | x) is the distribution of θ after observing the sample x, found by Bayes' theorem: π(θ | x) = L(x | θ)π(θ) / ∫ L(x | u)π(u) du. The prior is fixed in advance and does not depend on the sample; the posterior combines the prior with the likelihood and is the basis of Bayesian inference. For squared error loss L(θ, a) = (θ − a)², the posterior risk is E[(θ − a)² | x] = Var(θ | x) + (a − E[θ | x])². The first term does not depend on a, so the risk is least when a = E(θ | x). The Bayes estimator is the posterior mean, δ(x) = E(θ | x).