(a) Apply the method of link relatives to the following data and calculate the seasonal indices :

Price of Rice (in ₹ per 10 kg)

| Quarter | 2001 | 2002 | 2003 | 2004 |
|---------|------|------|------|------|
| 1 | 75 | 86 | 90 | 100 |
| 2 | 60 | 6

Question

(a) Apply the method of link relatives to the following data and calculate the seasonal indices :

Price of Rice (in ₹ per 10 kg)

| Quarter | 2001 | 2002 | 2003 | 2004 |
|---------|------|------|------|------|
| 1 | 75 | 86 | 90 | 100 |
| 2 | 60 | 65 | 72 | 78 |
| 3 | 54 | 63 | 66 | 72 |
| 4 | 59 | 80 | 82 | 93 |

10 marks

(b) Derive the means and variances of the sampling distributions of the OLS estimates of α and β in the two-variable linear model Y = α + βX + u. 10 marks

(c) Consider, in the usual notations, the equation y = Y₁β + X₁γ + u, where y is an (n × 1) vector, Y₁ is an (n × (g-1)) matrix, X₁ is an (n × k) matrix. Derive the equations for the two-stage least square method of estimation. 10 marks

(d) If the survivorship function l(x) in life table is linear between x and x+1, and complete expectations of life at ages 40 and 41 for a particular group of persons are 21·39 years and 20·91 years respectively and l(40) = 41176, find the number of persons that attain the age 41. 10 marks

(e) Compute the T-scores corresponding to test score x for the following frequency distribution :

| x | 1 | 2 | 3 | 4 | 5 |
|-----|---|---|---|---|---|
| f | 2 | 3 | 8 | 6 | 1 |

(Cumulative Normal Distribution Table is given in Page No. 9)

10 marks

UPSC Answer Check · Accepted Answer

(a) 
Method of link relatives: link relative for a quarter = (price in that quarter / price in preceding quarter) × 100. The prices are in ₹ per 10 kg, but link relatives and seasonal indices are pure numbers. Q1 of 2001 is omitted because its preceding quarter is not given.
2001: Q2 = 60/75 × 100 = 80.00; Q3 = 54/60 × 100 = 90.00; Q4 = 59/54 × 100 = 109.26.
2002: Q1 = 86/59 × 100 = 145.76; Q2 = 65/86 × 100 = 75.58; Q3 = 72/65 × 100 = 110.77; Q4 = 80/72 × 100 = 111.11.
2003: Q1 = 90/80 × 100 = 112.50; Q2 = 72/90 × 100 = 80.00; Q3 = 66/72 × 100 = 91.67; Q4 = 82/66 × 100 = 124.24.
2004: Q1 = 100/82 × 100 = 121.95; Q2 = 78/100 × 100 = 78.00; Q3 = 72/78 × 100 = 92.31; Q4 = 93/72 × 100 = 129.17.

Average link relatives, using unrounded values:
Q1 = (145.7627 + 112.5000 + 121.9512)/3 = 126.7380.
Q2 = (80.0000 + 75.5814 + 80.0000 + 78.0000)/4 = 78.3953.
Q3 = (90.0000 + 110.7692 + 91.6667 + 92.3077)/4 = 96.1859.
Q4 = (109.2593 + 111.1111 + 124.2424 + 129.1667)/4 = 118.4449.

Their sum is 419.7641. Adjust to make the four seasonal indices sum to 400 by multiplying each average by 400/419.7641.

Seasonal indices: Q1 = 120.77, Q2 = 74.70, Q3 = 91.66, Q4 = 112.87. They sum to 400.00 and are dimensionless percentages, not ₹ per 10 kg.

(b) 
Model: Yᵢ = α + βXᵢ + uᵢ, i = 1, ..., n. Assume E(uᵢ|X) = 0, Var(uᵢ|X) = σ², Cov(uᵢ, uⱼ|X) = 0 for i ≠ j, and define Xbar = (1/n)ΣXᵢ and Sxx = Σ(Xᵢ − Xbar)², with Sxx > 0. Let xᵢ = Xᵢ − Xbar.

The OLS estimators are:
βhat = ΣxᵢYᵢ / Sxx,
αhat = Ybar − βhatXbar.

Substitute Yᵢ = α + βXᵢ + uᵢ:
βhat = β + Σxᵢuᵢ / Sxx.

Mean:
E(βhat|X) = β + (1/Sxx)ΣxᵢE(uᵢ|X) = β.

Variance:
Var(βhat|X) = (1/Sxx²)Σxᵢ²Var(uᵢ|X) = (σ²/Sxx²)Σxᵢ² = σ²Sxx/Sxx² = σ²/Sxx.

For αhat:
αhat = α + ubar − XbarΣxᵢuᵢ/Sxx,
where ubar = (1/n)Σuᵢ.

Mean:
E(αhat|X) = α.

Let C = Σxᵢuᵢ/Sxx. Then:
Var(αhat|X) = Var(ubar) + Xbar²Var(C) − 2XbarCov(ubar, C).

Now:
Var(ubar) = σ²/n,
Var(C) = σ²/Sxx,
Cov(ubar, C) = (1/(nSxx))ΣᵢΣⱼxⱼCov(uᵢ, uⱼ) = (σ²/(nSxx))Σⱼxⱼ = 0,
because Σxᵢ = 0.

