(a) How does the concept of wholesale price index work? Describe the major components of wholesale price index. Explain the…

(a) Wholesale price index and agricultural indices

How the WPI works: it measures the average change in prices of goods at the first stage of bulk (factory-gate) trade, not retail prices or services. The Office of the Economic Adviser (DPIIT) compiles it monthly. Prices of a fixed basket are collected from ministries and departments and from selected manufacturing units; the series with base 2022-23 = 100 covers 957 items (697 in the 2011-12 series). Item indices are geometric means of price relatives (Jevons); higher levels use the Laspeyres form I = Σwᵢ Iᵢ, with weights from the base-year gross value of output. Inflation = (WPIₜ/WPIₜ₋₁₂ − 1) × 100.

Major components (2022-23 weights):

• Primary articles, 22.76% (129 items, such as food articles and minerals)
• Fuel and power, 14.11% (25 items; crude petroleum and natural gas now sit here)
• Manufactured products, 63.13% (803 items)

Agricultural indices (Directorate of Economics and Statistics, Ministry of Agriculture and Farmers Welfare), in the published series with base triennium ending 1993-94 = 100:

• The base is a three-year average, which removes year-to-year weather swings.
• Crop area index = (area in year j ÷ base area) × 100; crop production index = (production in year j ÷ base production) × 100.
• For a group such as foodgrains or non-foodgrains, the area index is total area ÷ base total area × 100; the production index is Σp₀qⱼ/Σp₀q₀ × 100, using base-period national average prices as weights, so that different crops are combined by value rather than tonnes.
• The yield index is derived, not measured: yield index = production index ÷ area index × 100.

(b) G/M/1 queue and the geometric arrival-point distribution

Model: one server; interarrival times are independent with a general distribution A(t), transform A*(s) and mean 1/λ; service times are exponential with rate μ; first-come first-served, unlimited waiting room; ρ = λ/μ < 1 for a steady state.

Embedded chain: let Yₙ be the number in the system just before the nth arrival. By the memoryless service, the number of services K completed during an interarrival time has aₖ = ∫₀^∞ e^−μt(μt)ᵏ/k! dA(t), Σaₖzᵏ = A*(μ(1 − z)). Then Yₙ₊₁ = max(0, Yₙ + 1 − K), so P(i, j) = aᵢ₊₁₋ⱼ for 1 ≤ j ≤ i + 1, and P(i, 0) = Σₖ≥i+1 aₖ.

Steady state: πⱼ = Σᵢ≥j−1 πᵢ aᵢ₊₁₋ⱼ, j ≥ 1. Try πⱼ = Cσʲ. The right side is Cσ^j−1 Σaₖσᵏ = Cσ^j−1 A*(μ(1 − σ)), which equals Cσʲ when σ = A*(μ(1 − σ)). The function f(z) = A*(μ(1 − z)) − z is convex with f(0) = a₀ > 0, f(1) = 0 and f′(1) = 1/ρ − 1 > 0, so exactly one root σ lies in (0, 1) when ρ < 1.

Normalising gives C = 1 − σ, so πⱼ = (1 − σ)σʲ, j = 0, 1, 2, …, a geometric distribution, with mean σ/(1 − σ) customers found by an arrival. The j = 0 equation then holds automatically because the probabilities sum to 1. The chain is irreducible and positive recurrent, so this is its unique stationary distribution. For M/M/1, A*(s) = λ/(λ + s) gives σ = ρ.

(c) Proof of q(x) = [1 − (e(x) − e(x + 1))]/[1 + e(x + 1)]

Here e(x) is the curtate expectation of life: a person alive at x completes year r only if alive at x + r, so e(x) = Σᵣ≥1 l(x + r)/l(x). Take out the term r = 1 and factor l(x + 1)/l(x): e(x) = [l(x + 1)/l(x)] 1 + Σᵣ≥2 l(x + r)/l(x + 1) = [l(x + 1)/l(x)][1 + e(x + 1)], since the sum is Σₛ≥1 l(x + 1 + s)/l(x + 1) = e(x + 1). With l(x + 1)/l(x) = 1 − q(x): e(x) = [1 − q(x)][1 + e(x + 1)] = 1 + e(x + 1) − q(x)[1 + e(x + 1)]. Hence q(x)[1 + e(x + 1)] = 1 + e(x + 1) − e(x) = 1 − (e(x) − e(x + 1)), and q(x) = [1 − (e(x) − e(x + 1))]/[1 + e(x + 1)]. ∎

Check with l₀ = 1000, l₁ = 900, l₂ = 700, l₃ = 300, l₄ = 0: e(0) = 1.9, e(1) = 1.111, so the formula gives (1 − 0.789)/2.111 = 0.1 = q(0).