(a) Consider the following structural equation model assuming Y's as endogenous and X's as pre-determined variables : Y₁ = α₁₀ +…

(a) Identification by the order and rank conditions (20 marks)

There are M = 4 endogenous variables (Y₁ to Y₄) and K = 3 predetermined variables (X₁, X₂, X₃). Let Gᵢ be the number of endogenous variables (including the dependent one) and kᵢ the number of predetermined variables in equation i.

Order condition (necessary): K − kᵢ ≥ Gᵢ − 1, that is, the number of predetermined variables excluded from the equation must be at least the number of endogenous variables included minus one; equality gives exact identification, and inequality over-identification.

• Equation 1 includes Y₁, Y₂, Y₃ (G = 3) and X₁ (k = 1): K − k = 2 ≥ 2, so the order condition holds with equality.
• Equation 2 includes Y₂, Y₃ (G = 2) and X₁, X₂ (k = 2): K − k = 1 ≥ 1, equality.
• Equation 3 includes Y₃, Y₁ (G = 2) and X₁, X₂ (k = 2): K − k = 1 ≥ 1, equality.
• Equation 4 includes Y₄, Y₁, Y₂ (G = 3) and X₃ (k = 1): K − k = 2 ≥ 2, equality.

So the order condition shows every equation to be just identified, but this condition is only necessary. The rank condition (necessary and sufficient) requires that the matrix formed from the coefficients of the variables excluded from equation i, taken from the other M − 1 = 3 equations, has rank M − 1 = 3.

Write the coefficient table (columns Y₁, Y₂, Y₃, Y₄, X₁, X₂, X₃; the intercept is in every equation and is ignored):

• Eq. 1: 1, −α₁₂, −α₁₃, 0, −β₁₁, 0, 0
• Eq. 2: 0, 1, −α₂₃, 0, −β₂₁, −β₂₂, 0
• Eq. 3: −α₃₁, 0, 1, 0, −β₃₁, −β₃₂, 0
• Eq. 4: −α₄₁, −α₄₂, 0, 1, 0, 0, −β₄₃

Equation 1: the excluded variables are Y₄, X₂, X₃. From equations 2, 3 and 4 the matrix is rows (0, −β₂₂, 0), (0, −β₃₂, 0), (1, 0, −β₄₃). Its first and third columns are proportional, so the rank is 2 < 3: equation 1 is not identified.

Equation 2: the excluded variables are Y₁, Y₄, X₃. From equations 1, 3 and 4: rows (1, 0, 0), (−α₃₁, 0, 0), (−α₄₁, 1, −β₄₃). The second and third columns are proportional, so the rank is 2 < 3: not identified.

Equation 3: the excluded variables are Y₂, Y₄, X₃. From equations 1, 2 and 4: rows (−α₁₂, 0, 0), (1, 0, 0), (−α₄₂, 1, −β₄₃). Rank is 2 < 3: not identified.

Equation 4: the excluded variables are Y₃, X₁, X₂. From equations 1, 2 and 3: rows (−α₁₃, −β₁₁, 0), (−α₂₃, −β₂₁, −β₂₂), (1, −β₃₁, −β₃₂). The determinant is −α₁₃(β₂₁β₃₂ − β₂₂β₃₁) + β₁₁(α₂₃β₃₂ + β₂₂), which is non-zero in general, so the rank is 3 and equation 4 is exactly identified.

Conclusion: only equation 4 is (exactly) identified; equations 1, 2 and 3 satisfy the order condition but fail the rank condition, and are unidentified. This shows that the order condition alone is not sufficient.

(b) Life table probabilities (15 marks)

Build lₓ from l₉₁ = 871 using lₓ₊₁ = lₓ − dₓ: l₉₁ = 871, l₉₂ = 575, l₉₃ = 366, l₉₄ = 222, l₉₅ = 129, l₉₆ = 71, l₉₇ = 37, l₉₈ = 19, l₉₉ = 9, l₁₀₀ = 4, l₁₀₁ = 1 (and l₁₀₂ = 0, so the d column sums to 871).

(i) A person aged 93 dies within three years: 1 − l₉₆/l₉₃ = 1 − 71/366 = 295/366 = 0.806 (equivalently (144 + 93 + 58)/366).

(ii) A person aged 92 survives to 96: l₉₆/l₉₂ = 71/575 = 0.1235.

(iii) Three persons aged 92, 93, 94 all survive four years, the lives being independent: (l₉₆/l₉₂)(l₉₇/l₉₃)(l₉₈/l₉₄) = (71/575)(37/366)(19/222) = 0.1235 × 0.1011 × 0.0856 = 0.00107.

(iv) Of three aged 93, 94, 95, exactly two alive in 4 years. The survival probabilities are p₁ = l₉₇/l₉₃ = 37/366 = 0.1011, p₂ = l₉₈/l₉₄ = 19/222 = 0.0856, p₃ = l₉₉/l₉₅ = 9/129 = 0.0698, with q's 0.8989, 0.9144, 0.9302. Then

P = p₁p₂q₃ + p₁q₂p₃ + q₁p₂p₃ = 0.1011 × 0.0856 × 0.9302 + 0.1011 × 0.9144 × 0.0698 + 0.8989 × 0.0856 × 0.0698 = 0.00805 + 0.00645 + 0.00537 = 0.0199.

(v) All three die within 4 years: q₁q₂q₃ = 0.8989 × 0.9144 × 0.9302 = 0.7646.

(c)(i) Parallel tests in psychological studies (5 marks)

In classical test theory an observed score is X = T + E. Two tests are parallel if they measure the same true score T for every person and have equal error variances (hence equal means, variances and correlations with any other variable). Then the correlation between their scores is the reliability, ρ_XX′ = σ_T²/σ_X². Uses: estimating reliability by the parallel-forms method without the memory and practice effects of test-retest; providing equivalent forms for pre- and post-testing and for repeated measurement; preventing cheating by using alternate forms; the split-half method, in which halves are treated as parallel and the Spearman-Brown formula raises the half-test reliability to that of the full test; and computing the standard error of measurement, σ_X√(1 − ρ_XX′).

(c)(ii) Combining the rankings given by several judges (10 marks)

Let m judges rank n subjects. First find the rank sum Rᵢ of each subject over the m judges. To test whether the judges agree, compute Kendall's coefficient of concordance:

W = 12 S/[m²(n³ − n)], with S = Σ(Rᵢ − R̄)² and R̄ = m(n + 1)/2.

W lies between 0 (no agreement) and 1 (complete agreement). With ties, average ranks are used and the denominator is corrected to m²(n³ − n) − m ΣT, T = Σ(t³ − t) over each judge's tied groups. To test significance, χ² = m(n − 1)W with n − 1 degrees of freedom (for n > 7; use tables of S for smaller n). It is related to the mean Spearman rank correlation between the judges by r̄ = (mW − 1)/(m − 1). If W is significant, the judges are in substantial agreement, and the combined ranking is obtained by ordering the subjects by their rank sums (smallest sum being rank 1), with ties averaged; this is the Borda-count method. If W is not significant, pooling is not justified.