Therefore:
Var(αhat|X) = σ²/n + Xbar²σ²/Sxx = σ²(1/n + Xbar²/Sxx).

Since ΣXᵢ² = Sxx + nXbar², this can also be written as:
Var(αhat|X) = σ²ΣXᵢ²/(nSxx).

If uᵢ are normally distributed, the sampling distributions are normal with these means and variances.

E(βhat|X) = β, Var(βhat|X) = σ²/Sxx; E(αhat|X) = α, Var(αhat|X) = σ²(1/n + Xbar²/Sxx) = σ²ΣXᵢ²/(nSxx).

(c) 
Let X be the n × m matrix of instruments. In the usual 2SLS setup, X must contain the exogenous regressors X₁ as columns; if the question’s X₁ already denotes the full instrument matrix, take X = X₁. Define P_X = X(X′X)⁻¹X′.

First stage: regress each column of Y₁ on X:
Y₁ = XΠ + V, with E(V|X) = 0.

The first-stage OLS normal equations are:
X′XΠhat = X′Y₁.

If X′X is nonsingular:
Πhat = (X′X)⁻¹X′Y₁.

The fitted endogenous regressors are:
Y₁* = XΠhat = P_XY₁.

Second stage: regress y on Y₁* and X₁:
y = Y₁*β + X₁γ + e.

The second-stage normal equations are:
Y₁*′Y₁*β + Y₁*′X₁γ = Y₁*′y,
X₁′Y₁*β + X₁′X₁γ = X₁′y.

Hence, with θ = (β′, γ′)′:
θhat = ([Y₁* X₁]′[Y₁* X₁])⁻¹[Y₁* X₁]′y.

Equivalently, let Z = [Y₁ X₁]. Since P_XX₁ = X₁ and P_XY₁ = Y₁*, the 2SLS estimator can be written compactly as:
θhat = (Z′P_XZ)⁻¹Z′P_Xy.

Identification requires the order condition m ≥ (g − 1) + k, or, if X₁ contains k included exogenous regressors, the number of excluded instruments m − k must be at least g − 1. The exact rank condition is:
rank(Z′P_XZ) = g − 1 + k,
equivalently:
rank([X₁, P_XY₁]) = k + g − 1.

If no excluded instruments are available and X = X₁, this rank condition fails when g > 1.

**2SLS estimator: θhat = ([Y₁* X₁]′[Y₁* X₁])⁻¹[Y₁* X₁]′y, where Y₁* = P_XY₁ and X contains X₁.

(d) 
Under the assumption that l(x) is linear between x and x + 1:
l(x + t) = l(x) − t[l(x) − l(x + 1)], for 0 ≤ t ≤ 1.

The complete expectation of life at age x is:
e°(x) = ∫ from 0 to ∞ l(x + t)/l(x) dt.

Split the integral at t = 1:
e°(x) = [∫ from 0 to 1 l(x + t)dt]/l(x) + [l(x + 1)/l(x)]e°(x + 1).

Because l(x + t) is linear on [0, 1], the integral is the trapezoidal area:
∫ from 0 to 1 l(x + t)dt = [l(x) + l(x + 1)]/2.

Let p = l(x + 1)/l(x). Then:
e°(x) = (1 + p)/2 + p e°(x + 1).

Solve for p:
p = [e°(x) − 1/2]/[e°(x + 1) + 1/2].

For x = 40:
p = (21.39 − 0.5)/(20.91 + 0.5) = 20.89/21.41.

Therefore:
l(41) = l(40)p = 41176 × 20.89/21.41.

Using exact fractions:
l(41) = 41176 × 2089/2141 = 40175.93.

Rounding to the nearest whole person:

Number of persons attaining age 41 = 40176 persons.

(e) 
Total frequency N = 2 + 3 + 8 + 6 + 1 = 20.

For a discrete score x, use the midpoint percentile rank:
P = 100(cf below x + f/2)/N.

Then convert P to a standard normal deviate z = Φ⁻¹(P/100), and compute the T-score:
T = 50 + 10z.
x = 1: cf below = 0, f = 2. P = 100(0 + 2/2)/20 = 5%. z = Φ⁻¹(0.05) = −1.645. T = 50 + 10(−1.645) = 33.55.
x = 2: cf below = 2, f = 3. P = 100(2 + 3/2)/20 = 17.5%. From the cumulative normal table, z ≈ −0.935. T = 50 + 10(−0.935) = 40.65.
x = 3: cf below = 5, f = 8. P = 100(5 + 8/2)/20 = 45%. z = Φ⁻¹(0.45) ≈ −0.126. T = 50 + 10(−0.126) = 48.74.
x = 4: cf below = 13, f = 6. P = 100(13 + 6/2)/20 = 80%. z = Φ⁻¹(0.80) ≈ 0.842. T = 50 + 10(0.842) = 58.42.
x = 5: cf below = 19, f = 1. P = 100(19 + 1/2)/20 = 97.5%. z = Φ⁻¹(0.975) ≈ 1.960. T = 50 + 10(1.960) = 69.60.

T-scores: x = 1: 33.55, x = 2: 40.65, x = 3: 48.74, x = 4: 58.42, x = 5: 69.60.**

(a) Apply the method of link relatives to the following data and calculate the seasonal indices : Price of Rice (in ₹ per 10…

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