(a) Explain the 2-stage least squares (2SLS) method of estimation for the following simultaneous equation systems : C_t = α +…

(a) Two-stage least squares for the Keynesian system (10 marks)

Here C and Y are endogenous and Z is predetermined; the second equation is an identity, so only the consumption function needs estimation. Substituting the identity gives the reduced form Y = α/(1 − β) + Z/(1 − β) + u/(1 − β). Hence Cov(Y, u) = σᵤ²/(1 − β) ≠ 0, and OLS on C = α + βY + u is inconsistent: plim β̂_OLS = β + (1 − β)σᵤ²/(σ_Z² + σᵤ²), an upward bias for 0 < β < 1.

The equation is identified: one exogenous variable (Z) is excluded and one endogenous variable (Y) appears on the right, so the order condition holds with equality (exact identification), and the rank condition holds since Z appears in the identity with a non-zero coefficient.

Stage 1: regress the endogenous regressor on all predetermined variables, Yₜ = π₀ + π₁Zₜ + vₜ, and obtain the fitted Ŷₜ = π̂₀ + π̂₁Zₜ, which is uncorrelated with u. Stage 2: regress Cₜ on Ŷₜ, which gives

β̂_2SLS = Σ(Ŷ − Ȳ)(C − C̄)/Σ(Ŷ − Ȳ)² = Σ(Z − Z̄)(C − C̄)/Σ(Z − Z̄)(Y − Ȳ), and α̂ = C̄ − β̂Ȳ.

The second form shows that 2SLS equals the instrumental-variable and indirect least squares estimators using Z as the instrument. The estimator is consistent but biased in small samples; the standard errors from the second-stage regression must be corrected by using residuals C − α̂ − β̂Y (actual Y), not those with Ŷ.

(b) Detecting heteroscedasticity and the Goldfeld-Quandt test (10 marks)

Heteroscedasticity means Var(uᵢ) = σᵢ² is not constant; OLS remains unbiased but is inefficient and its usual standard errors are wrong. Methods of detection:

• Informal: plot squared residuals against fitted values or against a suspected variable X; a systematic pattern (fan shape) indicates the problem.
• Park test: regress ln û² on ln X; a significant slope indicates heteroscedasticity.
• Glejser test: regress |û| on X^h (h = 1, ½, -1, -½); a significant slope indicates it.
• Spearman rank-correlation test between |û| and X, with t = r√(n − 2)/√(1 − r²).
• Goldfeld-Quandt, Breusch-Pagan-Godfrey (an LM test from the regression of û²/σ̃² on suspected variables) and White's general test (regress û² on regressors, their squares and cross-products; nR² ~ χ²).

Goldfeld-Quandt test: assume (1) normally distributed errors, (2) no autocorrelation, (3) a correctly specified model, and (4) that the error variance is positively related to one explanatory variable, σᵢ² = σ²Xᵢ². Procedure: order the observations by X; omit c central observations (about one-quarter to one-third of n, for example c = 8 when n = 30); run separate OLS regressions on the first and last (n − c)/2 observations; compute F = (RSS₂/df)/(RSS₁/df), with df = (n − c − 2k)/2 in each. Under homoscedasticity F follows the F distribution with (n − c − 2k)/2 and (n − c − 2k)/2 degrees of freedom; if F exceeds the critical value, reject homoscedasticity.

(c) MLE of β and the distribution of e′e/σ² (10 marks)

With u ~ N(0, σ²Iₙ), Y ~ N(Xβ, σ²Iₙ) and the log-likelihood is L = −(n/2)ln(2π) − (n/2)ln σ² − (Y − Xβ)′(Y − Xβ)/2σ². Setting ∂L/∂β = X′(Y − Xβ)/σ² = 0 gives the normal equations X′Xβ = X′Y, so β̂ = (X′X)⁻¹X′Y (rank of X is k, so X′X is invertible), and ∂L/∂σ² = 0 gives σ̂² = e′e/n.

Distribution: e = Y − Xβ̂ = (I − X(X′X)⁻¹X′)Y = MY = M(Xβ + u) = Mu, since MX = 0. Then e′e = u′M′Mu = u′Mu, because M is symmetric and idempotent (M² = M). Being idempotent, M has eigenvalues 0 or 1 only, and its rank equals its trace: tr M = tr Iₙ − tr[X(X′X)⁻¹X′] = n − tr[(X′X)⁻¹X′X] = n − k. So an orthogonal matrix P exists with P′MP = diag(Iₙ₋ₖ, 0). Let w = P′u/σ ~ N(0, Iₙ), since P is orthogonal. Then

e′e/σ² = u′Mu/σ² = w′P′MPw = w₁² + w₂² + … + w²ₙ₋ₖ,

a sum of n − k squares of independent standard normals, so e′e/σ² ~ χ² with (n − k) degrees of freedom.

(d) Vital events data, age errors and adjustment (10 marks)

Sources of vital-events data: the civil registration system (births and deaths registered under the Registration of Births and Deaths Act, 1969), the Sample Registration System (dual-record survey by the Registrar General), population censuses and their retrospective questions, national surveys (NFHS, NSS) and hospital or institutional records.

Errors in census age data: age heaping or digit preference (rounding to ages ending in 0 and 5, and preference for even numbers), age misstatement, with exaggeration of age by the elderly and understatement by young women, under-enumeration of children under five and of infants, and differential coverage by sex and age, giving irregular age-sex distributions.

Detection and adjustment: heaping is measured by Whipple's index (100 means no heaping, above about 175 indicates rough data) and Myers' blended index; the age-sex accuracy is judged by the United Nations Joint Score based on age and sex ratios. Errors are reduced by grouping into five- or ten-year intervals, which cancels heaping; by smoothing with Carrier-Farag, Karup-King, Sprague or Beers interpolation formulae; by applying stable- or quasi-stable-population methods; and by using post-enumeration surveys and reconciliation with registration data for children.

(e) Standard scores (10 marks)

Raw scores have no meaning by themselves: they depend on the difficulty and scale of the test. Converting to standard scores expresses each score as a distance from the mean in standard-deviation units, which (i) makes scores from different tests comparable, (ii) shows a person's relative position in the group, (iii) allows scores to be added or averaged across tests, and (iv) gives a standard scale (here mean 50 and SD 10, the T-score scale) that avoids negative values and decimals.

Computation. n = Σf = 62; Σfx = 10 + 36 + 48 + 70 + 48 + 65 + 72 + 6 = 355, so mean = 355/62 = 5.726. Σf(x − mean)² = 184.34, so σ = √(184.34/62) = √2.973 = 1.724. Standard score T = 50 + 10(x − 5.726)/1.724 = 50 + 5.80(x − 5.726):

• x = 10: 74.8; x = 9: 69.0; x = 8: 63.2; x = 7: 57.4
• x = 6: 51.6; x = 5: 45.8; x = 4: 40.0; x = 3: 34.2

The scores have mean 50 and SD 10, as required (with n − 1 as divisor the results differ only in the second decimal).