(a) Let X, Y, Z be three mutually independent standard exponential variates and W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ = X/(X + Y).
Then
(i) determine the joint distribution of W₁, W₂ and W₃.
(ii) find out the marginal probability density funct

Question

(a) Let X, Y, Z be three mutually independent standard exponential variates and W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ = X/(X + Y).
Then
(i) determine the joint distribution of W₁, W₂ and W₃.
(ii) find out the marginal probability density functions of W₁, W₂ and W₃.
(iii) examine the mutual independence of W₁, W₂ and W₃, and give your comment. (10+6+4=20 marks)

(b) Give an example to prove or disprove the following :
P(lim sup Aₙ) = 0 ⇒ Σₖ₌₁^∞ P(Aₖ) < ∞,
for any sequence {Aₙ, n ≥ 1} of events defined on a probability space (Ω, 𝓐, P). (15 marks)

(c) Let {Yₙ, n ≥ 1} be a sequence of random variables and Y be a degenerate random variable. Examine whether 'Yₙ converges in distribution to Y' implies 'Yₙ converges in probability to Y'. (15 marks)

UPSC Answer Check · Accepted Answer

(a)(i) Let S = W₁, U = W₂, V = W₃. Then the inverse transformation is X = S U V, Y = S U(1 − V), Z = S(1 − U). The support is s > 0, 0 < u < 1, 0 < v < 1. The Jacobian determinant is |∂(x,y,z)/∂(s,u,v)| = s² u. Since X, Y, Z are independent standard exponential variates, f(x,y,z) = exp(−(x+y+z)) = exp(−s). Therefore the joint density of W₁, W₂, W₃ is f(s,u,v) = s² u exp(−s), s > 0, 0 < u < 1, 0 < v < 1. This is the joint distribution. (a)(ii) The marginal densities are obtained by integrating the joint density. For W₁: f₁(s) = ∫₀¹∫₀¹ s² u exp(−s) du dv = s² exp(−s) ∫₀¹ u du = s² exp(−s)/2, s > 0. Thus W₁ ~ Gamma(3,1). For W₂: f₂(u) = ∫₀∞∫₀¹ s² u exp(−s) dv ds = u ∫₀∞ s² exp(−s) ds = u Γ(3) = 2u, 0 < u < 1. Thus W₂ ~ Beta(2,1). For W₃: f₃(v) = ∫₀∞∫₀¹ s² u exp(−s) du ds = (∫₀∞ s² exp(−s) ds)(∫₀¹ u du) = 2 × 1/2 = 1, 0 < v < 1. Thus W₃ ~ Uniform(0,1). Hence the marginal pdfs are f₁(s) = s² exp(−s)/2, s > 0; f₂(u) = 2u, 0 < u < 1; f₃(v) = 1, 0 < v < 1. (a)(iii) The joint density factorises as s² u exp(−s) = [s² exp(−s)/2] [2u] [1]. Therefore W₁, W₂, W₃ are mutually independent. Comment: the sum W₁ is independent of the ratios W₂ and W₃, and the two ratios W₂ and W₃ are also independent. (b) The statement is false. The first Borel–Cantelli lemma gives Σₖ₌₁^∞ P(Aₖ) < ∞ ⇒ P(lim sup Aₙ) = 0, but the converse need not hold. Counterexample: Take Ω = (0,1), with P equal to Lebesgue measure, and define Aₙ = (0, 1/n), n ≥ 1. Then P(Aₙ) = 1/n, so Σₙ₌₁^∞ P(Aₙ) = Σₙ₌₁^∞ 1/n = ∞. But {Aₙ} is decreasing, so lim sup Aₙ = ∩ₙ₌₁^∞ Aₙ = ∩ₙ₌₁^∞ (0, 1/n) = ∅. Therefore P(lim sup Aₙ) = P(∅) = 0, although the sum of probabilities diverges. Hence P(lim sup Aₙ) = 0 does not imply Σₖ₌₁^∞ P(Aₖ) < ∞. (c) Yes. If Y is degenerate, say Y ≡ c, then convergence in distribution of Yₙ to Y does imply convergence in probability of Yₙ to c. Let Y ≡ c. Its distribution function is F_Y(x) = 0 for x < c, and F_Y(x) = 1 for x ≥ c. Fix ε > 0. The points c − ε and c + ε/2 are continuity points of F_Y. Since Yₙ ⇒ Y, P(Yₙ ≤ c − ε) → F_Y(c − ε) = 0, and P(Yₙ ≤ c + ε/2) → F_Y(c + ε/2) = 1. Thus P(|Yₙ − c| ≥ ε) = P(Yₙ ≤ c − ε) + P(Yₙ ≥ c + ε) ≤ P(Yₙ ≤ c − ε) + P(Yₙ > c + ε/2) = P(Yₙ ≤ c − ε) + [1 − P(Yₙ ≤ c + ε/2)] → 0 + [1 − 1] = 0. Therefore Yₙ → c in probability. So convergence in distribution to a degenerate random variable implies convergence in probability to that degenerate value.

(a) Let X, Y, Z be three mutually independent standard exponential variates and W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ =…

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